91Ó°ÊÓ

Given in the question that we need to find the conditional probability that it contained $4$defective components

Let$i$be the defective lot.

The number of components in a lot is $10.$

It is known that the $3$components are randomly selected for inspection, if all $3$components are non-defective then lot will be accepted.

Find the probability that lot $1$is non-defective or accepted.

Here,$i=1$lot has $1$defective.

So,

$P($Accepted $)=\frac{\left(\begin{array}{l}1\\ 0\end{array}\right)\left(\begin{array}{l}9\\ 3\end{array}\right)}{\left(\begin{array}{l}10\\ 3\end{array}\right)}$

$=\frac{7}{10}$

Find the probability that lot $1$is defective.

Therefore,

$P($rejected $)=1-P($Accepted$)$

$=1-\frac{7}{10}$

$=\frac{3}{10}$

When $i=4$implies that lot has 4 defective components.

Now,

$\mathrm{P}($accepted $)=$$\frac{\left(\begin{array}{l}4\\ 0\end{array}\right)\left(\begin{array}{l}6\\ 3\end{array}\right)}{\left(\begin{array}{l}10\\ 3\end{array}\right)}$$=\frac{1}{6}$

Find the probability that lot $4$is defective.

Therefore,

$P($rejected $)=1-\frac{1}{6}$

$=\frac{5}{6}$

Find the conditional probability that it contained$4$defective components given that lot 1 and 4 are rejected.

Therefore, the required probability is,

$P(4$defective $âˆ\pounds$lot rejected$)$$=\frac{\left(\frac{5}{6}×\frac{3}{10}\right)}{\left(\frac{5}{6}×\frac{3}{10}\right)+\left(\frac{3}{10}×\frac{7}{10}\right)}$

$=\frac{75}{138}$

Required probability