Conditional probability  (Page 3/6)

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The formulation of conditional probability assumes the conditioning event C is well defined. Sometimes there are subtle difficulties. It may not be entirelyclear from the problem description what the conditioning event is. This is usually due to some ambiguity or misunderstanding of the information provided.

What is the conditioning event?

Five equally qualified candidates for a job, Jim, Paul, Richard, Barry, and Evan, are identified on the basis of interviews and told that they are finalists. Three ofthese are to be selected at random, with results to be posted the next day. One of them, Jim, has a friend in the personnel office. Jim asks the friend to tell him the name ofone of those selected (other than himself). The friend tells Jim that Richard has been selected. Jim analyzes the problem as follows.

ANALYSIS

Let ${A}_{i},1\le i\le 5$ be the event the i th of these is hired ( A 1 is the event Jim is hired, A 3 is the event Richard is hired, etc.). Now $P\left({A}_{i}\right)$ (for each i ) is the probability that finalist i is in one of the combinations of three from five. Thus, Jim's probability of being hired, before receiving the information about Richard, is

$P\left({A}_{1}\right)=\frac{1×C\left(4,2\right)}{C\left(5,3\right)}=\frac{6}{10}=P\left({A}_{i}\right),\phantom{\rule{0.277778em}{0ex}}\phantom{\rule{0.277778em}{0ex}}\phantom{\rule{0.277778em}{0ex}}1\le i\le 5$

The information that Richard is one of those hired is information that the event A 3 has occurred. Also, for any pair $i\ne j$ the number of combinations of three from five including these two is just the number of ways of picking one from the remaining three.Hence,

$P\left({A}_{1}{A}_{3}\right)=\frac{C\left(3,1\right)}{C\left(5,3\right)}=\frac{3}{10}=P\left({A}_{i}{A}_{j}\right),\phantom{\rule{0.277778em}{0ex}}\phantom{\rule{0.277778em}{0ex}}\phantom{\rule{0.277778em}{0ex}}i\ne j$

The conditional probability

$P\left({A}_{1}|{A}_{3}\right)=\frac{P\left({A}_{1}{A}_{3}\right)}{P\left({A}_{3}\right)}=\frac{3/10}{6/10}=1/2$

This is consistent with the fact that if Jim knows that Richard is hired, then there are two to be selected from the four remaining finalists, so that

$P\left({A}_{1}|{A}_{3}\right)=\frac{1×C\left(3,1\right)}{C\left(4,2\right)}=\frac{3}{6}=1/2$

Discussion

Although this solution seems straightforward, it has been challenged as being incomplete. Many feel that there must be information about how the friend chose to name Richard. Many would make an assumption somewhat as follows. Thefriend took the three names selected: if Jim was one of them, Jim's name was removed and an equally likely choice among the other two was made; otherwise, the friendselected on an equally likely basis one of the three to be hired. Under this assumption, the information assumed is an event B 3 which is not the same as A 3 . In fact, computation (see Example 5, below) shows

$P\left({A}_{1}|{B}_{3}\right)\phantom{\rule{4pt}{0ex}}=\frac{6}{10}=P\left({A}_{1}\right)\ne P\left({A}_{1}|{A}_{3}\right)$

Both results are mathematically correct. The difference is in the conditioning event, which corresponds to the difference in the information given (or assumed).

$\square$

Some properties

In addition to its properties as a probability measure, conditional probability has special properties which are consequences of the way it is related to the original probability measure $P\left(\cdot \right)$ . The following are easily derived from the definition of conditional probability and basic properties of the prior probabilitymeasure, and prove useful in a variety of problem situations.

(CP1) Product rule If $P\left(ABCD\right)>0$ , then $P\left(ABCD\right)=P\left(A\right)P\left(B|A\right)P\left(C|AB\right)P\left(D|ABC\right)$ .

Derivation

The defining expression may be written in product form: $P\left(AB\right)=P\left(A\right)P\left(B|A\right)$ . Likewise

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