Question
The probability that a certain kind of component will survive a check test is $0.6.$ Find the probability that exactly $2$ of the next $4$ tested components tested survive.
✓
Answer
Let $\mathrm{X}=$ number of tested components survive.
$p=$ probability that the component survives the check test
$ \therefore p=0.6=\frac{6}{10}=\frac{3}{5}$
$\therefore q=1-p=1-\frac{3}{5}=\frac{2}{5} $
Given : $n=4$
$\therefore X \sim B\left(4, \frac{3}{5}\right)$
The p.m.f. of $X$ is given as :
$P[X=x]={ }^n \mathrm{C}_x p^x q^{n-x}$
i.e. $p(x)={ }^4 C_x\left(\frac{3}{5}\right)^x\left(\frac{2}{5}\right)^{4-x}$
$P \text { (exactly } 2 \text { components survive) }$
$ =P[X=2]=p(2)$
$={ }^4 C_2\left(\frac{3}{5}\right)^2\left(\frac{2}{5}\right)^{4-2}$
$=\left(\frac{4 \times 3}{1 \times 2}\right) \times\left(\frac{3}{5}\right)^2\left(\frac{2}{5}\right)^2=\frac{6 \times 9 \times 4}{625}$
$=\frac{216}{625}=0.3456 $
Hence, the probability that exactly $2$ of the $4$ tested components survive is $0.3456 .$
$p=$ probability that the component survives the check test
$ \therefore p=0.6=\frac{6}{10}=\frac{3}{5}$
$\therefore q=1-p=1-\frac{3}{5}=\frac{2}{5} $
Given : $n=4$
$\therefore X \sim B\left(4, \frac{3}{5}\right)$
The p.m.f. of $X$ is given as :
$P[X=x]={ }^n \mathrm{C}_x p^x q^{n-x}$
i.e. $p(x)={ }^4 C_x\left(\frac{3}{5}\right)^x\left(\frac{2}{5}\right)^{4-x}$
$P \text { (exactly } 2 \text { components survive) }$
$ =P[X=2]=p(2)$
$={ }^4 C_2\left(\frac{3}{5}\right)^2\left(\frac{2}{5}\right)^{4-2}$
$=\left(\frac{4 \times 3}{1 \times 2}\right) \times\left(\frac{3}{5}\right)^2\left(\frac{2}{5}\right)^2=\frac{6 \times 9 \times 4}{625}$
$=\frac{216}{625}=0.3456 $
Hence, the probability that exactly $2$ of the $4$ tested components survive is $0.3456 .$
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