To solve the problem, we need to find the conditional probability of the event that the die shows a number greater than 4 given that there is at least one tail. Let's break this down step by step.
### Step 1: Identify the Sample Space
When tossing a coin, we have two outcomes: Heads (H) and Tails (T). If the coin shows Heads, we toss it again. If it shows Tails, we throw a die.
The possible outcomes are:
1. HH (Heads, Heads)
2. HT (Heads, Tails)
3. TH (Tails, Heads)
4. TT (Tails, Tails)
5. T1 (Tails, 1)
6. T2 (Tails, 2)
7. T3 (Tails, 3)
8. T4 (Tails, 4)
9. T5 (Tails, 5)
10. T6 (Tails, 6)
### Step 2: Calculate the Probabilities of Each Outcome
- The probability of getting Heads on the first toss is \( \frac{1}{2} \).
- The probability of getting Tails on the first toss is \( \frac{1}{2} \).
For the outcomes:
- HH: Probability = \( \frac{1}{2} \times \frac{1}{2} = \frac{1}{4} \)
- HT: Probability = \( \frac{1}{2} \times \frac{1}{2} = \frac{1}{4} \)
- TH: Probability = \( \frac{1}{2} \times \frac{1}{2} = \frac{1}{4} \)
- TT: Probability = \( \frac{1}{2} \times \frac{1}{2} = \frac{1}{4} \)
For the die outcomes:
- T1, T2, T3, T4, T5, T6: Each has a probability of \( \frac{1}{2} \times \frac{1}{6} = \frac{1}{12} \).
### Step 3: Define Events
Let:
- \( E \): Event that there is at least one tail.
- \( K \): Event that the die shows a number greater than 4 (i.e., T5 or T6).
### Step 4: Determine the Outcomes for Events E and K
- Outcomes for \( E \) (at least one tail): TH, TT, T1, T2, T3, T4, T5, T6
- Outcomes for \( K \) (die shows > 4): T5, T6
### Step 5: Calculate the Probability of Events E and K
- Probability of \( E \):
- Outcomes: TH, TT, T1, T2, T3, T4, T5, T6
- Probabilities: \( \frac{1}{4} + \frac{1}{4} + \frac{1}{12} + \frac{1}{12} + \frac{1}{12} + \frac{1}{12} + \frac{1}{12} + \frac{1}{12} \)
- Total: \( \frac{1}{4} + \frac{1}{4} + \frac{6}{12} = \frac{3}{4} \)
- Probability of \( K \):
- Outcomes: T5, T6
- Probabilities: \( \frac{1}{12} + \frac{1}{12} = \frac{2}{12} = \frac{1}{6} \)
### Step 6: Calculate the Probability of \( E \cap K \)
- Outcomes for \( E \cap K \): T5, T6
- Probability: \( \frac{1}{12} + \frac{1}{12} = \frac{2}{12} = \frac{1}{6} \)
### Step 7: Use Conditional Probability Formula
The conditional probability \( P(K|E) \) is given by:
\[
P(K|E) = \frac{P(E \cap K)}{P(E)}
\]
Substituting the values:
\[
P(K|E) = \frac{\frac{1}{6}}{\frac{3}{4}} = \frac{1}{6} \times \frac{4}{3} = \frac{4}{18} = \frac{2}{9}
\]
### Final Answer
The conditional probability that the die shows a number greater than 4 given that there is at least one tail is \( \frac{2}{9} \).
