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Probability Overdose

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Probability Overdose

by dtweah » Mon May 11, 2009 8:14 am
Team A plays team B in a series of games. If in each game the probability that A wins is 2/3, then what is the probability that A wins two games before B wins two games?

(a) 8/9
(b) 9/16
(c) 20/27
(d) 64/81
(e) 7/9
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by DeepakR » Mon May 11, 2009 8:39 am
A winning = 2/3 and B winning = 1/3

So we have the following scenario:
A B A B - i.e. A winning 1st and 3rd game so P= 2/3*1/3*2/3=4/27 --- (1)
A A B B - i.e. A winning the 1st and 2nd game so P = 2/3*2/3 = 4/9 --- (2)
B A A B - i.e. B winning 1st and A winning the next 2 so P = 1/3*2/3*2/3
= 4/27 --- (3)

Total P = 4/27 + 4/9 + 4/27 = 20/27 = C.)

-Deepak
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