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A die is thrown 4 times. If ‘getting an odd number’ is a success, find the probability of 2 successes - Mathematics and Statistics

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Question

A die is thrown 4 times. If ‘getting an odd number’ is a success, find the probability of 2 successes

Sum
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Solution

Let X denote the number of odd numbers.

P(getting and odd number) = p = `(3)/(6) = (1)/(2)`

∴ q = 1 – p = `1 - (1)/(2) = (1)/(2)`

Given, n = 4

∴ X ∼ B`(4, 1/2)`

The p.m.f. of X is given by

P(X = x) = `""^4"C"_x(1/2)^x(1/2)^(4 - x)`

= `""^4"C"_x(1/2)^4 , x` = 0, 1,...,4

P(2 successes) = P(X = 2)

= `""^4"C"_2(1/2)^4`

= `(4!)/(2! xx 2!)(1/2)^4`

= `(4 xx 3 xx 2!)/(2 xx 1 xx 2!)(1/16)`

= `(3)/(8)`

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Chapter 8: Probability Distributions - Exercise 8.3 [Page 150]

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Balbharati Mathematics and Statistics 2 (Commerce) [English] Standard 12 Maharashtra State Board
Chapter 8 Probability Distributions
Exercise 8.3 | Q 1.01 | Page 150
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