Question

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, find (a) P{X = 0} and (b) P{X > 2}

Short answer

(a).$P(X=0)\approx 0.522$

(b).$P(X>2)=0.0126$

Step-by-step solution

Step 1:Given information(part a)

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.

Step 2:Explanation(part a)

Observe that $X\in \{0,1,2,3,4,5,6\}$. Take any $k\in \{0,1,2,3,4,5,6\}$. 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

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

Step 3:Final answer(part a)

$P(X=0)$$\approx 0.522$

Step 4:Given information(part b)

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.

Step 5:Explanation(part b)

Observe that $X\in \{0,1,2,3,4,5,6\}$. Take any $k\in \{0,1,2,3,4,5,6\}$. 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,

$P(X>2)=\sum _{k=3}^{6}P(X=k)=0.0126$

Step 6:Final answer(part b)

$P(X>2)=0.0126$