68 ELEMENTARY PROBABILITY Ch. 3 



if an unbiased coin is tossed 5 times, there are 6 classes of events, and 

 the specific single events in the class 4#, IT are 



Toss 

 First Second Third Fourth Fifth 



5! 



It is seen that C 5? 4 = — — - = 5 = the number of single events in the 



class which includes exactly 4 heads. It also should be observed that 

 the outcome of each toss is an independent event relative to the out- 

 come of any other toss; hence the probability of the first result listed 

 above, THHHH, is (1/2) (1/2) ••• (1/2) = (1/2) 5 . With this un- 

 biased coin, that also is the probability for any of the other single 

 events in this class of events. Therefore, the probability of an event 

 in the class 4#, IT is C 5i 4 (l/2) 4 (l/2) 1 , in which the exponent 4 refers 

 to the number of heads and the exponent 1 refers to the number of 

 tails. The reader can verify the fact that for any specified number of 

 heads from to 5 the probability of exactly r heads is C 5> r (l/2) r 

 X (l/2) 5— r , where r takes any value from to 5. 



In general, if n unbiased coins are to be flipped (or one such coin be 

 flipped n times) the probability of the appearance of any specific 

 number of heads, say r, is 



(3.31) P[r heads, (n - r) tails] = C n , r (l/2) r (l/2) n ~ r . 



To extend this result a bit, let an event E have a constant prob- 

 ability, p, of occurrence on each of n repeated trials. Then the 

 probability that E will occur on exactly r of the trials [and fail on 

 the other (n — r) trials] is given by the following formula: 



(3.32) P[rE's, (n - r) nat-E'a] = C„, Ap) r (l - vT~ r - 



The student can verify that this formula becomes (3.31) if p = 1/2, 

 E = H, and (not-E) = T. 



One more generalization can be obtained regarding formulas 3.31 

 and 3.32 by considering the expansions of the two binomials 



