EX 2.24

• Complex components are assembled in a plant that uses two different assembly lines, $A$ and $A^{\prime }$.
• Line $A$ uses older equipment than $A^{\prime }$,
  • so it is somewhat slower and less reliable.
• Suppose
  • on a given day line $A$ has assembled 8 components,
    • of which 2 have been identified as defective $\left(B\right)$
    • 6 as nondefective $\left(B^{\prime }\right)$,
  • whereas $A^{\prime }$ has produced
    • 1 defective
    • 9 nondefective components.
• This information is summarized in the accompanying table.

	(B)	(B’)
(A)	2	6
(A’)	1	9

• Unaware of this information, the sales manager randomly selects
  • 1 of these 18 components for a demonstration.
• Prior to the demonstration

$$\begin{array}{rl}& P\left(\text{line A component selected}\right)\\ =& P\left(A\right)\\ =& \frac{N\left(A\right)}{N}\\ =& \frac{8}{18}=.44\end{array}$$

• However, if the chosen component turns out to be defective,
  • then the event $B$ has occurred,
  • so the component must have been 1 of the 3 in the $B$ column of the table.
• Since these 3 components are equally likely among themselves after $B$ has occurred,

$$P\left(A\mid B\right)=\frac{2}{3}=\frac{2/18}{3/18}=\frac{P\left(A\cap B\right)}{P\left(B\right)}\text{\hspace{1em}(2.2)}$$