To solve the problem of finding the probability that a sum of 8 occurs first when a pair of dice is thrown until either a sum of 4 or 6 appears, we can follow these steps:
### Step 1: Determine the probabilities of sums 4, 6, and 8.
1. **Calculate the probability of getting a sum of 4:**
- Possible combinations: (1,3), (2,2), (3,1)
- Total combinations for sum of 4 = 3
- Probability of sum of 4, \( P(4) = \frac{3}{36} = \frac{1}{12} \)
2. **Calculate the probability of getting a sum of 6:**
- Possible combinations: (1,5), (2,4), (3,3), (4,2), (5,1)
- Total combinations for sum of 6 = 5
- Probability of sum of 6, \( P(6) = \frac{5}{36} \)
3. **Calculate the probability of getting a sum of 8:**
- Possible combinations: (2,6), (3,5), (4,4), (5,3), (6,2)
- Total combinations for sum of 8 = 5
- Probability of sum of 8, \( P(8) = \frac{5}{36} \)
### Step 2: Calculate the probability of getting either a sum of 4 or 6.
- The probability of getting either a sum of 4 or 6 is given by:
\[
P(4 \text{ or } 6) = P(4) + P(6) = \frac{3}{36} + \frac{5}{36} = \frac{8}{36} = \frac{2}{9}
\]
### Step 3: Calculate the probability of not getting a sum of 4 or 6.
- The probability of not getting a sum of 4 or 6 is:
\[
P(\text{not } (4 \text{ or } 6)) = 1 - P(4 \text{ or } 6) = 1 - \frac{2}{9} = \frac{7}{9}
\]
### Step 4: Set up the probability for the event of getting a sum of 8 first.
- Let \( P \) be the probability that a sum of 8 occurs first. The event can occur in the following ways:
- Get a sum of 8 on the first throw.
- Get neither 4 nor 6, and then get a sum of 8 on the next throw.
- This continues indefinitely.
### Step 5: Create the equation for the probability \( P \).
- The probability can be expressed as:
\[
P = P(8) + P(\text{not } (4 \text{ or } 6)) \cdot P
\]
- Substituting the known probabilities:
\[
P = \frac{5}{36} + \frac{7}{9} P
\]
### Step 6: Solve for \( P \).
1. Rearranging the equation:
\[
P - \frac{7}{9} P = \frac{5}{36}
\]
\[
\frac{2}{9} P = \frac{5}{36}
\]
2. Multiply both sides by \( \frac{9}{2} \):
\[
P = \frac{5}{36} \cdot \frac{9}{2} = \frac{45}{72} = \frac{5}{8}
\]
### Final Answer:
The probability that a sum of 8 occurs first is \( \frac{5}{8} \).
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