Question 221915
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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:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \left(\frac{1}{2}\right)\left(\frac{1}{2}\right)\left(\frac{1}{2}\right)\ =\ \frac{1}{8}]


Since the probability that M will win is *[tex \Large \frac{1}{8}], the probability that M will lose, which is the same as the probability that J will win is


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ 1 - \frac{1}{8} = \frac{7}{8}]



John
*[tex \LARGE e^{i\pi} + 1 = 0]
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