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Probability

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Question
CBSEENMA12033575
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If a leap year is selected at random, what is the chance that it will contain 53 Tuesdays?

Solution

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.
 therefore space space space required space probability space equals space 2 over 7.

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Question
CBSEENMA12033576
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If each element of a second order determinant is either zero or one, what is the probability that the value of determinant is positive? (Assume that the individual enteries of the determinant are chosen independently, each value being assumed with probability 1 half right parenthesis.

Solution

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
open vertical bar table row 1 0 row 0 1 end table close vertical bar comma space open vertical bar table row 1 0 row 1 1 end table close vertical bar comma space open vertical bar table row 1 1 row 0 1 end table close vertical bar
therefore space space number of favourable cases  = 3
therefore the probability that the determinant is positive = 3 over 16.

Question
CBSEENMA12033577
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The probability of obtaining an even prime number on each die, when a pair of dice is rolled is

  • 0

  • 1 third
  • 1 over 12
  • 1 over 36

Solution

D.

1 over 36

S =open curly brackets table row cell left parenthesis 1 comma space 1 right parenthesis end cell cell left parenthesis 1 comma space 2 right parenthesis end cell cell left parenthesis 1 comma space 3 right parenthesis end cell cell left parenthesis 1 comma space 4 right parenthesis end cell cell left parenthesis 1 comma space 5 right parenthesis end cell cell left parenthesis 1 comma space 6 right parenthesis end cell row cell left parenthesis 2 comma space 1 right parenthesis end cell cell left parenthesis 2 comma space 2 right parenthesis end cell cell left parenthesis 2 comma space 3 right parenthesis end cell cell left parenthesis 2 comma space 4 right parenthesis end cell cell left parenthesis 2 comma space 5 right parenthesis end cell cell left parenthesis 2 comma space 6 right parenthesis end cell row cell left parenthesis 3 comma space 1 right parenthesis end cell cell left parenthesis 3 comma space 2 right parenthesis end cell cell left parenthesis 3 comma space 3 right parenthesis end cell cell left parenthesis 3 comma space 4 right parenthesis end cell cell left parenthesis 3 comma space 5 right parenthesis end cell cell left parenthesis 3 comma space 6 right parenthesis end cell row cell left parenthesis 4 comma space 1 right parenthesis end cell cell left parenthesis 4 comma space 2 right parenthesis end cell cell left parenthesis 4 comma space 3 right parenthesis end cell cell left parenthesis 4 comma space 4 right parenthesis end cell cell left parenthesis 4 comma space 5 right parenthesis end cell cell left parenthesis 4 comma space 6 right parenthesis end cell row cell left parenthesis 5 comma space 1 right parenthesis end cell cell left parenthesis 5 comma space 2 right parenthesis end cell cell left parenthesis 5 comma space 3 right parenthesis end cell cell left parenthesis 5 comma space 4 right parenthesis end cell cell left parenthesis 5 comma space 5 right parenthesis end cell cell left parenthesis 5 comma space 6 right parenthesis end cell row cell left parenthesis 6 comma space 1 right parenthesis end cell cell left parenthesis 6 comma space 2 right parenthesis end cell cell left parenthesis 6 comma space 3 right parenthesis end cell cell left parenthesis 6 comma space 4 right parenthesis end cell cell left parenthesis 6 comma space 5 right parenthesis end cell cell left parenthesis 6 comma space 6 right parenthesis end cell end table close curly brackets
Now favourable case is(2, 2)
therefore space space space required space probability space equals space 1 over 36

Question
CBSEENMA12033578
Wired Faculty App

If straight P left parenthesis straight A right parenthesis space equals space 7 over 13 comma space straight P left parenthesis straight B right parenthesis space equals space 9 over 13 comma space space straight P left parenthesis straight A intersection straight B right parenthesis space equals 4 over 13 comma space evaluate space straight P left parenthesis straight A space left enclose straight B right parenthesis.

Solution

Here, straight P left parenthesis straight A right parenthesis space equals 7 over 13 comma space space straight P left parenthesis straight B right parenthesis space equals space 9 over 13 comma space space straight P left parenthesis straight A intersection straight B right parenthesis space equals space 4 over 13                        ...(1)
Now,     straight P left parenthesis straight A space left enclose straight B right parenthesis space equals space fraction numerator straight P left parenthesis straight A intersection straight B right parenthesis over denominator straight P left parenthesis straight B right parenthesis end fraction space equals space fraction numerator begin display style 4 over 13 end style over denominator begin display style 9 over 13 end style end fraction space space space space space space space space space space space space space space open square brackets because space space of space space left parenthesis 1 right parenthesis close square brackets
                    equals space 4 over 13 cross times 13 over 9 space equals 4 over 9.

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