# A die is thrown 6 times. If ‘getting an odd number’ is a success, find the probability of at least 5 successes. - Mathematics and Statistics

Sum

A die is thrown 6 times. If ‘getting an odd number’ is a success, find the probability of at least 5 successes.

#### Solution

Let X = number of successes, i.e. number of odd numbers.

p = probability of getting an odd number in a single throw of a die

∴ p = 3/6 = 1/2 and q = 1 - "p" = 1 - 1/2 = 1/2

Given: n = 6

∴ X ∼ B(6, 1/2)

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

p("X = x") = "^nC_x  p^x  q^(n - x)

i.e. p(x) = "^6C_x (1/2)^x (1/2)^(6 - x)

 = "^6C_x (1/2)^6, x = 0, 1, 2, ...,6

P(at least 5 successes) = P[X ≥ 5]

= p(5) + p(6)

= ""^6C_5 (1/2)^6 + "^6C_6 (1/2)^6

= (""^6C_5 + "^6C_6) (1/2)^6

= (6 + 1) 1/64 = 7/64

Hence, the probability of at least 5 successes is 7/64.

Concept: Binomial Distribution
Is there an error in this question or solution?