swerve wrote:Romeo and Juliet play a dice game in which the two participants take turns rolling a single fair six-sided die. The first player to roll 6 wins. If Rome rolls first, what is the probability that Juliet will win?
A. \(\frac{1}{4}\)
B. \(\frac{1}{3}\)
C. \(\frac{4}{9}\)
D. \(\frac{5}{11}\)
E. \(\frac{1}{2}\)
Let P(R) = the probability that Romeo wins and P(J) = the probability that Juliet wins.
Since there are only two possible outcomes -- either Romeo wins or Juliet wins -- we get:
P(R) + P(J) = 1
Probability that Romeo wins on the first roll = 1/6
Probability that Romeo wins on the third roll = 1/6 + (5/6)(5/6)(1/6)
Probability that Romeo wins on the fifth roll = 1/6 + (5/6)(5/6)(1/6) + (5/6)(5/6)(5/6)(5/6)(1/6)
And so on.
Thus:
P(R) =
1/6 + (5/6)(5/6)(1/6) + (5/6)(5/6)(5/6)(5/6)(1/6)...
Probability that Juliet wins on the second roll = (5/6)(1/6)
Probability that Juliet wins on the fourth roll = (5/6)(1/6) + (5/6)(5/6)(5/6)(1/6)
Probability that Juliet wins on the sixth roll = (5/6)(1/6) + (5/6)(5/6)(5/6)(1/6) + (5/6)(5/6)(5/6)(5/6)(5/6)(1/6)
And so on.
Thus:
P(J) = (5/6)
(1/6) + (5/6)
(5/6)(5/6)(1/6) + (5/6)
(5/6)(5/6)(5/6)(5/6)(1/6)...
The blue portions common to P(R) and P(J) indicate the following:
P(J) = (5/6)P(R)
P(R) = (6/5)P(J)
Substituting P(R) = (6/5)P(J) into P(R) + P(J) = 1, we get:
(6/5)P(J) + P(J) = 1
(11/5)P(J) = 1
P(J) = 5/11
The correct answer is
D.
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