Multiple choice

A man alternatively tosses a coin and throws a die. The probability of getting a head on the coin before he gets 4 on the die is

  1. $ \displaystyle \frac{6}{7} $
  2. $ \displaystyle \frac{2}{3} $
  3. $ \displaystyle \frac{3}{4} $
  4. $ \displaystyle \frac{1}{2} $
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A Correct answer
Explanation

Let H be getting a head (p=1/2) and D be getting a 4 (p=1/6). The man wins if he gets H before D. The sequence of events is: H (win), D (continue), H (win), D (continue)... The probability of winning is P(H) + P(not H and not D) * P(H) + ... This is a geometric series. P(win) = (1/2) + (1/2 * 5/6) * (1/2) + (1/2 * 5/6)^2 * (1/2) = (1/2) / (1 - 5/12) = (1/2) / (7/12) = 6/7.

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