SOLUTION: Manu and Janani are playing a coin toss game with a fair penny. Manu gets a point if the penny lands on heads, and Janani gets a point if the penny lands on tails. The score is Jan
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Question 221915: Manu and Janani are playing a coin toss game with a fair penny. Manu gets a point if the penny lands on heads, and Janani gets a point if the penny lands on tails. The score is Janani 9, Manu 7, in a game to 10 points. What is the probability that Janani will win the game? Express your answer as a common fraction.
Answer by solver91311(24713) (Show Source): You can put this solution on YOUR website!
The probability that J will win is one minus the probability that he will lose because there are only two possible outcomes. In order for J to lose, M must win. In order for M to win, the next three coin tosses in a row must result in heads. The probability that three consecutive tosses of a fair coin will be heads is:
\left(\frac{1}{2}\right)\left(\frac{1}{2}\right)\ =\ \frac{1}{8})
Since the probability that M will win is 
, the probability that M will lose, which is the same as the probability that J will win is
John
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