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

Sum

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(at most 2 successes)
= P(X ≤ 2)

= P(X = 0 or X = 1 or X = 2)

= P(X = 0) + P(X = 1) + P(X = 2)

= `(11)/(16)`

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

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

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