2025 AMC 12B Problem 18

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Concepts:expected valuegeometric distribution

Difficulty rating: 1770

18.

Awnik repeatedly plays a game that has a probability of winning of 13.\dfrac{1}{3}. The outcomes of the games are independent. What is the expected value of the number of games he will play until he has both won and lost at least once?

52\dfrac{5}{2}

33

165\dfrac{16}{5}

72\dfrac{7}{2}

154\dfrac{15}{4}

Solution:

The first game produces one outcome. If it was a win (probability 13\tfrac{1}{3}), the expected wait for a loss is 12/3=32;\tfrac{1}{2/3} = \tfrac{3}{2}; if it was a loss (probability 23\tfrac{2}{3}), the expected wait for a win is 11/3=3.\tfrac{1}{1/3} = 3. So the expected total is 1+1332+233=1+12+2=72.1 + \tfrac{1}{3}\cdot\tfrac{3}{2} + \tfrac{2}{3}\cdot 3 = 1 + \tfrac{1}{2} + 2 = \tfrac{7}{2}.

Thus, the correct answer is D.

Problem 18 in Other Years