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# Answer to Question #5013 in Statistics and Probability for Ainy

Question #5013
each time Mae and Richard play chess, Mae has probability of 4/5 that she wins. If they play 5 games, determin the probability that a) Mae wins 3 games b) Mae wins either 4 or 5 games.
1
2011-11-15T09:59:46-0500
Each time Mae and Richard play chess, Mae has probability of 4/5 that she wins. If they play 5 games, determine the probability that
a) Mae wins 3 games

It means Mae wins three games and loose two games, so
P = (4/5)&sup3;(1-4/5)&sup2; = (4/5)&sup3;(1/5)&sup2; = 0.02048

b) Mae wins either 4 or 5 games

The probability Mae wins 4 games (win four games and loose one):
P1 = (1-4/5)(4/5)^4 = 0.08192.
The probability Mae wins 5 games:
P1 = (4/5)^5 = 0.32768.
So, the desired probability is
P = P1+P2 = 0.08192 + 0.32768 = 0.4096.

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