Probability

More Topic from Mathematics

Question 1

A box contains 15 green and 10 yellow balls. If 10 balls are randomly drawn, one–by–one, with replacement, then the variance of the number of green balls drawn is

  • 6/25

  • 12/5

  • 6

  • 4

Solution

B.

12/5

We can apply binomial probability distribution Variance = npq
5equals space 10 space straight x space 3 over 5 space straight x space 2 over 5 space equals 12 over 5

Question 2

A man X has 7 friends, 4 of them are ladies and 3 are men. His wife Y also has 7 friends,3 of them are ladies and 4 are men. Assume X and Y have no common friends. Then the total number of ways in which X and Y together can throw a party inviting 3 ladies and 3 men, so that 3 friends of each of X and Y are in this party, is

  • 484

  • 485

  • 468

  • 469

Solution

B.

485


Total number of ways
4C0 · 3C3 · 3C3 · 4C0 + 4C1 · 3C2 · 3C2 · 4C1 + 4C2 · 3C1 · 3C1 · 4C2 + 4C3 · 3C0 · 3C0 · 4C3= 485

Question 3

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

Solution

D.

8/243

Probability of getting score 9 in a single throw = 4/36 = 1/9
Probability of getting score 9 exactly twice = straight C presuperscript 3 subscript 2 space straight x space open parentheses 1 over 9 close parentheses squared space straight x 8 over 9 space equals space 8 over 243

Question 4

A random variable X has Poisson distribution with mean 2. Then P(X > 1.5) equals

  • 2/e2

  • 0

  • 1 minus 3 over straight e squared
  • 3/e2

Solution

C.

1 minus 3 over straight e squared straight P space left parenthesis straight x space equals straight k right parenthesis space equals space straight e to the power of negative straight lambda end exponent space fraction numerator straight lambda to the power of straight k over denominator straight k factorial end fraction
straight P left parenthesis straight x space greater or equal than space 2 right parenthesis space space equals 1 space minus space straight P space left parenthesis straight x space equals 0 right parenthesis space minus space straight P space left parenthesis straight x space equals 1 right parenthesis
space equals space 1 minus straight e to the power of negative straight lambda end exponent space minus straight e to the power of negative straight lambda end exponent space open parentheses fraction numerator straight lambda over denominator 1 factorial end fraction close parentheses
space equals space 1 minus 3 over straight e squared
Question 5

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

For the events E = {X is a prime number} and F = {X < 4}, the probability P (E ∪ F) is
  • 0.87

  • 0.77

  • 0.35

  • 0.50

Solution

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