Multiple choice

Suppose a girl throws a die. If she gets a 5 or 6, she tosses a coin three times and notes the number of heads. If she gets a 1, 2, 3 or 4, she tosses a coin once and notes whether a head or tail is obtained. If she obtained exactly one head, what is the probability of getting a 1, 2, 3 or 4 on the die?

  1. 8/11

  2. 1/11

  3. 3/11

  4. 2/3

  5. 4/11
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A Correct answer
Explanation

Let E be the event of getting exactly one head. P(E) = P(E|5,6)P(5,6) + P(E|1,2,3,4)P(1,2,3,4). P(E|5,6) = 3C1 * (0.5)^3 = 3/8. P(E|1,2,3,4) = 1C1 * 0.5 = 1/2. P(E) = (3/8 * 2/6) + (1/2 * 4/6) = 1/8 + 1/3 = 11/24. P(1,2,3,4|E) = (1/2 * 4/6) / (11/24) = (1/3) / (11/24) = 8/11.

AI explanation

Using Bayes' theorem, the probability of rolling a 1, 2, 3 or 4 given exactly one head is found by comparing the weighted probability of that die roll producing the outcome to the total probability of getting exactly one head. The probability of rolling 1, 2, 3 or 4 is 4/6 and tossing exactly one head is 1/2, yielding 2/6. The probability of rolling 5 or 6 is 2/6 and tossing exactly one head in three tosses is 3/8, yielding 1/8 or 0.75/6. The total probability of one head is 2/6 + 0.75/6 = 2.75/6. The final conditional probability is (2/6) divided by (2.75/6), which equals 2/2.75 or 8/11. The result is 8/11.

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