Sheldon Ross chapter 4 Random variables Qno-72

Question

Two athletic teams play a series of games; the first team to win 4 games is declared the overall winner .Suppose that one of the team is stronger than the other and wins each game with prob 0.6,independently of the outcomes of other games.Find the prob for i=4,5,6,7,,,that the sronger team wins the series in exactly i games.Compare the prob that the stronger team wins with the prob that it would win a 2-out-3 series.

Answer 1

Let us do for one value of 'i' and same can be done for other values of 'i' as well..Let us take i = 6 here..

So it means that the 6th game must result to a win in order to win the series..Hence we need to ensure 3 wins in preceding 5 matches..Hence now this problem reduces to binomial distribution B(5 , 0.6) as we are considering the preceding 5 matches(trials) and probability of individual trial = 0.6..

Hence 

P(series is won by strong team)   =   P(3 wins in 5 matches) * P(win in 6th match)

                                                 =  B(5,0.6)  *  0.6

                                                 =  ⁵C₃ (0.6)³ (0.4)² * 0.6

                                                 =  0.20736

                                                 =  0.21 (correct to 2 decimal places)

Similarly we can find for i = 4 , 5  and 7 as well.

P(win a 2 - out of 3 series)          =  ³C₂ (0.6)² (0.4)

                                                =  0.432

Comments

okay..I got it.Thanks :)

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Ans given in instructer manual is P{wins in i games} = C(i-1 ,3) *(.6)⁴ *(.4)^i-4

@ Habibkhan your approach is really nice.

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Ya on simplification it will come out to be the same..:)