54 ELEMENTARY PROBABILITY Ch. 3 



marble. Then n x = 10, n 2 = 15, n 3 = 5, and n 4 = 20; hence 

 P(#i, # 2 , or Ei) = (10 + 15 + 20) /50 = 10/50 + 15/50 + 20/50 

 = .90. 



In the discussion leading to the Law of Total Probability the events 

 E 1} E 2 , . . . , E r were assumed to be mutually exclusive, that is, only 

 one of those events could occur on any one random trial. Suppose 

 now that E x and E 2 are independent events in the sense that each can 

 occur simultaneously on one trial without interfering or helping with 

 the occurrence of the other in any way. For example, the obtaining 

 of a 6 on one die and a 5 on another die on a single throw of the pair 

 of dice is an illustration of independent events. If E x and E 2 are 

 independent events, they can occur together in n 1 -n 2 combinations of 

 single events because each of the n x single events in E x can occur 

 with each of the n 2 single events in E 2 . The classes of events, E x 

 and E 2 , will each belong to a general class of events, which will be 

 supposed to contain JVi and N 2 single events, respectively. There- 

 fore, the total number of combinations of single events possible on 

 random trials now is N x -N 2 . Of those possible single events, n x -n 2 

 will belong to both E x and E 2 . Therefore, the probability that an 

 even in E x will occur simultaneously with an event in E 2 is given by 



P(E X smdE 2 ) = (n x -n 2 )/{N x -N 2 ) = {nJN^-in^N^ 



= P(E 1 )-P(E 2 ). 



As an illustration of the above discussion and results, suppose that 

 a game consists in throwing a penny and an unbiased die simul- 

 taneously, with the thrower winning if he throws a head on the 

 penny along with a 5 or a 6 on the die. Let E x represent throwing a 

 head on the coin, and E 2 throwing a 5 or a 6 on the die. A single 

 event now consists of a particular result on the coin plus a particular 

 result of throwing the die. The coin can turn up either of two ways, 

 the die any of six ways; hence there are 2(6) = 12 combinations of 

 events, each equally likely to occur on any one trial. In these cir- 

 cumstances, n x = 1, n 2 = 2, Ni — 2, and N 2 = Q; hence 



(1)(2) 

 P(H and a 5 or a 6) = — — - = (1/2) (2/6) = 1/6. 

 (2) (6) 



The reasoning and algebra above can be extended easily to prove 

 that, if the occurrence of events in classes E x , E 2 , . . . , and E r are 

 independent and can occur in n* out of Ni ways, respectively {i = 1 

 to r) , the probability of the simultaneous occurrence of these r events 



