Inclusive ProbabilitiesDate: 11/06/98 at 13:41:57 From: Dana Africa Subject: Inclusive/exclusive probabilities Dear Dr. Math, I am helping my tenth grade daughter with probability and am lost with the formula for inclusive probability. Her worksheet reads: P(A or B) = P(A) + P(B) - P(A and B) I don't know how to use this or explain it to her. Thanks for your time and an answer. Dana Africa Date: 11/06/98 at 16:38:05 From: Doctor Anthony Subject: Re: Inclusive/exclusive probabilities An example will show how this works. Find the probability of drawing a red card or an ace when one card is taken from a pack of 52. Let P(A) = probability of a red card = 26/52 = 1/2 Let P(B) = probability of an ace = 4/52 = 1/13 Obviously P(A and B) = probability of the ace of hearts or diamonds = 2/52 = 1/26 By the above equation: P(A or B) = P(A) + P(B) - P(A and B) = 26/52 + 4/52 - 2/52 = 28/52 = 7/13 Thinking about it, there 28 cards satisfying the requirement. The 26 red cards plus the aces of spades and clubs. We had to subtract 2/52 because the ace of hearts and the ace of diamonds had been counted in twice, once as red cards and a second time as aces. Another way to visualize the relation is to think of two sets A and B in a Venn diagram. (A or B) is the total area given by the union of the sets. (A and B) is the area given by the intersection of the sets. Then (A or B) = A + B - (A and B) We subtract the intersection (A and B) because it was counted twice: once as part of set A and again as part of set B. It should of course only be counted once. - Doctor Anthony, The Math Forum http://mathforum.org/dr.math/ |
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