Example 210 suppose that we accept that when drawing

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Example 2.10: Suppose that we accept that when drawing a single card from a deck of cards that the probability that we get an Ace is ( ) 4 / 52 P Ace . Then the probability of not getting an Ace is ( ) 48 / 52 P Not an Ace . Also, based on the definition of a probability function, we know that if sets A and B are disjoint, then ( ) ( ) ( ) P A B P A P B . Example 2.11: Suppose that we accept that when drawing a single card from a deck of cards that the probability that we get an Ace is ( ) 4 / 52 P Ace and the same is true for a King ( ) 4 / 52 P King . Then the probability of getting an Ace or a King is ( ) 4 / 52 4 / 52 8 / 52 P Ace King . Is it always true that ( ) ( ) ( ) P A B P A P B ? That is not part of our definition of a probability function, but maybe it is always true and we should say so in a theorem. Let’s consider some examples. Example 2.12: We will be drawing a card from a deck of cards. Let A be the set of Red cards and let B be the set of Diamonds. We will assign ( ) .5 P A and ( ) .25 P B . Since B is a subset of A, it should be clear that ( ) ( ) P A B P A . We already know that ( ) .5 P A . So if ( ) ( ) ( ) P A B P A P B were always true, we would have ( ) .5 .25 .75 .5 P A B . Learning to do this sort of investigative thinking will serve you well in any mathematics type course. Example 2.13: Let the sets A and B both be (the sample space). ( ) ( ) ( ) 1 1 2 P P P . This contradicts our theorem that states all probabilities are less than or equal to 1. Clearly, we need to work on ( ) P A B when A and B are not disjoint. We now define the set A-B to be those outcomes in A that are not in B and similarly for the set B-A. We can now write the union of two sets as three disjoint sets as follows: ( ) ( ) ( ) A B A B A B B A . This gives
23 ( ) ( ) ( ) ( ) P A B P A B P A B P B A ( ) ( ) ( ) ( ) ( ) P A B P A B P B A P A B P A B (Adding 0 is a beautiful thing) ( ) ( ) ( ) P A P B P A B (Since ( ) ( ) A A B A B and the last 2 sets are disjoint.) Now we have our general rule for the probability of the union of two events. Theorem 2.3: ( ) ( ) ( ) ( ) P A B P A P B P B A (We just did the proof above) Example 2.14: If ( ) .4 P A , ( ) .5 P B and ( ) .1 P A B , determine ( ) P A B . ( ) .4 .5 .1 P A B Exercise 2.5: If ( ) .25 P A , ( ) .37 P B and ( ) .12 P A B , determine ( ) P A B . At this point in time, some of you may be considering a new rule for intersections based on things from the past that are partially remembered: ( ) ( ) ( ) P A B P A P B Is this statement true? Do some investigative thinking! Theorem 2.4: If the events A and B are disjoint, then ( ) 0 P A B Proof: ( ) ( ) ( ) ( ) P A B P A P B P A B by the theorem above. Also by Rule 3, we have ( ) ( ) ( ) P A B P A P B . We see that ( ) P A B is equal to two different quantities and we will set those two quantities equal to each other. ( ) ( ) ( ) ( ) ( ) ( ) 0 P A P B P A B P A P B P A B Theorem 2.5: If A B , then ( ) ( ) P A B P A Proof: Since A B , the set A B B . Using our rule for union: ( ) ( ) ( ) ( ) P A B P A P B P A B .

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