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Sum

The probability that a certain kind of component will survive a check test is 0.6. Find the probability that exactly two of the next four components tested will survive.

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#### Solution

Let X = number of tested components survive.

p = probability that the component survives the check test

∴ p = 0.6 = `6/10 = 3/5`

∴ q = 1 - p = `1 - 3/5 = 2/5`

Given: n = 4

∴ X ~ B`(4, 3/5)`

The p.m.f. of X is given as:

P[X = x] = `"^nC_x p^x q^(n - x)`

i.e. p(x) = `"^4C_x (3/5)^x (2/5)^(4 - x)`

P (exactly 2 components survive)

= P[X = x] = p(2)

`= "^4C_2 (3/5)^2 (2/5)^(4 - 2)`

`= ((4 xx 3)/(1 xx 2)) xx (3/5)^2 (2/5)^2 = (6 xx 9 xx 4)/625`

`= 216/625`

= 0.3456

Hence, the probability that exactly 2 of the 4 tested components survive is 0.3456.

Concept: Binomial Distribution

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