91Ó°ÊÓ

Given in the question that suppose that a batch of $100$items contains $6$that are defective and $94$that are not defective. If $X$is the number of defective items in a randomly drawn sample of $10$ items from the batch.

Observe that $X∈\{0,1,2,3,4,5,6\}$. Take any $k∈\{0,1,2,3,4,5,6\}$. The event $X=k$means that in our sample, we have taken $k$defective items and $10-k$non-defective items. Hence

$P(X=k)=\frac{\left(\begin{array}{l}6\\ k\end{array}\right)·\left(\begin{array}{c}94\\ 10-k\end{array}\right)}{\left(\begin{array}{c}100\\ 10\end{array}\right)}$

Now, we have that

$P(X=0)=\frac{\left(\begin{array}{c}94\\ 10\end{array}\right)}{\left(\begin{array}{c}100\\ 10\end{array}\right)}≈0.522$

$P(X=0)$$≈0.522$

Given in the question that,suppose that a batch of $100$items contains $6$that are defective and $94$that are not defective. If $X$is the number of defective items in a randomly drawn sample of $10$ items from the batch.

Observe that $X∈\{0,1,2,3,4,5,6\}$. Take any $k∈\{0,1,2,3,4,5,6\}$. The event $X=k$means that in our sample, we have taken $k$defective items and $10-k$non-defective items. Hence

$P(X=k)=\frac{\left(\begin{array}{c}6\\ k\end{array}\right)·\left(\begin{array}{c}94\\ 10-k\end{array}\right)}{\left(\begin{array}{c}100\\ 10\end{array}\right)}$

Now we have that,

$P(X>2)=\underset{k=3}{\overset{6}{∑}}P(X=k)=0.0126$

$P(X>2)=0.0126$