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and
This pattern may be extended to the intersection of any finite number of events. Also, the events may be taken in any order.
— $\square $
An electronics store has ten items of a given type in stock. One is defective. Four successive customers purchase one of the items. Each time, the selection is on anequally likely basis from those remaining. What is the probability that all four customes get good items?
SOLUTION
Let E _{i} be the event the i th customer receives a good item. Then the first chooses one of the nine out of ten good ones, the second chooses one of the eight out of ninegoood ones, etc., so that
Note that this result could be determined by a combinatorial argument: under the assumptions, each combination of four of ten is equally likely; the number of combinationsof four good ones is the number of combinations of four of the nine. Hence
Three items are to be selected (on an equally likely basis at each step) from ten, two of which are defective. Determine the probability that the first and third selected aregood.
SOLUTION
Let ${G}_{i},1\le i\le 3$ be the event the i th unit selected is good. Then ${G}_{1}{G}_{3}={G}_{1}{G}_{2}{G}_{3}\bigvee {G}_{1}{G}_{2}^{c}{G}_{3}$ . By the product rule
(CP2) Law of total probability Suppose the class $\{{A}_{i}:\phantom{\rule{0.277778em}{0ex}}1\le i\le n\}$ of events is mutually exclusive and every outcome in E is in one of these events. Thus, $E={A}_{1}E\bigvee {A}_{2}E\bigvee \cdots \bigvee {A}_{n}E$ , a disjoint union. Then
Let A _{i} be the event the i th card is drawn on the first selection and let B _{i} be the event the card numbered i is drawn on the second selection (from the box). Determine $P\left({B}_{5}\right)$ , $P\left({A}_{1}{B}_{5}\right)$ , and $P\left({A}_{1}\right|{B}_{5})$ .
SOLUTION
From [link] , we have $P\left({A}_{i}\right)=6/10$ and $P\left({A}_{i}{A}_{j}\right)=3/10$ . This implies
Now we can draw card five on the second selection only if it is selected on the first drawing, so that ${B}_{5}\subset {A}_{5}$ . Also ${A}_{5}={A}_{1}{A}_{5}\bigvee {A}_{1}^{c}{A}_{5}$ . We therefore have ${B}_{5}={B}_{5}{A}_{5}={B}_{5}{A}_{1}{A}_{5}\bigvee {B}_{5}{A}_{1}^{c}{A}_{5}$ . By the law of total probability (CP2) ,
Also, since ${A}_{1}{B}_{5}={A}_{1}{A}_{5}{B}_{5}$ ,
We thus have
Occurrence of event B _{1} has no affect on the likelihood of the occurrence of A _{1} . This condition is examined more thoroughly in the chapter on "Independence of Events" .
Often in applications data lead to conditioning with respect to an event but the problem calls for “conditioning in the opposite direction.”
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