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Question
A die is thrown 4 times. If ‘getting an odd number’ is a success, find the probability of at least 3 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 least 3 successes)
= P(X ≥ 3)
= 1 – P(X < 3)
= 1 – P(X = 0 or X = 1 or X = 2)
= 1 – [P(X = 0) + P(X = 1) + P(X = 2)]
= `1 - [""^4"C"_0(1/2)^4 + ""^4"C"_1(1/2)(1/2)^3 + ""^4"C"_2 (1/2)^2(1/2)^2]`
= `1 - [1/16 + 4/16 + 6/16]`
= `1 - (11)/(16)`
= `(5)/(16)`
= 0.3125
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