Advertisement
Advertisement
Advertisement
Sum
A die is thrown 4 times. If ‘getting an odd number’ is a success, find the probability of at most 2 successes
Advertisement
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)`
Is there an error in this question or solution?