# 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

Sum

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

#### 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)

Concept: Random Variables and Its Probability Distributions
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#### APPEARS IN

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