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Consider a seven-game world series between team A and B, where for each game
P(A wins)=0.6

a) Find P(A wins series in x game)
b) You hold a ticket for the seventh game. What is the probability that you will get to use it? .answer 0.2765
c) if P(A wins a game)=p, what value of p maximizes your chance in b)?answer p=1/2
d) what is the most likely number of games to be played in the series for p=0.6?
answer x=6

I have the answer but I do not understand the process. Can you explain this with details and step by step What does most likely means?

Solution Preview

From the condition, the probability that A wins a game is P(A)=0.6, then the probability that B wins a game is P(B)=1-P(A)=0.4.

(a) If A wins the series in x games, since it is a seven-game series, then A actually wins 4 games. The winner is the one who wins 4 games first. We also note A must win the last game. So A wins 3 games in the first (x-1) games. We select 3 from x-1, and there are C(x-1,3) ways. For each way, it has probability 0.6^3*0.4^(x-1-3)=0.216*0.4^(x-4).
Therefore, P(A wins series in x ...

Solution Summary

Probability problems are solved. The solution is detailed and well presented.