Sec. 3.1 THE DETERMINATION OF PROBABILITIES 55 



on one future random trial can be obtained from the following 

 formula : 



(3.16) P(E lf E 2 , ■ ■ ■, and E r ) = P(E 1 )-P(E 2 ) • • • P(E r ). 



This result is known as the Law of Compound Probability for Inde- 

 pendent Events. 



A similar and more general law than (3.16) can be established for 

 situations involving dependent, rather than independent, events. 

 Suppose that the occurrence of event E 2 depends on another event 

 E-l having occurred previously, or that it is useful to regard E 2 's oc- 

 currence as depending on the prior happening of E X , perhaps for the 

 sake of convenient computation. For example, suppose that a bridge 

 deck is to be well shuffled, and then two cards drawn successively 

 and at random without replacing the first card drawn. What is the 

 probability that the second card to be drawn will be an ace? It 

 should be apparent that the answer depends somehow on the out- 

 come of the first draw from the deck so that the second event is 

 dependent upon the first event. 



To attack the problem rather generally, suppose that n single 

 events are possible under a given set of circumstances and that an 

 event E x is associated with n x of these n single events. Assume also 

 that an event E 2 occurs on n 12 of the events on which E x also occurs. 

 Then the probability that both E x and E 2 will occur on one trial is 

 P{E X E 2 ) = n 12 /n, which can be rewritten in the following way: 



(3.17) P(E 1 E 2 ) = _ = (-)(_) = P(E 1 )-P(E 2 /E 1 ), 



n \n / \ni / 



where P{E 2 /E X ) is the probability that E 2 will occur after it is known 

 that E x has occurred. 



The probability law expressed in (3.17) actually includes the law 

 of (3.16) as a special case. If E 2 is independent of E lt the number 

 of single events on which E 2 can occur will be the same regardless 

 of the prior occurrence or the non-occurrence of E x ; hence the prob- 

 ability P{E 2 /E X ) will be just P{E 2 ), and the formula 3.17 becomes 

 (3.16). 



Problem 3.11. What is the probability that two successive aces will be drawn 

 from a well-shuffled bridge deck if the first card drawn is not replaced before 

 the second draw is made? 



The number of aces available for the first draw is 4, and any of 

 52 cards might be drawn; hence P(E X ) = 4/52. On the second draw 



