Multiple choice

A man throws a die and tosses a coin alternately, starting with the coin. Find the probability that he gets head before $5$ or $ 6$ on the die.

  1. $\displaystyle\frac{3}{4}$
  2. $\displaystyle\frac{3}{7}$
  3. $\displaystyle\frac{1}{2}$

  4. None of these
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

The probability of getting a head is 1/2. The probability of getting 5 or 6 on a die is 2/6 = 1/3. The man wins if he gets a head (H) or if he fails to get 5/6 on the die (D') and then gets a head on the next turn. The probability is P(H) + P(D')P(T)*P(H) + ... which is a geometric series: (1/2) + (2/3)(1/2)(1/2) + (2/3)(1/2)(2/3)(1/2)(1/2) + ... = (1/2) / (1 - (2/3)(1/2)) = (1/2) / (2/3) = 3/4.

AI explanation

Let p be the probability of getting a head on the coin, which is 1/2, and q be the probability of not getting a 5 or 6 on the die, which is 4/6 or 2/3. Using the sum of an infinite geometric series where the man gets a head before the specific die outcomes, the probability is p / (1 - q*p). Plugging in the values gives (1/2) / (1 - (2/3)*(1/2)) = (1/2) / (2/3) = 3/4.

Keep practicing — related questions