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2017 AMC 12B Problem 17

Problem 17 of 25IntermediateCounting & Probability

A coin is biased in such a way that on each toss the probability of heads is 23\dfrac{2}{3} and the probability of tails is 13.\dfrac{1}{3}. The outcomes of the tosses are independent. A player has the choice of playing Game A or Game B. In Game A she tosses the coin three times and wins if all three outcomes are the same. In Game B she tosses the coin four times and wins if both the outcomes of the first and second tosses are the same and the outcomes of the third and fourth tosses are the same. How do the chances of winning Game A compare to the chances of winning Game B?

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Solution

Let p=23.p = \dfrac23. Game A is won when all three tosses match: p3+(1−p)3.p^3 + (1-p)^3. Game B needs the first pair to match and the second pair to match, each with probability p2+(1−p)2,p^2 + (1-p)^2, so the win probability is (p2+(1−p)2)2.\bigl(p^2 + (1-p)^2\bigr)^2. With p=23,p = \tfrac23, Game A gives (23)3+(13)3=927=13,\left(\tfrac23\right)^3 + \left(\tfrac13\right)^3 = \tfrac{9}{27} = \tfrac13, and Game B gives (49+19)2=(59)2=2581.\left(\tfrac49 + \tfrac19\right)^2 = \left(\tfrac59\right)^2 = \tfrac{25}{81}. The difference is 2781−2581=281,\tfrac{27}{81} - \tfrac{25}{81} = \tfrac{2}{81}, so Game A is 281\tfrac{2}{81} more likely. Thus, the correct answer is D.

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Concepts: independent events · basic probability

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