Mathematical Methods for Physics and Engineering - Matematica.NET

PROBABILITYThis is particularly useful for problems in which evaluating the probabilityof the complement is easier than evaluating the probability of the eventitself.◮Calculate the probability of drawing an ace or a spade from a pack of cards.Let A be the event that an ace is drawn and B the event that a spade is drawn. Itimmediately follows that Pr(A) = 4 = 113and Pr(B) = = 1 . The intersection of A and52 13 52 4B consists of only the ace of spades and so Pr(A ∩ B) = 1 . Thus, from (30.9)52Pr(A ∪ B) = 1 + 1 − 1 = 4 .13 4 52 13In this case it is just as simple to recognise that there are 16 cards in the pack that satisfythe required condition (13 spades plus three other aces) and so the probability is 16 . ◭ 52The above theorems can easily be extended to a greater number of events. Forexample, if A 1 ,A 2 ,...,A n are mutually exclusive events then (30.10) becomesPr(A 1 ∪ A 2 ∪ ···∪ A n )=Pr(A 1 )+Pr(A 2 )+···+Pr(A n ). (30.12)Furthermore, if A 1 ,A 2 ,...,A n (whether mutually exclusive or not) exhaust S, i.e.are such that A 1 ∪ A 2 ∪ ···∪ A n = S, thenPr(A 1 ∪ A 2 ∪ ···∪ A n )=Pr(S) =1. (30.13)◮A biased six-sided die has probabilities 1 p, p, p, p, p, 2p of showing 1, 2, 3, 4, 5, 62respectively. Calculate p.Given that the individual events are mutually exclusive, (30.12) can be applied to givePr(1 ∪ 2 ∪ 3 ∪ 4 ∪ 5 ∪ 6) = 1 13p + p + p + p + p +2p = p.2 2The union of all possible outcomes on the LHS of this equation is clearly the samplespace, S, andsoPr(S) = 13 p. 2Now using (30.7),132p =Pr(S) =1 ⇒ p = . ◭2 13When the possible outcomes of a trial correspond to more than two events,and those events are not mutually exclusive, the calculation of the probability ofthe union of a number of events is more complicated, and the generalisation ofthe addition law (30.9) requires further work. Let us begin by considering theunion of three events A 1 , A 2 and A 3 , which need not be mutually exclusive. Wefirst define the event B = A 2 ∪ A 3 and, using the addition law (30.9), we obtainPr(A 1 ∪ A 2 ∪ A 3 )=Pr(A 1 ∪ B) =Pr(A 1 )+Pr(B) − Pr(A 1 ∩ B).(30.14)1126