The probability that a certain kind of component will survive a check test is 0.5. Find the probability that exactly two of the next four components tested will survive.

Explanation

Let X = number of tested components survive.
p = probability that the component survive the check test.
`therefore" "p=0.5 =(5)/(10)=(1)/(2)`
`therefore" "q=1-p=1-(1)/(2)=(1)/(2)`
Given n = 4
`therefore X ~ B(4, 1//4)`
The probability mass function as :
`P(X=x)=.^(n)C_(x)p^(x).q^(n-x)`
i.e., `p(x)=.^(4)C_(x)((1)/(2))^(2x)((1)/(2))^(4-x)`
P(exactly 2 components survive)
`=P(X=2)=p(2)`
`=.^(4)C_(2)((1)/(2))^(2)((1)/(2))^(4-2)`
`=(4!)/(2!.2!).((1)/(2))^(2).((1)/(2))^(2)`
`=(4.3.2!)/(2.1.2!)xx((1)/(2))^(4)`
`=(6)/(16)=0.375`
Hence, the probabillity that exactly 2 of the 4 tested components survive is 0.375.