Conditional probability  (Page 4/6)

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$P\left(ABC\right)=P\left(A\right)\frac{P\left(AB\right)}{P\left(A\right)}\cdot \frac{P\left(ABC\right)}{P\left(AB\right)}=P\left(A\right)P\left(B|A\right)P\left(C|AB\right)$

and

$P\left(ABCD\right)=P\left(A\right)\frac{P\left(AB\right)}{P\left(A\right)}\cdot \frac{P\left(ABC\right)}{P\left(AB\right)}\cdot \frac{P\left(ABCD\right)}{P\left(ABC\right)}=P\left(A\right)P\left(B|A\right)P\left(C|AB\right)P\left(D|ABC\right)$

This pattern may be extended to the intersection of any finite number of events. Also, the events may be taken in any order.

$\square$

Selection of items from a lot

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

$P\left({E}_{1}{E}_{2}{E}_{3}{E}_{4}\right)=P\left({E}_{1}\right)P\left({E}_{2}|{E}_{1}\right)P\left({E}_{3}|{E}_{1}{E}_{2}\right)P\left({E}_{4}|{E}_{1}{E}_{2}{E}_{3}\right)=\frac{9}{10}\cdot \frac{8}{9}\cdot \frac{7}{8}\cdot \frac{6}{7}=\frac{6}{10}$

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

$P\left({E}_{1}{E}_{2}{E}_{3}{E}_{4}\right)=\frac{C\left(9,4\right)}{C\left(10,4\right)}=\frac{126}{210}=3/5$

A selection problem

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

$P\left({G}_{1}{G}_{3}\right)=P\left({G}_{1}\right)P\left({G}_{2}|{G}_{1}\right)P\left({G}_{3}|{G}_{1}{G}_{2}\right)+P\left({G}_{1}\right)P\left({G}_{2}^{c}|{G}_{1}\right)P\left({G}_{3}|{G}_{1}{G}_{2}^{c}\right)=\frac{8}{10}\cdot \frac{7}{9}\cdot \frac{6}{8}+\frac{8}{10}\cdot \frac{2}{9}\cdot \frac{7}{8}=\frac{28}{45}\approx 0.62$

(CP2) Law of total probability Suppose the class $\left\{{A}_{i}:\phantom{\rule{0.277778em}{0ex}}1\le i\le n\right\}$ 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

$P\left(E\right)=P\left(E|{A}_{1}\right)P\left({A}_{1}\right)+P\left(E|{A}_{2}\right)P\left({A}_{2}\right)+\cdots +P\left(E|{A}_{n}\right)P\left({A}_{n}\right)$

A compound experiment.

Five cards are numbered one through five. A two-step selection procedure is carried out as follows.

1. Three cards are selected without replacement, on an equally likely basis.
• If card 1 is drawn, the other two are put in a box
• If card 1 is not drawn, all three are put in a box
2. One of cards in the box is drawn on an equally likely basis (from either two or three)

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 $P\left({A}_{1}|{B}_{5}\right)$ .

SOLUTION

From [link] , we have $P\left({A}_{i}\right)=6/10$ and $P\left({A}_{i}{A}_{j}\right)=3/10$ . This implies

$P\left({A}_{i}{A}_{j}^{c}\right)=P\left({A}_{i}\right)-P\left({A}_{i}{A}_{j}\right)=3/10$

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) ,

$P\left({B}_{5}\right)=P\left({B}_{5}|{A}_{1}{A}_{5}\right)P\left({A}_{1}{A}_{5}\right)+P\left({B}_{5}|{A}_{1}^{c}{A}_{5}\right)P\left({A}_{1}^{c}{A}_{5}\right)=\frac{1}{2}\cdot \frac{3}{10}+\frac{1}{3}\cdot \frac{3}{10}=\frac{1}{4}$

Also, since ${A}_{1}{B}_{5}={A}_{1}{A}_{5}{B}_{5}$ ,

$P\left({A}_{1}{B}_{5}\right)=P\left({A}_{1}{A}_{5}{B}_{5}\right)=P\left({A}_{1}{A}_{5}\right)P\left({B}_{5}|{A}_{1}{A}_{5}\right)=\frac{3}{10}\cdot \frac{1}{2}=\frac{3}{20}$

We thus have

$P\left({A}_{1}|{B}_{5}\right)=\frac{3/20}{5/20}=\frac{6}{10}=P\left({A}_{1}\right)$

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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