Sponsor Area
Leap year contains 366 days.
∴ there are 52 complete weeks and two days other. The following are the possibilities for these two ‘over’ days:
(i) Sunday and Monday (ii) Monday and Tuesday (iii) Tuesday and Wednesday
(iv) Wednesday and Thursday (v) Thursday and Friday (vi) Friday and Saturday (vii) Saturday and Sunday.
Now there will be 53 Tuesdays in a leap year when one of the two over days is a Tuesday.
∴ out of 7 possibilities, two are favourable to this event.
There are four entries in determinant of 2 × 2 order. Each entry may be filled up in two ways with 0 or 1.
∴ number of determinants that can be formed = 24 = 16
∴ total number of cases = 16
The value of determinant is positive in the cases
number of favourable cases = 3
the probability that the determinant is positive = 
0
D.
S =
Now favourable case is(2, 2)
...(1)
Here, P(A) = 0.8, P(B) = 0.5, P(B | A) = 0.4
Now, 
Here, P(A) = 0.8, P(B) = 0.5, P(B | A) = 0.4
Now, 
Here, P(A) = 0.8, P(B) = 0.5, P(B | A) = 0.4
Now, P(A ∪ B) = 
= 0.98
A fair die is rolled. Consider events E = {1, 3, 5}, F = {2, 3} and G = {2, 3, 4, 5}. Find P(E|F) and P(F|E).
Since fair die is rolled
A fair die is rolled. Consider events E = {1, 3, 5}, F = {2, 3} and G = {2, 3, 4, 5}. Find P(E|G) and P(G|E).
Since fair die is rolled
A fair die is rolled. Consider events E = {1, 3, 5}, F = {2, 3} and G = {2, 3, 4, 5}. Find P((E ∪ F) | G) and P((E ∩ F) | G)
Since fair die is rolled
and 
and 
Sponsor Area
and 
A die is thrown three times. Events A and B are defined as below:
A: 4 on the third throw
B: 6 on the first and 5 on the second throw
Find the probability of A given that B has already occurred.
Let the events be
E : number 4 appears at least once.
F : the sum of the numbers appearing is 6.
The elementary events favourable to the occurrence of E are
(4, 1), (4, 2), (4, 30, (4, 4), (4, 5), (4, 6), (1, 4), (2, 4), (3, 4), (5, 4), (6, 4).
The elementary events favourable to the occurrence of F are
(1, 5), (2, 4), (3, 3), (4, 2), (5, 1).
The elementary events favourable to the occurrence of both E and F are (2, 4), (4, 2)
A die is thrown and the sum of the numbers appearing is observed to be 7. What is the conditional probability that the number 2 has appeared at least once.
Let the events be
E : number 2 appears at least at least once
F : the sum of the numbers appearing is 7.
The elementary events favorable to the occurence of E are
(1, 2), (2, 1), (2, 2), (2, 3), (2, 4), (2, 5), (2, 6), (3, 2), (4, 2), (5, 2). (6, 2).
The elementary events favorable to the occurance of F are
(1, 6), (2, 5), (3, 4), (4, 3), (5, 2), (6, 1).
The elementary events favorable to the occurence of both E and F are (2, 5),(5. 2).
Let the events be
E : sum is 10 or greater
F : 5 appears on the first die.
Required probability = 
Find the probability that a single throw of a die results in a number less than 6 if
(i) no other information is given
(ii) it is given that the throw resulted in an odd number
Let S be the sample space when a die is thrown.
Let E denote the event of getting number less than 6 and F denote the event of getting an odd number.
A pair of dice is thrown . If the sum is seven, find the probability that one of the dice shows 3.
We have
Let A = 
and 
Required probability = P(A/B) = 
E : The card drawn is king or queen
F : The card drawn is queen or ace.
A die thrown three times,
E : 4 appears on the third toss,
F : 6 and 5 appears respectively on First two tosses
Determine P(E | F).
sample space has 6 x 6 x 6 = 216 outcomes.
A coin is tossed three times, where
E : head on third toss, F : heads on first two tosses
Here S = {HHH, HHT, HTH, THH, TTH, THT, HTT, TTT}
E : head on third toss
E = {HHH, HTH, THH, TTH}
and F : heads on first two tosses
A coin is tossed three times, where
E : at least two heads , F : at most two heads
Here S = {HHH, HHT, HTH, THH, TTH, THT, HTT, TTT}
E : at least two heads
F: at most two heads
A coin is tossed three times, where
E : at most two tails , F : at least one tail
Here S = {HHH, HHT, HTH, THH, TTH, THT, HTT, TTT}
E: at most two tails
F: at least one tail
Here 
Number of elements (outcomes) of the above sample space is 6 × 6 = 36.
Let E : sum greater than 9
Let E: black dice resultant in a 5.
Required probability = 
Here 
Number of elements (outcomes) of the above sample space is 6 × 6 = 36.
Let E: a total of 8
Here 
Number of elements (outcomes) of the above sample space is 6 × 6 = 36.
Let E : the sum of numbers on the dice is 4
∴ E = {(1, 3), (2, 2), (3, 1)}
and F : numbers appearing on the two dice are different
∴ F contains all the points of S except (1, 1), (2, 2), (3, 3), (4, 4), (5, 5), (6, 6).
∴ F contains 30 elements.
Also E ∩ F = {(1, 3), (3, 1)}
Required probability = 
The given data is
|
Easy |
Difficult |
Total |
|
|
True/False |
300 |
200 |
500 |
|
Multiple Choice |
500 |
400 |
900 |
|
Total |
800 |
600 |
1400 |
Let E : Easy questions, D : Difficult questions
T : True/false questions and M : Multiple choice questions
Number of easy multiple choice questions = 500
Total number of questions = 1400
= Probability of selecting an easy and multiple choice question
Total number of multiple choice questions
= 500 + 400 = 900
P(M) = Probability of selecting one multiple choice question
Number of cards = 52
Number of kings = 5
Number of face cards = 16
Let E denote the event of drawing a king and F the even of drawing face card.
Required probability = P(E/F) = 
Two coins are tossed once, where
E : tail appears on one coin, F : one coin shows head
Determine P(E | F).
Here S = {HH, HT, TH, TT}
E: tail appears on one coin
Sponsor Area
Two coins are tossed once, where
E : no tail appears, F : no head appears
Determine P(E | F).
Here S = {HH, HT, TH, TT}
E : no tail appears
Let the first child be denoted by capital letter and the second i.e. younger one by a smaller letter.
∴ S = {Bb, Bg, Gb, Gg}
Let E : both children are girls
∴ E = {Gg}
Let F : the youngest is a girl
Required proability = 
Let the first child be denoted by capital letter and the second i.e. younger one by a smaller letter.
∴ S = {Bb, Bg, Gb, Gg}
Let E : both children are girls
∴ E = {Gg}
Let F : at least one is a girl
Required probability = 
A family has two children. What is the probability that both the children are boys given that at least one of them is a boy?
Let b stand for boy and g for girl.
∴ S = {(b, b), (g, b), (b, g), (g, g)}
Let E and F denote the following events:
E : ‘both the children are boys’
F : ‘at least one of the child is a boy’
∴ E = {(b, b)} and F = {(b, b), (g, b), (b, g)}
Now, 
A couple has two children,
Find the probability that both children are males, if it is known that at least one of the children is male.
S = {MM, MF, FM, FF}
where M stands for male and F stands for female
Let A: both the children are males
and B: at least one of the children is male.
A = {MM}, B = {MM, MF, FM}
Required probability = 
A couple has two children,
Find the probability that both children are females, if it is known that the elder child is a female.
S = {MM, MF, FM, FF}
where M stands for male and F stands for female
Let A: both the children are female
and B: elder child is a a female.
An electronic assembly consists of two subsystems, say A and B. From the previous testing procedures, the following probabilities are assumed to be known:
P( A fails) = 0.2
P(B fails alone) = 0.15
P(A and B fail) = 0.15
Evaluate the following probabilities:
(i) P( A fails 
) (ii) P(A fails alone)
Consider the following events:
E : A fails
F : B fails.
Now P(A fails) = 0.2 
P(E) = 0.2
and P(A and B fails) = 0.15 
Also P(B fails alone) = 0.15
(i) P(A fails| B has failed) = P(E | F) = 
(ii) P(A fails alone) = 
= 0.2 - 0.15 = 0.05
S = {(3, 1), (3, 2), (3, 3), (3, 4), (3, 5), (3, 6), (6, 1), (6, 2), (6, 3), (6, 4), (6, 5), (6, 6), (1, H), (1, T), (2, H), (2, T), (4, H), (4, T), (5, H), (5, T)}
The outcomes of S are not equally likely. First 12 outcomes are equally likely and are such that the sum of their probabilities is 
So each of the first 12 outcomes has a probability equal to 
Remaining eight outcomes are equally likely and are such that the sum of their probabilities is 
So, each of these has a probability equal to 
Let E: 'the coin shows a tail'
and F: 'at least one die shows up a 3',
and 
Required probability = 
the probabilities assigned to the 8 elementary events (H, H), (H, T), (T, 1), (T, 2), (T, 3), (T, 4), (T, 5), (T, 6) are 
respectively.
D.
P(A) = P(B)
A.
A ⊂ B 
C.
P(B|A) > P(B)P(A|B) > P(A)
(C) is correct answer.
B.
P(A|B) = 1
None of these
C.
When A ⊂ B, then 
∴ (C) is correct answer.
If 
find P(A ∩ B) if A and B are independent events.
Here, 
...(1)
Since A and B are independent events
Here, P(A) = 0.38, 
Let P(B) = p
Since A and B are independent.
...(1)
Now 
The events E and F are given to be independent. Find P(F) if it is given that P(E) = 0.35 and P(E ∪ F) = 0.60.
Here, P(E) = 0.35, 
Now, 
P(A) = 0.40, P(A ∪ B) = 0.70
Now 
P(not B) = 0.65
Now, 
For two events A and B, let P(A) = 0.4 and P(B) = p and P(A∪ B) = 0.6
(i) Find p so that A and B are independent events.
(ii) Find p so that A and B are mutually exclusive events.
...(1)
...(2)
Find p if they are mutually exclusive.
...(1)
Find p if they are independent.
Let A and B be independent events with P(A) = 0.3 and P(B) = 0.4. Find P(A ∩ B)
P(A) = 0.3, P(B) = 0.4
Let A and B be independent events with P(A) = 0.3 and P(B) = 0.4. Find P(A ∪ B)
P(A) = 0.3, P(B) = 0.4
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= 0.58
Let A and B be independent events with P(A) = 0.3 and P(B) = 0.4. Find P(A|B)
P(A) = 0.3, P(B) = 0.4
Let A and B be independent events with P(A) = 0.3 and P(B) = 0.4. Find P(B | A)
P(A) = 0.3, P(B) = 0.4
P(B | A) = 
State whether A and B are independent?
If P(A) = 
and 
find:
(i) P(A or B), if A and B are mutually exclusive events.
(ii) P (A and B), if A and B are independent events.
Here 
(i) 
(ii) Since A and B are independent events
Here P(A) = 0.3, P(B) = 0.6
P(A and B) = P(A) x P(B) [
A and B are independent]
= 0.3 x 0.6 = 0.18
Here P(A) = 0.3, P(B) = 0.6
P(A and not B) = 
Here P(A) = 0.3, P(B) = 0.6
P(A or B) = P(A)+P(B) - 
= 0.3 + 0.6 - 0.18 = 0.72
Here P(A) = 0.3, P(B) = 0.6
P (neither A nor B) = 
S = {1, 2, 3, 4, 5, 6}
E = {3, 6}, F = {2, 4, 6}, 
Since P(E and F) = 
∴ the events E and F are independent events.
Sponsor Area
S = {(H, 1), (H, 2), (H, 3), (H, 4), (H, 5), (H, 6), (T, 1), (T, 2), (T, 3), (T, 4), (T, 5), (T, 6)}
Now A: 'head appears on the coin'
and B: '3 appears on the dice'.
and 
Also, 
and 
Now, 
∴ A and B are independent.
A die marked 1, 2, 3 in red and 4, 5, 6 in green is tossed. Let A be the event 'the number is even', and B be the event, 'the number is red'. Are A and B independent?
S = {1, 2, 3, 4, 5, 6}
Now A: 'number is even'
and B: 'number is red'
Also, 
Now, 
∴ A and B are not independent.
Total number of outcomes = 6 × 6 = 36
Number of outcomes favourable to E = 2. [i.e., (5, 6), (6, 5)]
Let F be the event that the number 5 does not appear on the first die.
Number of outcomes favourable to F = 30
Now, 
Therefore, E and F are not independent events.
One card is drawn at random from a pack of well-shuffled deck of 52 cards. In which of the following cases are the events E and F are independent?
E : “the card drawn is a spade”
F : “the card drawn is an ace”.
P(E) = P(the card drawn is spade) = 
P(F) = P (the card drawn is an ace) = 
Now, 
One card is drawn at random from a pack of well-shuffled deck of 52 cards. In which of the following cases are the events E and F are independent?
E : “the card drawn is a king or queen”
F : “the card drawn is a queen or jack”.
P(E) = P (card drawn is a king or queen) = 
P(F) = P (card drawn is queen or jack) = 
A coin is tossed thrice. In which of the following cases are the events E and F independent?
E : the first throw results in head”.
F : “the last throw results in tail”.
Here, S = {HHH, HHT, HTH, THH, TTH, THT, HTT, TTT}
P(E) = P(HHH, HHT, HTH, HTT) = 
P(F) = P(HHT, THT, HTT, TTT) = 
A coin is tossed thrice. In which of the following cases are the events E and F independent?
E : “the number of heads is two”.
F : “the last throw results in head”.
Here, S = {HHH, HHT, HTH, THH, TTH, THT, HTT, TTT}
A coin is tossed thrice. In which of the following cases are the events E and F independent?
E : “the number of heads is odd”.
F : “the number of tails is odd”.
Here, S = {HHH, HHT, HTH, THH, TTH, THT, HTT, TTT}
P(E) = P(HHH, TTH, THT, TTH) = 
S = {HHH, HHT, HTH, THH, HTT, THT, TTH, TTT}
E = {HHH, TTT}, F = {HHH, HHT, HTH, THH}
and G = {HHT, HTH, THH, HTT, THT, TTH, TTT}
and 
Also, 
and 
and 
∴ the events (E and F) are independent, and the events (E and G) and (F and G) are dependent.
Prove that if E and F are independent events, then so are the events E and F'.
Since E and F are independent.
∴ P(E ∩ F) = P(E). P(F)
From the Venn diagram in the figure, it is clear that E ∩ F and E ∩ F' are mutually exclusive events and also
∴ E and F' are independent.
Tips: -
Note: In a similar manner, it can be shown that if the events E and F are independent, then
(a) E' and F are independent,
(b) E' and F' are independent.
Since A and B are independent events
∴ P(A ∩ B) = P(A) P(B) ...(1)
P(at least one of A and B) = P(A ∪ B)
= P(A) + P(B) - P(A ∩ B)
= P( A) + P(B) - P(A) P(B) [∵ of(1)]
= P(A) + P(B) [1 - P(A)]
= P(A) + P(B). P(A') = 1 - P(A') + P(B) P(A')
= 1 - P(A') [1 - P(B)] = 1 - P(A') P(B')
Two events A and B will be independent, if
B.
P(A'B') = [1 - P(A)] [1 - P(B)] P(A'B') = (P(A' ∩ B') = P(A ∪ B)'
= 1 - P(A ∪ B)
= 1 - [P(A) + P(B) - P(A ∩ B)]
= 1 - P(A) - P(B) + P(A ∩ B)
= 1 - P(A) - P(B) + P(A) P(B) [A and B are independent]
= [1 - P(A)] [1 - P(B)]
∴ (B) is correct answer.
Number of black balls = 10
Number of white balls = 5
∴ total number of balls = 10 + 5 = 15
Let E and F denote respectively the events that first and second ball drawn are black.
Now P(E) = P(black ball in first draw) = 
Since there is no replacement
By multiplication rule of probability,
Total number of cards = 52
Number of black cards = 26
Let E and F denote respectively events that first and second balls drawn are black
Required probability = P(E) P(F|E)
Here, 
P(head and number 6) = P(H) P(6) = 
Here, 
P (head and even number) = P(H) P (even) = 
A bag contains 8 marbles of which 3 are blue and 5 are red. One marble is drawn at random, its colour is noted and the marble is replaced in the bag. A marble is again drawn from the bag and its colour is noted. Find the probability that the marbles will be
(i) blue followed by red (ii) blue and red in any order (iii) of the same colour.
Number of blue balls = 3
Number of red balls = 5
Total number of balls = 3 + 5 = 8
(i) P(blue followed by red) = P (BR) = P(B) P(R)
(ii) P(blue and red in any order) = P (BR) + P (RB)
= P(B) P(R) + P(R) P(B)
(iii) P(of the same colour) = P(BB) + P(RR) = P(B) P(B) + P(R) P(R)
P(red card is first draw and black card in second draw) = 
P(black card is first draw and red card in second draw) = 
required probability = 
Let A and B denote the two events
Required probability = 
A bag contains 4 white balls and 2 black balls. Another bag contains 3 white balls and 5 black balls. If one ball is drawn from each bag, find the probability that both are black.
Total number of balls = 25
∴ total number of cases = 25
Let E denote the event of drawing even numbered ball and O denote the event of drawing odd numbered ball.
Number of balls numbered odd = 13
Number of balls numbered even = 12
Since the balls are replaced back before the next draw
Total number of balls = 25
∴ Total number of cases = 25
Let E denote the event of drawing even numbered ball and O denote the event of drawing odd numbered ball.
Number of balls numbered odd = 13
Number of balls numbered even = 12
Total number of balls = 25
∴ Total number of cases = 25
Let E denote the event of drawing even numbered ball and O denote the event of drawing odd numbered ball.
Number of balls numbered odd = 13
Number of balls numbered even = 12
Probaility of getting both the balls even numbered
= P(EE) = P(E) P(E) = 
P(at least one odd) = 1 - P(EE) = 1 - 
Total number of balls = 25
∴ Total number of cases = 25
Let E denote the event of drawing even numbered ball and O denote the event of drawing odd numbered ball.
Number of balls numbered odd = 13
Number of balls numbered even = 12
Probaility of getting both the balls even numbered
= P(EE) = P(E) P(E) = 
P(at least one odd) = 1 - P(EE) = 1 - 
P(no. odd number) = 1 - P(at least one odd) = 
A box of oranges is inspected by examining three randomly selected oranges drawn without replacement. If all the three oranges are good, the box is approved for sale, otherwise, it is rejected. Find the probability that a box containing 15 oranges out of which 12 are good and 3 are bad ones will be approved for sale.
Number of good oranges = 12
Number of bad oranges = 3
∴ total number of oranges = 12 + 3 = 15
Required probability = P(3 good oranges taken without replacement)
Two balls are drawn at random with replacement from a box containing 10 black and 8 red balls. Find the probability that both balls are red.
Number of black balls = 10
Number of red balls = 8
Total number of balls = 10 + 8 = 18
P(both balls are red) = P(RR) = P(R) P(R) = 
Two balls are drawn at random with replacement from a box containing 10 black and 8 red balls. Find the probability that first ball is black and second is red.
Number of black balls = 10
Number of red balls = 8
Total number of balls = 10 + 8 = 18
P(first ball is black and second is red) = P(BR)
Two balls are drawn at random with replacement from a box containing 10 black and 8 red balls. Find the probability that one of them is black and other is red.
Number of black balls = 10
Number of red balls = 8
Total number of balls = 10 + 8 = 18
P(one of them is black and other red) = P(BR) + P(RB) = P(B) P(R) + P(R) P(B)
Number of red balls = 7
Number of blue balls = 4
P(2 red balls) =
Number of red balls = 7
Number of blue balls = 4
P(2 blue balls) =
Number of red balls = 7
Number of blue balls = 4
P (one red and one blue ball) = 
An urn contains 4 red and 7 blue balls. Two balls are drawn at random with replacement. Find the probability of getting 2 red balls.
Number of red balls = 4
Number of blue balls = 7
Total number of balls = 4 + 7 = 11
P(both balls are red) = P(RR) = P(R) P(R) = 
Sponsor Area
An urn contains 4 red and 7 blue balls. Two balls are drawn at random with replacement. Find the probability of getting 2 blue balls.
Number of red balls = 4
Number of blue balls = 7
Total number of balls = 4 + 7 = 11
P(both balls are blue) = P(BB) = P(B) P(B)
= 
An urn contains 4 red and 7 blue balls. Two balls are drawn at random with replacement. Find the probability of getting one red and one blue ball.
Number of red balls = 4
Number of blue balls = 7
Total number of balls = 4 + 7 = 11
P(one of them is red and other blue) = P(RB) + P(BR) = P(R) P(B) + P(B) P(R)
= 
Number of red balls = 6
Number of blue balls = 4
Total number of balls = 6 + 4 = 10
P(both balls are red) = P(RR) = P(R) P(R)
Number of red balls = 6
Number of blue balls = 4
Total number of balls = 6 + 4 = 10
P(both balls are blue) = P(BB) = P(B) P(B)
Number of red balls = 6
Number of blue balls = 4
Total number of balls = 6 + 4 = 10
P(one of them is red and other blue) = P(RB) + P(BR) = P(R) + P(B) + P(B) P(R)
Group I Group II Group III
Boys 1 2 2
Girls 3 2 1
P (1 girl and two boys) = P(GBB) + P(BGB) + P(BBG)
= P(G) P(B) P(B)+ P(B) P(G) P(B) + P(B) P(B) P(G)
Number of white balls = 5
Number of black balls = 3
∴ total number of balls = 5 + 3 = 8
P (balls are of alternatively of different colours)
In first bag, there are 3 red and 5 black ball. In second bag, there are 6 red and 4 black balls.
P (one red and one black ball)= P (‘one red from first bag and one black from second bag’ or ‘one black from first bag and one red from second bag’)
= P (one red from first bag and one black from second bag) + P (one black from first bag and one red from second bag).
In first bag, there are 4 white and 2 black balls ; in second bag there are 3 white and 5 black balls. P (one white ball and one black ball) = P (‘one white ball from first bag and one black ball from second bag’ or ‘one black ball from first bag and one white ball from second bag’)
= P (one white ball from first bag and one black ball from second bag) + P (one black ball from first bag and one white ball from second bag)
In first bag. there are 4 red and 5 black balls and in second bag, there are 3 red and 7 black balls.
P(two red and one black ball) = P(‘one red ball from first bag and one red, one black from second bag’ or ‘one black ball from first bag and two red balls from second bag’ )
= P(one red ball from first bag and one red. one black from second bag) + P(one black ball from first bag and two red balls from second bag).
Fatima and John appear in an interview for two vacancies in the same post. The probability of Fatima's selection is 
that of John's selection is 
what is the probability that both of them will be selected?
Let A : Fatima is selected
B : John is selected
Clearly ‘A’ and ‘B’ are independent.
P(both of them will be selected)
Fatima and John appear in an interview for two vacancies in the same post. The probability of Fatima's selection is 
that of John's selection is 
What is the probability that only one of them is selected?
Let A : Fatima is selected
B : John is selected
Clearly ‘A’ and ‘B’ are independent.
P(only one of them will be selected)
= P('A and not B' or 'B and not A')
= P(A and not B) + P(B and not A)
= P(A) P(not B) + P(B) P(not A)
= P(A) [1 - P(B)] + P(B) [1 - P(A)]
Fatima and John appear in an interview for two vacancies in the same post. The probability of Fatima's selection is 
that of John's selection is 
What is the probability that none of them will be selected.
P(none of them will be selected)
= P(not A and not B)
= P(not A). P(not B)
= [1 - P(A)] [1 - P(B)]
=
and that of Tarun's selection is 
Find the probability that only one of them will be selected.
and that of Monica’s selection is 
Find the probability that only one of them will be selected.
respectively. Find the probability both selected.
Let A : Surjit is selected
B ; Sachdev is selected
Clearly A and B are independent.
P(both selected) = 
respectively. Find the probability only one of them selected.
Let A : Surjit is selected
B ; Sachdev is selected
P(only one is selected) = P('A and not B' or 'B and not A')
= P(A and not B) + P(B and not A)
= P(A) P(not B) + P(B) P(not A)
= P(A) [1 - P(B)] + P(B) [1 - P(A)]
respectively. Find the probability neither of them selected.
Let A : Surjit is selected
B ; Sachdev is selected
Clearly A and B are independent.
P(neither selected) = P(not A and not B) = P(not A) P(not B) = [1 - P(A)] [1 - P(B)]
= 
respectively. Find the probability both selected.
Let A : Balwant is selected
B : Kulwant is selected
Clearly A and B are independent.
P(both selected) = 
respectively. Find the probability only of them selected.
Let A : Balwant is selected
B : Kulwant is selected
P(only one is selected) = P('A and not B' or B' and not A')
= P (A and not B) + P(B and not A)
= P(A) P(not B) + P(B) P(not A)
= P(A) [1 - P(B)] + P(B) [1 - P(A)]
= 
respectively. Find the probability neither of them selected.
Let A : Balwant is selected
B : Kulwant is selected
P(neither selected) = P(not A and not B) = P(not A) P(not B) = [1 - P(A)] [1 - P(B)]
respectively. Find the probability ofLet A : Navdeepika is selected
B : Meenakshi is selected
Clearly A and B are independent.
(i) P(both selected) = 
(ii) P(only one is selected) = P('A and not B' or 'B and not A')
= P(A and not B) + P(B and not A)
= P(A) P(not B) + P(B) P(not A)
= P(A) [1 - P(B)] + P(B) [1 - P(A)]
(iii) P(neither selected) = P(not A and not B) = P(not A) P(not B)
The probability of A solving a problem is 
and that of B solving it is 
What is the probability that
(i) at least one of them will solve the problem?
(ii) Only one of them will solve the problem?
Let P(A), P(B) be the probability of A and B are speaking the truths. Then
Now, P(A and B contradict) = 
respectively. Find the probability that one and only one of them will hit the target when they fire simultaneously.
probability that at least one shot hit the plane.
respectively. What is the probability that exactly one of them may solve it?
respectively. Find the probability that exactly one of them may solve it.
respectively. If all the three try to solve the problem simultaneously find the probability that exactly one of them will solve it.
respectively, what is the probability that the problem will be solved?
respectively.
respectively.
and that of B hitting it is 
They both fire at the target. Find the probability that 
respectively. If both try to solve the problem independently, find the probability:
respectively. 
the probabilities of A and B solving the problems are 
respectively.
probability that problem is not solved by A and B = 
probability that the problem is solved = 
and B will solve it is 
. If both try the problem independently, find the probability that the problem will be solved.
The probabilities of A and B solving the problem are 
respectively.
the probabilities of A and B not solving the problems are 
i.e., 
respectively.
probabilities that problem is not solved by A and B 
probability that the problem will be solved = 
What is the probability that the problem will be solved?
The probabilities of three students solving the problem are 
respectively.
proabilites of three students not solving the problem are
the probability that the problem is not solved by any one of them
probability that the problem will be solved by atleast one of them
Let A and B be two events such that
A : a student reads Hindi newspaper
B : a student reads English newspaper.
P(student reads neither Hindi nor English newspaper)
Let A and B be two events such that
A : a student reads Hindi newspaper
B : a student reads English newspaper.
P(she reads English newspaper when it is given that she reads Hindi newspaper) = P(B | A)
Let A and B be two events such that
A : a student reads Hindi newspaper
B : a student reads English newspaper.
P(she reads Hindi newspaper when it is given that she reads English newspaper) = P(A | B)
Let W , L and T denote the events that the team wins, loses and ties respectively.
∴ P (W) = 0.7, P (L) = 0.2, P (T) = 0.1
Required probability = P (team wins at least two games but not loses)
= P (WWT) + P (WTW) + P (TWW) + P (WWW)
= P (W) P (W) P (T) + P (W) P (T) P (W) + P (T) P (W) P (W) + P(W) P(W) P(W)
= 3 [P (W) P (W) P (T)] + P (W) P (W) P (W)
= 3 [(0.7) (0.7) (0.1) + (0.7) (0.7) (0.7)
= 3 × 0.049 + 0.343 = 0.147 + 0.343
= 0.49
A can hit a target 4 times in 5 shots, B 3 times in 4 shots, and C 2 times in 3 shots. Calculate the probability that
(i) A B, C all may hit (ii) B, C may hit and A may lose.
A and B toss a coin alternatively till one of them gets a head and wins the game. If A starts the game, find the probability of his winning at his third toss.
Let P(A), 
be the probabilities of A's getting the head and not getting the head respectively.
Then 
Similarly, P(B) = 
Now A starts the game and wins in his third toss i.e., in 5th row
required probability = 
Let S denote the success (getting a '6') and F denote the failure (not getting a '6').
P(A wins in the first throw) = P(S)= 
A gets the throw, when the first throw by A and second throw by B result into failures. Therefore,
P(B wins) = 1 - P(A wins) = 
Let P(A), 
be the probabilities of A's getting the head and not getting the head respectively, then
Similarly, 
Let A start the game. He can win in the first throw, 3rd throw, 5th throw and so on.
Probability of A's winning in first throw = P(A) = 
Probability of A's winning in 3rd throw
Since all these cases are mutually exclusive
probability of A's winning the game first is
Since either A or B wins
probability of B's winning the game first = 
Let E be the event of getting sum 9 in a throw of two dice.
Now A wins in first throw or third row or 5th throw or................
proability of winning of A = P(E) + 
+ 
probability of winning of B = 
required ratio = 
Any person who starts the game can win in the first throw, 3rd throw, 5th throw and so on.
probability of winning A
Since either of two person wins
probability of winning B = 
required probabilities are 
Urn A Urn B Urn C
White 4 3 3
Blue 5 4 6
P(two third and one blue) = P(WWB) + P(WBW) + P(BWW)
= P(W) P(W) P(B) + P(W) P(B) P(W) + P(B) P(W) P(W) 
There are three urns A, B and C. Urn A contains 4 white balls and 5 blue balls. Urn B contains 4 white balls and 3 blue balls. Urn C contains 2 white balls and 4 blue balls. One half is drawn from each of these urns. What is the probability that out of these three balls drawn, two are white balls and one is a blue ball?
Urn A Urn B Urn C
White 4 4 2
Blue 5 3 4
P(two white and one blue) = P(WWB) + P(WBW) + P(BWW)
= P(W) P(W) P(B) + P(W) P(B) P(W) + P(B) P(W) P(W)
Bag A Bag B Bag C
White 5 7 6
Red 8 6 5
P(all three balls of the same colour) = P(all white or all red)
= P(all white) + P(all red)
= P(WWW) + P(RRR)
= P(W) P(W) P(W) + P(R) P(R) P(R)
Let A be the event that the construction job will be completed on time, and B be the event that there will be a strike.
Now P(B) = 0.65
P(no strike) = P(B’) = 1 - P(B) = 1 - 0.65 = 0.35
P(A | B) = 0.32, P(A | B') = 0.80
Since events B and B' form a partition of the sample space S,
∴ By theorem on total probability, we have
P(A) = P(B) P(A | B) + P(B') P(A | B’)
= 0.65 × 0.32 + 0.35 × 0.8
= 0.208 + 0.28 = 0.488
∴ the probability that the construction job will be completed in time is 0.488.
An urn contains 5 red and 5 black balls. A ball is drawn at random, its colour is noted and is returned to the urn. Moreover, 2 additional balls of the colour drawn are put in the urn and then a ball is drawn at random. What is the probability that the second ball is red?
Number of red balls = 5
Number of black balls = 5
∴ total number of balls = 10
Let A be the event that second drawn ball is red, and B be the event of drawing first ball as red and adding two red balls to urn.
Required probability = P(A)= P(B) P(A | B) + P(B') P(A | B')
= P (a red ball is drawn and returned along with 2 red balls and then a red ball is drawn) + P(a black ball is drawn and returned along with 2 black balls and then a red ball is drawn)
Bag I, Bag II
Number of white balls = 3, Number of white balls = 5
Number of red balls = 2, Number of red balls = 4
Total number of balls = 5, Total number of balls = 9
Let E1 and E2 be the events of the ball being drawn from bag I and bag II respectively and E be the event that the drawn ball is red.
Since both the bags are equally likely to be selected
P(the red ball is drawn from bag II) = 
|
Bag I |
Bag II |
|
Number of white balls = 4 |
Number of white balls = 6 |
|
Number of red balls = 3 |
Number of red balls = 5 |
|
Total number of balls = 7 |
Total number of balls = 11 |
Let E1 and E2 be the events of the ball being drawn from bag I and bag II respectively and E be the event that the drawn ball is red.
Since both the bags are equally likely to be selected
|
Bag I |
Bag II |
|
Number of white balls = 2 |
Number of white balls = 5 |
|
Number of red balls = 4 |
Number of red balls = 3 |
|
Total number of balls = 6 |
Total number of balls = 8 |
Let E1 and E2 be the events of the ball being drawn from bag I and bag II respectively and E be the event that the drawn ball is red.
Since both the bags are equally likely to be selected
Bag I contains 3 red and 4 black balls.
Bag II contain 4 red and 5 black balls.
Let E1 : Event that a red ball is drawn from bag I
E2 : Event that a black ball is drawn from bag I
After transferring a red ball from bag I to bag II, the bag II will have 5 red and 5 black balls.
Let A be the event of drawing red ball
Further when a black ball is transferred from bag I to bag II, it will contain 4 red and 6 black balls.
Required probability = 
Let E1, E2, E be the events
E1 : ‘A student is residing in hostel’
E2 : ‘A student is day scholar’,
E : ‘A student gets A grade,’
Now, 
Required probability = 
Let E1 and E2 be the events of the ball being drawn from bag A and bag B respectively and E be the event that the drawn ball is red.
Since both the bags are equally likely to be selected i.e., E1 and E2 are mutually exclusive and exhaustive events.
be the probability that he knows the answer and 
be the probability that he guesses. Assuming that a student who guesses at the answer will be correct with probability 
What is the probability that the student knows the answer given that he answered it correctly?
Let E1, E2, E be the events
E1 : ‘the student knows the answer’
E2 : ‘the student guesses the answer’,
E : ‘the answer is correct’
(By Baye's Theorem)
Let E1, E2, E be the events
E1 : ‘the person has the disease’
E2 : ‘the person is healthy’,
E : ‘test is positive’,
Required probability = 
(By Bare's Theorem)
Let E1, E2, E be the events
E1 : ‘selected person is a male’
E2 : ‘selected person is a female’,
E : ‘selected person is grey haired’
and 
Required probaility = 
Let A be the event that the machine produces 2 acceptable items.
Again, let B1 represent the event of correct setup and B2 represent the event of incorrect setup.
Now, 
Let E1 and E2 be the events
E1 : Treatment of yoga and meditation
E2 : Treatment of prescription of certain drugs
Let A denotes that a person has heart attack
∴ P(A) = 40% = 0.40
Yoga and mediation reduces the heart risk by 30%
i.e. inspite of getting yoga and meditation treatment heart risk is 70% of the 0.40
∴ P(A | E1) = 0.40 × 0.70 = 0.28
Drug prescription reduces the heart risk by 25%.
Even after adopting the drug prescription heart risk is 75% of the 0.40
∴ P(A | E2) = 0.40 × 0.75 = 0.30
Let E1, E2, E be the events
E1 : ‘First group wins’,
E2 : ‘Second group wins’,
E : ‘New product is introduced’
Required probability = 
Let E1, E2, E be the events
E1 : ‘1, 2, 3 or 4 is shown on die’,
E2 : ‘5 or 6 is shown on die’,
E : ‘exactly one head shows up’
and 
Required probability = 
Let E1, E2, E he the events
E1 : ‘lost card is a diamond’,
E2 : ‘lost card is not a diamond’,
E : ‘two cards drawn from the remaining pack are diamonds’
Also, 
and 
Required probability = 
Suppose that the reliability of HIV test is specified as follows:
Of people having HIV, 90% of the test detect the disease but 10% go undetected. Of people free of HIV, 99% of the test are judged HIV -ve but 1% are diagonsed as showing HIV +ve. From a large population of which only 0.1% have HIV, one person is selected at random, given the HIV test, and the pathologist reports him/her as HIV +ve. What is the probability that the person actually has HIV?
by Baye's Law
Let E1, E2, E3, E be the events
E1 : ‘coin chosen is two headed’,
E2 : ‘coin chosen is biased’,
E3 : ‘coin chosen is unbiased’,
E : ‘tossed coin shows up a head’
Required probability = 
Now. the probability that the other coin in the box is of gold
= the probability that gold coin is drawn from the box I
= P(E1 | A)
By Bayes’ theorem, we know that
Let E1, E2, E3, E be the events
E1 : ‘Item is produced by operator A’,
E2 : ‘Item is produced by operator B’,
E3 : ‘Item is produced by operator C’,
E : ‘Item chosen is found defective’
Required probability = 
(By Baye's Theorem)
respectively. The probabilities that he will be late are 
if he comes by train, bus or scooter respectively but by taxi he will not be late. When he arrives he is late What is the probability that he came by bus?
The probability that he will be late are 
if he comes by train, bus and scooter respectively, but if he comes by other means of transport, then he will not be late. When he arrives, he is late. What is the probability that he comes by train ?
, since he is not late if he comes by other means of transport.
and the probability that he copies the answer is 
The probability that his answer is correct (given that he copied it) is 
Find the probability that he knew the answer to the question (given that he correctly answered it).
Let E1 be the event that the letter has come from TATANAGAR and E2 be the event that it has come from CALCUTTA.
Let A denote the event that the two consecutive letters on the envelope are TA.
∴ E1 and E2 are mutually exclusive and exhaustive events
P(A/E1) = Probability that the consecutive letters TA visible on the envelope belong to TATANAGAR.
Similarly P(A/E2) = Probability that the consecutive letters TA are visible belong to CALCUTTA
Let A be the event that the drawn balls are, one red and one white.
Since, the urns are equally likely
∴ P( A/E1) = Probability that one red and one white ball are drawn from first urn
Let E1 the event that the missing card is black and E2 be the event that the missing card is red.
Let A be the event the first 13 cards which are examined are all red.
P(A/E1) = Probability of selecting 13 red cards, when the missing card is black
P(A/E2) = Probability of selecting 13 red cards, when the missing card is red
Coloured balls are distributed in four boxes as shown in the following table:
|
Box |
Colour |
|||
|
Black |
White |
Red |
Blue |
|
|
I |
3 |
4 |
5 |
6 |
|
II |
2 |
2 |
2 |
2 |
|
III |
1 |
2 |
3 |
1 |
|
IV |
4 |
3 |
1 |
5 |
Let A, E1, E2, E3 and E4 be the events as defined below:
A : a black ball is selected E1 : box I is selected
E2 : box II is selected E3 : box II is selected
E4 : box IV is selected
Since the boxes are chosen at random.
Also, 
P(box III is selected, given that the drawn ball is black) = P(E3 | A).
Suppose we have four boxes A, B, C and D containing coloured marbles as given below:
|
Box |
Marble Colour |
||
|
Red |
White |
Black |
|
|
A |
1 |
6 |
3 |
|
B |
6 |
2 |
2 |
|
C |
8 |
1 |
1 |
|
D |
0 |
6 |
4 |
One of the boxes has been selected at random and a single marble is drawn from it. If the marble is red. what is the probability that it was drawn from box A? box B? box C?
Let E, E1, E2, E3 and E4 be the events as defined below:
E : ‘ball drawn is red’, E1 : ‘box A is selected’, E2: ‘box B is selected’,
E3 : ‘box C is selected ‘ and E4 : ‘box D is selected’,
and 
and 
P(ball is drawn from box A) = 
P(ball is drawn from box C) = 
and P(ball is drawn from box A) = 
A coin is tossed. A reports that a head appears. The probability that actually there was head is
A.
Let E1, E2, E be the events
E1 : ‘coin comes up with a head’,
E2 : ‘coin comes up with a tail’,
E : ‘A reports that a head appears’
and 
∴ (A) is correct answer.
X is a number whose values are defined on the outcomes of a random experiment.
∴ X is a random variable.
Now S = {HHH, HHT, HTH, THH, HTT, THT, TTH, TTT}
Then X(HHH) = Rs. (2 × 3) = Rs. 6 X(HHT) = X(HTH) = X(THH)
= Rs. (2 × 2 - 1 × 1.50) = Rs. 2.50
X(HTT) = X(THT) = (TTH) = Rs. (1 × 2 - 2 × 1.50)
= - Re 1
and X(TTT) = - Rs. (3 × 1.50) = - Rs. 4.50
where, minus sign shows the loss to the player. Thus, for each element of the sample space, X takes a unique value, hence, X is a function on the sample space whose range is {- 1, 2.50, - 4.50, 6}.
Number of red balls = 5
Number of black balls = 2
These two balls may be selected as RR, RB, BR, BB, where R represents red and B represents black ball.
Variable X has the value 0, 1, 2 i.e., there may be no black balls, may be one black ball, or both the balls are black.
Yes, X is a random variable.
Let a denote the number of heads and b, the number of tails when a coin is tossed 6 times, then
X = difference between a and b = |a - b|.
Here, both a and b can take values 0, 1, 2, 3, 4, 5, 6 but a + b is always equal to 6.
∴ we have the table:
Table
|
a |
0 |
1 |
2 |
3 |
4 |
5 |
6 |
|
b |
6 |
5 |
4 |
3 |
2 |
1 |
0 |
|
X |
6 |
4 |
2 |
0 |
2 |
4 |
6 |
∴ X takes values 0, 2, 4 and 6.
Let the balls in the bag be denoted by w1, w2, r.
∴ S = {w1 w1 , w1 w2 , w2 w2, w2 w1, w1 r, w2 r, r w1, r w2, r r}
Now, for ω ∈ S
X(ω) = number of red balls
∴ X({w1 w1}) = X({w1 w2}) = X({w2 w2}) = X({w2 w1}) = 0
X({w1 r}) = X({w2 r}) = X({r w1}) = X({r w2}) = 1
and X({r r}) = 2
∴ X is a random variable which can take values 0, 1 or 2.
|
X |
0 |
1 |
2 |
|
P(X) |
0.4 |
0.4 |
|
|
X |
0 |
1 |
2 |
|
P(X) |
0.4 |
0.4 |
0.2 |
Now 0 ≤ P(X) ≤ 1
and Σ P(X) = 0.4 + 0.4 + 0.2 = 1
∴ given distribution is a probability distribution of a random variable.
|
X |
0 |
1 |
2 |
3 |
4 |
|
P(X) |
0.1 |
0.5 |
0.2 |
-0.1 |
0.3 |
|
X |
0 |
1 |
2 |
3 |
4 |
|
P(X) |
0.1 |
0.5 |
0.2 |
-0.1 |
0.3 |
Here P(X) = -0.1 for X = 3
This is not possible as 0 ≤ P(X) ≤ 1
∴ given distribution is not a probability distribution of a random variable
|
Y |
-1 |
0 |
1 |
|
P(Y) |
0.6 |
0.1 |
0.2 |
|
Y |
-1 |
0 |
1 |
|
P(Y) |
0.6 |
0.1 |
0.2 |
ΣP(Y) = 0.6 + 0.1 + 0.2 = 0.9 ≠ 1
This is not possible as ΣP(Y) = 1
∴ given distribution is not a probability distribution of a random variable.
|
Z |
3 |
2 |
1 |
0 |
-1 |
|
P(Z) |
0.3 |
0.2 |
0.4 |
0.1 |
0.05 |
|
Z |
3 |
2 |
1 |
0 |
-1 |
|
P(Z) |
0.3 |
0.2 |
0.4 |
0.1 |
0.05 |
ΣP(Z) = 0.3 + 0.2 + 0.4 + 0.1 + 0.04 = 1.05 > 1
This is not possible as Σ P(Z) = 1
∴ given distribution is not a probability distribution of a random variable.
(a) Find the value of k.
(b). What is the probability that you study at least two hours ? Exactly two hours ? At most two hours?
The probability distribution of X is
|
X |
0 |
1 |
2 |
3 |
4 |
|
P(X) |
0.1 |
k |
2 k |
2k |
k |
(a) We know that
(b) P(you study at least two hours) = P(X ≥ 2)
= P(X = 2) + P(X = 3) + P(X = 4)
= 2k + 2k + k = 5k = 5 × 0.15 = 0.75
P(you study exactly two hours) = P(X = 2)
= 2k = 2 × 0.15 = 0.3
P(you study at most two hours) = P(X ≤ 2)
= P(X = 0) + P(X = 1) + P(X = 2)
= 0.1 + k + 2k = 0.1 + 3k
= 0.1 + 3 × 0.15 = 0.55
|
X |
0 |
1 |
2 |
3 |
4 |
5 |
6 |
1 |
|
P(X) |
0 |
k |
2k |
2k |
3k |
k2 |
2k2 |
7k2 + k |
(i) Since Σ P(X) = 1
∴ P(0) + P(1) + P(2) + P(3) + P(4) + P(5) + P(6) + P(7) = 1
⇒ 0 + k + 2k + 2k + 3k + k2 + 2k2 + 7 k2 + k = 1
But k cannot be negative as probability is always non-negative.
∴ probability distribution of X is
(a) Determine the value of k.
(b) Find P(X < 2), P(X ≤ 2), P(X ≥ 2).
The probability distribution of X is
|
X |
0 |
1 |
2 |
|
P(X) |
k |
2k |
3k |
Here X denotes the number of heads obtained in three tosses of a coin. Thus X can take the values 0, 1, 2, 3.
Let p be the probabilities of getting a head and q be the probability of not getting a head,
∴ probability distribution table is
Here X denotes the number of heads obtained in three tosses of a coin. X can take the values 0, 1, 2, 3, 4.
Let p be the probability of getting a head and q be the probability of not getting a head.
P(X = 2) = p p q q + p q p q + q p p q + p q q p + q p q p + q q p p
P(X = 3) = p p p q + p p q p + p q p p + q p p p
P(X = 4) = p p p p = 
∴ probability distribution of X is
Here X denotes the number of heads obtained in five tosses of a coin. X can take the values 0, 1, 2, 3, 4, 5.
Let p be the probability of getting a head and q be the probability of not getting a head.
∴ probability distribution of X is
Here X denotes the number of tails obtained in three tosses of a coin. Thus X can take the values 0, 1, 2, 3.
Let p be the probability of getting tail and q be the probability of not getting a tail.
∴ probability distribution table is
Let X denote the random variable ‘number of red balls’. X can take the values 0, 1, 2, 3.
Let p be the probability of getting red ball.
∴ probability distribution is
Let X denote the random variable ‘number of green balls’. X can take the values 0, 1, 2, 3.
Let p be the probability of getting first green ball and q the probability of not getting first green ball.
∴ probability distribution is
Let X denote the random variable ‘number of white balls’. X can take the values 0, 1, 2, 3.
Let p be the probability of getting first white ball and q the probability of not getting first white ball.
∴ probability distribution is
Number of white balls = 2
Number of red balls = 3
Number of blue balls = 4
Total number of balls = 2 + 3 + 4 = 9
Let X denote the event ‘the number of white balls’. X can take the values 0, 1, 2.
Let p be the probabilities of getting white ball and q be the probabilities of not getting white ball.
∴ probability distribution is
Total number of eggs = 12
Number of good eggs = 10
∴ number of bad eggs = 2
Let X denote the random variable. X takes the values 0, 1, 2.
∴ probability distribution table is
Let X denote the number of doublets. X can take the value 0, 1, 2, or 3.
Possible doublets are (1, 1), (2, 2), (3, 3), (4, 4), (5, 5), (6, 6)
Probability of getting a doublet = 
Probability of not getting a doublet = 
Now, P(X = 0) = P(no doublet) = 
and 
∴ the required probability distribution is
Let p be the probability of obtaining a head when a coin is tossed once and q that of obtaining a tail
From given condition,
p = 3 q, p + q = 1
Let X denote the number of tails in two tosses of the coin, then X can take values 0, 1 and 2.
and 
∴ probability distribution of X is
respectively.
When a dice is rolled once, probability of obtaining six = 
and probability of obtaining a non-six = 
Now X can take values 0, 1, 2.
P(X = 0) = P(six does not appear a die) = P(non-six on both dice)
P(X = 1) = P(six appear exactly on one die)
= P(six on first die and non-six on the second die)
+ P(six on the second die and non-six on the first die)
P(X = 2) = P(six on both the dice) = 
∴ probability distribution of X is
∴ Expectation of X = mean of the variable X
Here X denotes the number of heads obtained in three tosses of a coin. Thus X can take the values 0, 1, 2, 3.
Let p be the probabilities of getting a head and q be the probability of not getting a head,
∴ probability distribution table is
Since 1 cannot be greater than the other selected number.
∴ X can take values 2, 3, 4, 5, 6.
∴ P(X = 2) = P(2 and a number less than 2 are selected)
= P(1 and 2 are selected)
P(X = 3) = P(3 and a number less than 3 are selected)
[∵ (1, 3) and (2, 3) can be selected]
P(X = 4) = P(4 and a number less than 4 are selected)
[∵ (1, 4), (2, 4) and (3, 4) can be selected]
P(X = 5) = P(5 and a number less than 5 are selected)
[∵ (1, 5), (2, 5), (3, 5) and (4, 5) can be selected]
P(X = 6) = P(6 and a number less than 6 are selected)
[∵ (1, 6), (2, 6), (3, 6), (4, 6) and (5, 6) can be selected]
∴ probability distribution table is
Expectation of X = E(X) = 
Table
|
Age X |
Frequency |
|
14 |
2 |
|
15 |
1 |
|
16 |
2 |
|
17 |
3 |
|
18 |
1 |
|
19 |
2 |
|
20 |
3 |
|
21 |
1 |
Probability Distribution Table
Mean X = 
Variance X = 
Here S = {1, 2, 3, 4, 5, 6}.
Let X denote the number obtained on the throw. Then X is a random variable which can take values 1, 2, 3, 4, 5 or 6.
Also, P(1) = P(2) = P(3) = P(4) = P(5) = P(6) = 
∴ the probability distribution of X is
Now, 
Also, 
X takes the values 0, 1, 2 with probabilities 
respectively.
Number of good eggs = 7
Number of bad eggs = 3
∴ total number of eggs = 7 + 3 = 10
Let X denote the random variable ‘bad egg is obtained’. X takes the values 0, 1, 2, 3.
X takes the values 0, 1, 2, 3 with probabilities 
Let X denote the number of aces obtained in two draws. X takes the values 0, 1, 2.
Let p denote probability of first ace.
∴ probability distribution is
When X = 1, the man gains Re 1.
When X = 2, the man does not gain anything.
[∵ on first throw he loses Re 1 and on second throw he gains Re 1]
When X = 3,
(i) the man may lose Re 3 when all the three throws show a non-six, which happens with probability 
(ii) the man may lose Re 1 when first two throws show a non-six and third shows a six, which happens with probability 
If Y is the amount gained or lost, then Y takes values 1, 0, - 3, - 1.
∴ probability distribution of Y is
Expected value of Y = 
Hence, the man is expected to lose Rs. 
1
2
5
B.
2
Let X denote the number on the die, then X takes values 1, 2 and 5 only.
Six balls are drawn successively from an urn containing 7 red and 9 black balls. Tell whether or not the trials of drawing balls arc Bernoulli trials when after each drawn the ball drawn is (i) replaced (ii) not replaced in the urn.
which is same for all six trials (draws).
in 2nd trial is 
if the first ball drawn is red or 
if the first ball drawn is black and so on. Clearly, the probability of success is not same for all trials, hence the trials are not Bernoulli trials.
Here n = 10
Let p be the probability of success
Here n = 4
Let p be the probability that doublet is obtained
Here n = 10
Let p be the probability that doublet is obtained
Required probability = P(4) = 
Here, n = 4, p = 
(i) P(none is white) = P(0) = 
(ii) P(all white) = P(4) = 
(iii) P(only 2 are white) = P(2) = 
(iv) P(at least one is white) = P(1) + P(2) + P(3) + P(4)
Five cards are drawn successively with replacement from a well shuffled deck of 52 cards. What is the probability that
(i) all the five cards are spades?
(ii) only 3 cards are spades?
(iii) none is a spade?
Here n = 5
p = P(a spade in a single draw) = 
(i) P(all the five cards are spades)
(ii) P (only 3 cards are spades)
(iii) P (none is a spade)
Here, n = 7
p = P(5) = 
P(exactly twice) = 
Here n = 5
p = P(spade) = 
P(4 spades) = 
Here n = 8
Required probability = 
Here n = 10
Required probability = 
Here n = 45
p = P (ball drawn is marked with 0) = 
P(none is marked with the digit 0) = P(0) = 
Here n = 5
P(four or more correct answers) = P(4) + P(5)
Here 
Required probability = P(no defective) + P(1 defective)
Here n = 8
P(less than 2 defective )
= P(0) + P(1)
Here n = 5
P(not more than 2 defective items) = P(0) + P(1) + P(2)
Here n = 8
P(not more than one defective item) = P(0) + P(1)
Here n = 8
P(less than 2 defective item) = P(0) + P(1)
Here n = 10
P(not more than 3 defective items) = P(0) + P(1) + P(2) + P(3)
Here n = 5
p = P(spade) = 
(i) P(5 spades) = 
(ii) P(3 spades) = 
(iv) P(none is a spade) = 
Here n = 6
p = P(6) = 
P(at most 2 sixes) = P(0) + P(1) + P(2)
He fires 7 times. What is the probability of his hitting at least twice the target?
Here n = 7
P(hitting at least twice) = 1 - [P(0) + P(1)]
If he fires 10 rounds, what is the probability of his hitting the target at least twice?
Here n = 10, 
P(hitting at least twice) = 1 - [P(0) + P(1)]
What is the probability that the will knock down fewer than 2 hurdles?
Here n = 10
Probability of clearing each hurdle = 
Let p be the probability of knocking each hurdle
P(fewer than 2 hurdles) = P(0) + P(1)
Here n = 6, p = 0.6 
P(dacoit is still alive) = P(0) = 
Here n = 8, 
P (passing the examination) =P(6) + P(7) + P(8)
Here n = 6
Now, p + q = 1 and p = 2q
P(at least 4 successes in 6 trials) = 
Here n = 10
P(at least one defective egg) = 1 - P(0) = 
Here n = 6
p = P(getting an odd number) = P(1, 3, 5) = 
(i) P(five successes) = P(5) = 
(ii) P(at least 5 successes) = P(5) + P(6)
(iii) P(atmost 5 successes) = 
Here n = 5
P(not more than one bulb will fuse after 150 days)
Here n = 5
P(at least one bulb wil fuse after 150 days)
Here n = 5, p = 0.06 = 
P(at most one bulb will fuse) = P(0) + P(1)
We get a doublet when we get
(1, 1), (2, 2), (3, 3), (4, 4), (5, 5), (6, 6)
Let p be probability of getting a doublet
P(at least two doublets) = 1 - [P(0) + P(1)]
A pair of dice is thrown. Therefore various outcomes are
|
11 |
12 |
13 |
14 |
15 |
16 |
|
21 |
22 |
23 |
24 |
25 |
26 |
|
31 |
32 |
33 |
34 |
35 |
36 |
|
41 |
42 |
43 |
44 |
45 |
46 |
|
51 |
52 |
53 |
54 |
55 |
56 |
|
61 |
62 |
63 |
64 |
65 |
66 |
∴ total number of outcomes = 36
Favorable outcomes are 36, 45, 54 and 63.
∴ number of favorable outcomes = 4
Here n = 10
We get total of 9 when we get, (3, 6), (4, 5), (5, 4), (6, 3)
P(getting sum 9 at least twice) = 1 - [P(0) + P(1)]
Here n = 6
In a throw of pair of dice, we get total of 7 when we get
(1, 6) (2, 5), (3, 4), (4, 3), (5, 2), (6, 1)
Here n = 6
In a throw of pair dice, we get total of 7 when get
(1, 6), (2, 5), (3, 4), (4, 3), (5, 2), (6, 1)
P(at least 5 successes) = P(5) + P(6)
Let p be the probability of throwing 4
P(at least one die shows up) = 1 - P(0)
Here n = 6
Let p be the probability of getting a head
P(at least 2 heads) = 1 - 
Here n = 6
Let p be the probabilities of getting a head
P(at least four heads) =P(4) + +P(5) + P(6)
Here n = 5
Let p be the proability of getting a tail
P(at least 2 tails) = 1 - 
Here n = 7
Let p be the probability of getting a tail
P(at least 5 tails) = P(5) + P(6) + P(7)
Here n = 3
Let p be the probability of getting a head.
P(at least 2 heads) = P(2) + P(3)
Here n = 6
Let p be the probability of getting 5 or 6
P(at least two successes) = 1 - P(0) + P(1)
Here n = 3
p = P(2 or 5) = 
q = 1 - p = 1 - 
P(at least two successes) = P(2) + P(3)
n = 5
p = P(3 or 4 or 5) = 
P(at least three successes) = P(3) + P(4) + P(5)
Here n = 5
P(head appearing an even number of times)
Here n = 7
p = P(tail in one toss) = 
P(tails appears on odd number of times) = P(1) + P(3) + P(5) + P(7)
Here n = 5
p = P(head) = 
P(head appears an odd number of times) = P(1) + P(3) + P(5)
Here n = 4
P(at least three successful operations) = P(3) + P(4)
Here n = 6
P(at least three numbers are busy) = P(3) + P(4) + P(5) + P(6)
Here n = 4
Let p be the probability of drawing a white ball
P(at least one ball is white) = 1 - P(0)
Here n = 7
In a throw of pair of dice, we get total of 9 when we get (3, 6), (4, 5), (5, 4), (6, 3)
P(atmost 6 successes) = 1 - P(7) = 1 - 
Here n = 5, p = P(1, 3, 5) = 
q = 1 - p = 1 - 
P(atmost three successes) = 1 - [P(4) + P(5)]
Here n = 6, p = P(3, 6) = 
P(atmost 5 successes) = 1 - P(6) = 
Here n = 5, p = P(getting head) = 
, q = 1 - p = 1 - 
(i) P(at least 3 heads) = P(3) + P(4) + P(5)
(ii) P(at most 2 heads) = 
Here n = 6
p = P{(2, 6), (3, 5), (4, 4), (5, 3), (6, 2) = 
q = 1 - p = 1 - 
(i) P(at least 4 successes) = P(4) + P(5) + P(6)
(ii) P(at atmost 5 successess) = 1 - P(6) = 1 - 
= 
Here n = 6
p = Probability of head = 
P (exactly 4 heads) = P (4)
Here n = 6
p = Probability of head = 
P (less than 3 heads) = P (0) + P (1) + P (2)
Here n = 6
p = Probability of head = 
P (more than 4 heads) = P (5) + P (6)
Here n = 6
p = Probability of head = 
P(more than 4 heads and less than 6 heads) = P(5)
P(more than 6 heads) = P (impossible event) = 0
Here n = 6
p = Probability of head = 
P (at least 4 heads) = P (4) + P (5) + P (6)
Here n = 6
p = Probability of head = 
P (at most 4 heads) = 1 - [P (5) + P (6)]
Here n = 6
p = Probability of head = 
P(2 heads) = P(2) = 
Here n = 6
p = Probability of head = 
Here n = 5
Let p be the probability of getting defective bulbs
(i) Required probability = P(0)
(ii) Required probability = P (2)
Here n = 5
Let p be the probability of getting 6
Here n = 5
p = (Probability of getting 5 or 6) = 
A pair of dice is thrown 7 times. If getting a total of 7 is considered a success, what is the probability of
(i) no success? (ii) 6 successes? (iii) at least 6 successes? (iv) at most 6 successes?
Here n = 7
Let p be the probability of getting a total of 7.
(i) P(no success) = P(0) = 
(ii) P(6 successes) = P(6) = 
(iii) P(at least 6 (successes) = P(6) + P(7)
(iv) P(at most 6 successes) = P(not 7 success)
Show that X = 3 is the most likely outcome. Since 
is binomial distribution
Since 
is maximum of all the above values
Let X be the random variable whose probability distribution is 
Now 
∴ the distribution of X is
mean = 
How many minmum number of times must he/she fire so that the probability of hitting the target at least once is more than 0.99?
Let the shooter fire n times.
Now, 
Now, P(hitting the target at least once) > 0.99 (given)
or 
...(1)
The minimum value of n to satisfy the inequality (1) is 4.
∴ the shooter must fire 4 times.
Here n = 10
P = Probability of head = 
(i) P(exactly six heads) = P(6)
(ii) P(at least six heads) = P(6) + P(7) + P(8) + P(9) + P(10)
(iii) P(at most sixes) = P(0) + P(1) + P(2) + P(3) + P(4) + P(5) + P(6)
What is the probaility that he will win a prize exactly once (a) at least once (b) exactly once (c) at least twice?
Here n = 50
(a) P(at least once) = 1 - P(0) = 
(b) P(exactly once) = P(1) = 
(c) P(at least twice) = 1 - [P(0) + P(1)]
An urn contains 25 balls of which 10 balls bear a mark ‘X’ and the remaining 15 bear a mark ‘Y’ . A ball is drawn at random and it is replaced. If 6 balls are drawn in this way, find the probability that
(i) all will bear mark X
(ii) not more than 2 balls will bear 'Y' mark.
(iii) at least one ball will bear 'Y' mark.
(iv) the number of balls with 'X' mark and 'Y' mark will be equal.
Here n = 6
p = (Probability that a ball marked 'X' is drawn) = 
(i) P(all balls bear X mark) = P(6) = 
(ii) P(not more than 2 will bear Y mark)
= P(not less than 4 will bear X mark) = P(4) + P(5) + P(6)
(iii) P(number of balls with X mark and Y mark equal)
(iv) P(at least one ball bear Y mark)
= P(not more than 5 balls bear mark X)
Here n = 20
p = P(answers true) = P(head) = 
P(answers at least 12 questions correctly)
= P(12) + P(13) + P(14) + P(15) + P (16) + P(17) + P(18) + P(19) + P(20)
Here n = 10
Required probability 
C.
Here n = 5
P(none is defective) = P(0) = 
∴ (C) is correct answer.
Then the probability that out of five students, four are swimmers is
None of these
A.
Here n = 5
∴ (A) is correct answer.
F = {(1, 1), (2, 2), (3, 3), (4, 4), (5, 5), (6, 6)}
B wins in second throw or fourth throw or 6th throw or ..........
probability of winning of B = 
An experiment succeeds thrice as often as it fails. Find the probability that in the next five trials, there will be at least 3 successes.
An experiment succeeds as often as it fails.
Therefore, there are 3 successes and 1 failures.
Thus the probability of success 
And the probability of failure = 
We need to find the probability of atleast 3 successes in the next five trials.
Required Probability = P(X= 3) + P(X = 4) + P(X = 5)
Two numbers are selected at random (without replacement) from the first six positive integers. Let X denote the larger of the two numbers obtained. Find the probability distribution of the random variable X, and hence find the mean of the distribution.
If 1 is the smallest number,
the other numbers are:2,3,4,5,6
If 2 is the smallest number, the other numbers are:3,4,5,6
If 3 is the smallest number, the other numbers are:4,5,6
If 4 is the smallest number, the other numbers are: 5, 6
If 5 is the smallest number, the other number is:6
Thus, there are 15 set of numbers in the sample space.
Let X be
X: 2 3 4 5 6
1/15 2/15 3/15 4/15 5/15
We know that,
A speaks truth in 60% of the cases, while B in 90% of the cases. In what percent of cases are they likely to contradict each other in stating the same fact? In the cases of contradiction do you think, the statement of B will carry more weight as he speaks truth in more number of cases than A?
Let the probability that A and B speak truth be P(A) and P(B) respectively.
Therefore, 
A and B can contradict in stating a fact when one is speaking the truth and other is not speaking the truth.
Case 1: A is speaking the truth and B is not speaking the truth.
Case 2: A is not speaking the truth and B is speaking the truth.
Required Probability = 
Percentage of cases in which they are likely to contradict in stating the same fact =
From case 1, it is clear that is not necessary that the statement of B will carry more weight as he speaks truth in more number of cases than A.
Assume that the chances of a patient having a heart attack is 40%. Assuming that a meditation and yoga course reduces the risk of heart attack by 30% and prescription of certain drug reduces its chance by 25%. At a time a patient can choose any one of the two options with equal probabilities. It is given that after going through one of the two options, the patient selected at random suffers a heart attack. Find the probability that the patient followed a course of meditation
Let A, E1, and E2, respectively denote the events that a person has a heart attack, the selected person followed the course of yoga and meditation, and the person adopted the drug prescription.
Probability that the patient suffering a heart attack followed a course of meditation and yoga = 
Now, calculate 
Since 
, the course of yoga and meditation is more beneficial for a patient.
A black and a red die are rolled together. Find the conditional probability of obtaining the sum 8, given that the red die resulted in a number less than 4.
Here
A number of elements (outcomes) of the above sample space is 6 × 6 = 36.
Let E: a total of 8
Two numbers are selected at random (without replacement) from the first five positive integers. Let X denote the larger of the two numbers obtained. Find the mean and variance of X.
X can take values as 2, 3, 4,5 such that,
P (X =2) = probability that the larger of two number 2.
= prob. of getting 1 in first selection and 2 in second selection getting 2 in first selection and 1 in second selection.
| x | 2 | 3 | 4 | 5 |
| P(X) |
Suppose a girl throws a die. If she gets 1 or 2 she tosses a coin three times and notes the number of tails. If she gets 3,4,5 or 6, she tosses a coin once and notes whether a ‘head’ or ‘tail’ is obtained. If she obtained exactly one ‘tail’, what is the probability that she threw 3,4,5 or 6 with the ride ?
Let A be the event that girl will get 1 or 2
Let B be the event that girl will get 3,4, or 6
A pair of dice is thrown 4 times. If getting a doublet is considered a success, find the probability distribution of the number of successes.
Total number of outcomes = 36
The possible doublets are ( 1, 1 ), ( 2, 2 ), ( 3, 3 ), ( 4, 4 ), ( 5, 5 ), and ( 6, 6).
Let p be the probability of success, therefore,
Since the dice is thrown 4 times, n = 4
Let X denote the number of times of getting doublets in the experiment of throwing two dice simultaneously four times.
Therefore X can be take the values 0, 1, 2, 3, or 4.
Thus the probability distribution is:
| X | 0 | 1 | 2 | 3 | 4 |
| P(X) |
An insurance company insured 2000 scooter drivers, 4000 car drivers and 6000 truck drivers. The probability of an accident involving a scooter, a car and a truck are 0.01, 0.03 and 0.15 respectively. One of the insured persons meets with an accident. What is the probability that he is a scooter driver?
Let E1, E2 and E3 be the events of a driver being a scooter driver, car driver and truck driver respectively. Let A be the event that the person meets with an accident.
There are 2000 insured scooter drivers, 4000 insured car drivers and 6000 insured truck drivers.
Total number of insured scooter drivers vehicle drivers = 2000 + 4000 + 6000 = 12000
Also, we have:
Now, the probability that the insured person who meets with an accident is a scooter driver is
Using Baye's theorem, we obtain:
A die is thrown again and again until three sixes are obtained. Find the probability of obtaining the third six in the sixth throw of the die.
p = probability of success = , q = probability of failure =
Third six comes at the 6 th throw so the remaining two sixes can appear in any of the previous 5 throws.
Probability of obtaining 2 sixes in 5 throws
6 th throw definitely gives six with probability =
Required probability
Two groups are competing for the position on the Board of Directors of a corporation. The probabilities that the first and the second groups will win are 0.6 and 0.4 respectively. Further, if the first group wins, the probability of introducing a new product is 0.7 and the corresponding probability is 0.3 if the second group wins. Find the probability that the new product was introduced by the second group.
Let E1 be the event of the first group winning and E2 be the event of the second group winning and S be the event of introducing a new product.
P ( E1 ) = probability that the first group wins the competition = 0.6
P ( E2 ) = probability that the second group wins the competition = 0.4
p ( SE1 ) = Probability of introducing a new product if the first group wins = 0.7
p ( SE2 ) = Probability of introducing a new product if the second group wins = 0.3
The probability taht the new product is introduced by the second group is given by P ( E2S ).
By using Bayes' theorem, we obtain
P ( E2S =
=
P ( E2S ) = .
On a multiple choice examination with three possible answers (out of which only one is correct) for each of the five questions, what is the probability that a candidate would get four or more correct answers just by guessing?
Let X denote the number of questions answered correctly by guessing in multiple choice examinations.
Probability of getting a correct answer by guessing, p=
Therefore, q, the probability of an incorrect answer by guessing = 1 -
There are in 5 questions in all.
So X follows binomial distribution with n = 5, p =
A card form a pack of 52 cards is lost. From the remaining cards of the pack, two cards are drawn at random and are found to be both clubs. Find the probability of the lost card being of clubs.
Let the events E1, E2, E3, E4 and A be defined as follows:
E1: Missing card is a diamond
E2: Missing card is a spade
E3: Missing card is a ciub
E4: Missing card is a heart
A: Drawing two club cards
P ( E1 ) = P ( E2 ) = P ( E3 ) = P ( E4 ) =
From a lot of 10 bulbs, which includes 3 defectives, a sample of 2 bulbs is drawn at random. Find the probability distribution of the number of defective bulbs.
Total number of bulbs = 10
Number of defective bulbs = 3
Number of non-defective bulbs = 7
P ( drawing a defective bulb ), P =
P ( drawing a non-defective bulb ), q =
Two bulbs are drawn.
Let X denote the number of defective bulbs, then X can take values
0, 1, and 2.
P ( X = 0 ) = P ( drawing both non-defective bulb ) =
P ( x = 1 ) = P ( drawing one defective bulb and one non-defective bulbs )
= P ( drawing a non-defective bulb and a defective bulb ) + P ( drawing a defective bulb and a non-defective bulb )
Required probability distribution is
| X | 0 | 1 | 2 |
| P ( x ) |
Probabilities of solving problem independently by A and B are and respectively. If both try to solve the problem independently, find the probability that
(i) the problem is solved
(ii) exactly one of them solves the problem.
The probability of solving the problem independently by A and B are given as respectively.
[ Since the events corresponding to A and B are independent ]
( i ) Probability that the problem is solved
( ii ) Probability that exactly one of them solves the problem
Suppose 5% of men and 0.25% of women have grey hair. A grey haired person is selected at random. What is the probability of this person being male? Assume that there are equal number of males and females.
Let the events M, F and G be defined as follows:
M: A male is selected
F: A female is selected
G: A person has grey hair
It is given that the number of males = the number of females
A grey haired person is selected at random, the probability that this person is a male
From 6 different novels and 3 different dictionaries, 4 novels and 1 dictionary are to be selected and arranged in a row on a shelf so that the dictionary is always in the middle. Then the number of such arrangements is
less than 500
at least 500 but less than 750
at least 750 but less than 1000
at least 1000
D.
at least 1000
N1N2N3 – N6, D1D2D3
The number of ways = 6c4 × 3c1 × 24
= 15 × 3 × 24 = 1080
In a shop, there are five types of ice-creams available. A child buys six ice-creams.
Statement -1: The number of different ways the child can buy the six ice-creams is 10C5.
Statement -2: The number of different ways the child can buy the six ice-creams is equal to the number of different ways of arranging 6 A’s and 4 B’s in a row.
Statement −1 is false, Statement −2 is true
Statement −1 is true, Statement −2 is true, Statement −2 is a correct explanation for Statement −1
Statement −1 is true, Statement −2 is true; Statement −2 is not a correct explanation for Statement −1.
Statement − 1 is true, Statement − 2 is false.
A.
Statement −1 is false, Statement −2 is true
x1 + x2 + x3 + x4 + x5 = 6
5 + 6 – 1C5 – 1 = 10C4.
A pair of fair dice is thrown independently three times. The probability of getting a score of exactly 9 twice is
1/729
8/9
8/729
8/243
D.
8/243
Probability of getting score 9 in a single throw = 4/36 = 1/9
Probability of getting score 9 exactly twice = 
Two aeroplanes I and II bomb a target in succession. The probabilities of I and II scoring a hit correctly are 0.3 and 0.2, respectively. The second plane will bomb only if the first misses the target. The probability that the target is hit by the second plane is
0.06
0.14
0.2
0.3
B.
0.14
The desired probability
= 0.7 × 0.2 + (0.7) (0.8) (0.7) (0.2) + (0.7) (0.8) (0.7) (0.8) (0.7) (0.2) + ….
= 0.14 [ 1 + (0.56) + (0.56)2 + … ]
Three houses are available in a locality. Three persons apply for the houses. Each applies to one house without consulting others. The probability that all the three apply for the same house is
2/9
1/9
8/9
7/9
B.
1/9
For a particular house being selected
Probability = 1/3
Prob(all the persons apply for the same house) =
A random variable X has the probability distribution:
| X: | 1 | 2 | 3 | 4 | 5 | 6 | 7 | 8 |
| P(X): | 0.15 | 0.23 | 0.12 | 0.10 | 0.20 | 0.08 | 0.07 | 0.05 |
0.87
0.77
0.35
0.50
B.
0.77
E = {x is a prime number} = {2, 3, 5, 7}
P(E) = P(X = 2) + P(X = 3) + P(X = 5) + P(X = 7)
P(E) = 0.23 + 0.12 + 0.20 + 0.07 = 0.62
F = {X < 4}= {1, 2, 3}
P(F) = P(X = 1) + P(X = 2) + P(X = 3)
P(F) = 0.15 + 0.23 + 0.12 = 0.5
E ∩ F = {X is prime number as well as < 4 }
= {2, 3}
P (E ∩ F) = P(X = 2) + P(X = 3)
= 0.23 + 0.12 = 0.35
∴ Required probability
P (E∪ F) = P(E) + P(F) - P(E ∩ F)
P (E∪ F) = 0.62 + 0.5 - 0.35
P (E ∪ F) = 0.77
The mean and the variance of a binomial distribution are 4 and 2 respectively. Then the probability of 2 successes is
37/256
219/ 256
128/256
28/256
D.
28/256
Given that mean = 4
np = 4 and variance = 2
npq = 2 ⇒ 4q = 2
⇒ q = 1/2
∴ p = 1 - q = 1 - 1/2 = 1/2
also n = 8 Probability of 2 successes
A bag contains 4 red and 6 black balls. A ball is drawn at random from the bag, its colour is observed and this ball along with two additional balls of the same colour are returned to the bag. If now a ball is drawn at random from the bag, then the probability that this drawn ball is red, is:
3/4
3/10
2/5
1/5
C.
2/5
E1: Event that first ball drawn is red.
E2: Event that first ball drawn is black.
E: Event that second ball drawn is red.
Sponsor Area
Sponsor Area
.
...(1)
(given)
.
.
.
is
Are E and F independent?
find P(not A and not B).
A and B are independent
find 
find 
then P(B) is 
Then a possible value of n among the following is