SOURH7WK wrote:
A and B decide to play a game based on probabilities. They throw a fair dice alternately, whoever gets a 5 on his throw wins. What is the probability that A wins the game, if it is known that B starts the game?
A. 7/12
B. 5/12
C. 5/11
D. 7/11
E. 1/2
The probability that A wins the game after the first round is
\frac{5}{6}*\frac{1}{6} (B must get anything but 5 and A must get 5);
The probability that A wins the game after the second round is
(\frac{5}{6}*\frac{5}{6})*(\frac{5}{6}*\frac{1}{6}) (both must get anything but 5 in the first round and A must win in the second round);
The probability that A wins the game after the third round is
(\frac{5}{6}*\frac{5}{6})*(\frac{5}{6}*\frac{5}{6})*(\frac{5}{6}*\frac{1}{6});
and so on...
The overall probability that A wins, will be the sum of the above probabilities:
\frac{5}{6}*\frac{1}{6}+(\frac{5}{6}*\frac{5}{6})*(\frac{5}{6}*\frac{1}{6})+(\frac{5}{6}*\frac{5}{6})*(\frac{5}{6}*\frac{5}{6})*(\frac{5}{6}*\frac{1}{6})+...Notice that we have the sum of the terms of geometric progression with the first term equal to
\frac{5}{6}*\frac{1}{6}=\frac{5}{36} and the common ratio equal to
\frac{5}{6}*\frac{5}{6}=\frac{25}{36}.
Now, the sum of
infinite geometric progression with
common ratio |r|<1, is
sum=\frac{b}{1-r}, where
b is the first term.
So, the above sum equals to
\frac{\frac{5}{36}}{1-\frac{25}{36}}=\frac{5}{11}.
Answer: C.
P.S. Not a GMAT type of question.
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52% (06:01) correct
