EX 3.8 defective components in a box

Six boxes of components are ready to be shipped by a certain supplier. The number of defective components in each box is as follows:

Box	1	2	3	4	5	6
Number of defectives	0	2	0	1	2	0

• Scenario:
  • One of several boxes is randomly selected for shipment to a customer
• Let $X$ = number of defectives in the selected box
  • Possible $X$ values:
    • 0, 1, 2
• Simple events:
  • Six equally likely simple events
  • Three result in $X=0$
  • One results in $X=1$
  • Two result in $X=2$
• Then

$$\begin{array}{rl}p(0)& =P(X=0)=P(\text{box 1, 3, or 6 is sent})=\frac{3}{6}=0.500\\ p(1)& =P(X=1)=P(\text{box 4 is sent})=\frac{1}{6}=0.167\\ p(2)& =P(X=2)=P(\text{box 2 or 5 is sent})=\frac{2}{6}=0.333\end{array}$$

• Probability distribution for $X$:
  • $X=0$: probability = 0.500
  • $X=1$: probability = 0.167
  • $X=2$: probability = 0.333
• These values collectively specify the probability mass function (pmf)
• Long-run behavior if the experiment is repeated:
  • $X=0$ occurs one-half of the time
  • $X=1$ occurs one-sixth of the time
  • $X=2$ occurs one-third of the time