Answer to Question #104030 in Statistics and Probability for Justin Cabantac

Question

A shipment of five computers contains two that are slightly defective. If a retailer receives three of these computers, let Z be the number of slightly defective computers

Answer 1

Let $Z$ be the number of slightly defective computers. List the elements of the sample using the letters D and N for defective and non-defective computers, respectively.​

Since there are 2 slightly defective computers and a retailer receives three computers, $Z$ can take values $0,1$ and $2.$

If $Z=0$ then there are no defective units. All received computers are among 3 non-defective computers : $\{NNN\}.$ There is only 1 scenario for the first case.

$$P(Z=0)={\binom{2}{0}\binom{3}{3} \over \binom{5}{3}}={1\cdot1\over 10}={1\over 10}$$

If $Z=1$ then there is 1 slightly defective computer. That means that 2 of 3 received computers are among 3 non-defective computers : $\{DNN, NDN, NND\}.$

$$P(Z=1)={\binom{2}{1}\binom{3}{2} \over \binom{5}{3}}={2\cdot3\over 10}={3\over 5}$$

If $Z=2$ then there is 2 slightly defective computer. That means that 1 of 3 received computers are among 3 non-defective computers : $\{DDN, DND, NDD\}.$

$$P(Z=2)={\binom{2}{2}\binom{3}{1} \over \binom{5}{3}}={1\cdot3\over 10}={3\over 10}$$

Sample space:

$$S=\{NNN,DNN, NDN, NNDDDN, DND, NDD\}$$

$$\begin{matrix} z & 0 & 1 & 2 \\ f(z) & 0.1 & 0.6 & 0.3 \end{matrix}$$