Question

A sample of 3 items is selected at random from a box containing 20 items of which 4 are defective. Find the expected number of defective items in the sample.

Short answer

The expected number of defective items in the sample is$E\left(X\right)=0.6$

Step-by-step solution

Given Information

$X$is the random variable which denotes the number of items in the sample.

we selected 3 items for sample ,then we can assume the value of $X$is $0,1,2,3$

probability of $X$is k$k$

Explanation

Here,

$P(X=k)=\frac{\left(\begin{array}{l}4\\ k\end{array}\right)·\left(\begin{array}{c}16\\ 3-k\end{array}\right)}{\left(\begin{array}{c}20\\ 3\end{array}\right)}$

Using the formula for expectation,we have that

$E\left(X\right)=\sum _{k=0}^{3}k·P(X=k)=\sum _{k=0}^{3}k·\frac{\left(\begin{array}{l}4\\ k\end{array}\right)·\left(\begin{array}{c}16\\ 3-k\end{array}\right)}{\left(\begin{array}{c}20\\ 3\end{array}\right)}$

$=\frac{1}{\left(\begin{array}{c}20\\ 3\end{array}\right)}\left[1·\left(\begin{array}{l}4\\ 1\end{array}\right)·\left(\begin{array}{c}16\\ 2\end{array}\right)+2·\left(\begin{array}{l}4\\ 2\end{array}\right)·\left(\begin{array}{c}16\\ 1\end{array}\right)+3·\left(\begin{array}{l}4\\ 3\end{array}\right)·\left(\begin{array}{c}16\\ 1\end{array}\right)\right]$

$=0.6$

Step 3:Final Answer

The expected number of defective items in the sample is$E\left(X\right)=$$0.6$