Question

6. A shipment of five computers contain two that are slightly defective. If a retailer receives three of these computers. How many random variables X representing the number of slightly defective computers? a. 1 b. 2 c. 3 d. 4

Answer 1

Let $X=$ the number of slightly defective computers

$3N, 0D$

$X=0$

This combination does not exist, because $5-3=2<3.$ A retailer can receive at most 2 normal computers.

$2N, 1D$

$X=1$

$$P(2N \ \&\ 1D)={\binom{5-3}{2}\binom{3}{1} \over \binom{5}{3}}={1\cdot3 \over {5! \over 3!(5-3)!}}={3 \over 10}$$

$1N, 2D$

$X=2$

$$P(1N \ \&\ 2D)={\binom{5-3}{1}\binom{3}{2} \over \binom{5}{3}}={2\cdot3 \over 10}={3 \over 5}$$

$0N, 3D$

$X=3$

$$P(0N \ \&\ 3D)={\binom{5-3}{0}\binom{3}{3} \over \binom{5}{3}}={1\cdot1 \over 10}={1 \over 10}$$

Sample space:

$$S=\{{NND, NDN,NDD,DNN, DND, DDN,DDD}\}$$ $$S_X=\{{1,2,3}\}$$