Suppose a girl throws a die. If she gets 5 or 6, she tosses a coin three times and notes the number of heads. If she gets 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 that she threw 1, 2, 3, or 4 with the die ?

Question

Accepted Answer

To solve this problem, we need to determine the probability that the girl threw a 1, 2, 3, or 4 on the die, given that she obtained exactly one head.

Let's break down the problem:

1. **Die Roll Outcomes:**
   - If the die shows 5 or 6, she tosses a coin three times.
   - If the die shows 1, 2, 3, or 4, she tosses a coin once.

2. **Coin Toss Outcomes:**
   - For a single coin toss (when the die shows 1, 2, 3, or 4), the outcomes are either Head (H) or Tail (T).
   - For three coin tosses (when the die shows 5 or 6), the possible outcomes for exactly one head are: HTT, THT, TTH.

3. **Probability Calculations:**
   - The probability of rolling a 1, 2, 3, or 4 on the die is $\frac{4}{6} = \frac{2}{3}$.
   - The probability of rolling a 5 or 6 on the die is $\frac{2}{6} = \frac{1}{3}$.

4. **Probability of Obtaining Exactly One Head:**
   - If the die shows 1, 2, 3, or 4, the probability of getting exactly one head (H) in one toss is $\frac{1}{2}$.
   - If the die shows 5 or 6, the probability of getting exactly one head in three tosses is calculated as follows:
     - There are 3 favorable outcomes (HTT, THT, TTH) out of 8 possible outcomes (HHH, HHT, HTH, HTT, THH, THT, TTH, TTT).
     - Thus, the probability is $\frac{3}{8}$.

5. **Using Bayes' Theorem:**
   - We need to find $P(\text{1, 2, 3, or 4} \mid \text{exactly one head})$.
   - Using Bayes' theorem:
     $$
     P(\text{1, 2, 3, or 4} \mid \text{exactly one head}) = \frac{P(\text{exactly one head} \mid \text{1, 2, 3, or 4}) \cdot P(\text{1, 2, 3, or 4})}{P(\text{exactly one head})}
     $$
   - Where:
     - $P(\text{exactly one head} \mid \text{1, 2, 3, or 4}) = \frac{1}{2}$
     - $P(\text{exactly one head} \mid \text{5 or 6}) = \frac{3}{8}$
     - $P(\text{exactly one head}) = P(\text{exactly one head} \mid \text{1, 2, 3, or 4}) \cdot P(\text{1, 2, 3, or 4}) + P(\text{exactly one head} \mid \text{5 or 6}) \cdot P(\text{5 or 6})$
     - $P(\text{exactly one head}) = \frac{1}{2} \cdot \frac{2}{3} + \frac{3}{8} \cdot \frac{1}{3} = \frac{1}{3} + \frac{1}{8} = \frac{11}{24}$

6. **Final Calculation:**
   - Substituting back into Bayes' theorem:
     $$
     P(\text{1, 2, 3, or 4} \mid \text{exactly one head}) = \frac{\frac{1}{2} \cdot \frac{2}{3}}{\frac{11}{24}} = \frac{\frac{1}{3}}{\frac{11}{24}} = \frac{8}{11}
     $$

Therefore, the probability that she threw a 1, 2, 3, or 4 given that she obtained exactly one head is $\frac{8}{11}$, which corresponds to Option 4.

**Explanation of Incorrect Options:**

- **Option 1: $\frac{11}{24}$**: This is the probability of obtaining exactly one head overall, not the conditional probability we are looking for.
- **Option 2: $\frac{2}{3}$**: This is the probability of rolling a 1, 2, 3, or 4 on the die, not the conditional probability given exactly one head.
- **Option 3: $\frac{1}{8}$**: This does not correspond to any relevant probability in this context.

Thus, Option 4 is the correct answer.