Question

If two fair dice are thrown twice, find the probability of obtaining a product of six in the first throw and a sum of eight in the second throw

Answer 1

To solve this problem, we need to find two separate probabilities: the probability of obtaining a product of six on the first throw of two dice, and the probability of obtaining a sum of eight on the second throw of the same two dice. Finally, we will combine the probabilities since these events are independent.

### Step 1: Product of 6 on the first throw

Let's find all the combinations of the two dice that give a product of 6:

- Possible pairs (Die 1, Die 2):
- (1, 6) → Product = 6
- (2, 3) → Product = 6
- (3, 2) → Product = 6
- (6, 1) → Product = 6

Thus, the successful outcomes yielding a product of 6 are:
- (1, 6)
- (2, 3)
- (3, 2)
- (6, 1)

Counting these combinations, we have 4 successful outcomes.

The total number of outcomes when throwing two dice is:
\[
6 \times 6 = 36
\]

So, the probability of obtaining a product of 6 is:
\[
P(\text{Product of 6}) = \frac{4}{36} = \frac{1}{9}
\]

### Step 2: Sum of 8 on the second throw

Next, we find all the combinations of the two dice that give a sum of 8:

- Possible pairs (Die 1, Die 2):
- (2, 6) → Sum = 8
- (3, 5) → Sum = 8
- (4, 4) → Sum = 8
- (5, 3) → Sum = 8
- (6, 2) → Sum = 8

Thus, the successful outcomes yielding a sum of 8 are:
- (2, 6)
- (3, 5)
- (4, 4)
- (5, 3)
- (6, 2)

Counting these combinations, we have 5 successful outcomes.

As before, the total number of outcomes when throwing two dice is still:
\[
6 \times 6 = 36
\]

So, the probability of obtaining a sum of 8 is:
\[
P(\text{Sum of 8}) = \frac{5}{36}
\]

### Step 3: Combine the probabilities

Since the two events (product of 6 on the first throw and sum of 8 on the second throw) are independent, we can multiply their probabilities:

\[
P(\text{Product of 6 in first throw and Sum of 8 in second throw}) = P(\text{Product of 6}) \times P(\text{Sum of 8})
\]
\[
P(\text{Combined}) = \left(\frac{1}{9}\right) \times \left(\frac{5}{36}\right)
\]
\[
= \frac{1 \times 5}{9 \times 36} = \frac{5}{324}
\]

### Conclusion

Therefore, the probability of obtaining a product of six in the first throw and a sum of eight in the second throw is:
\[
\boxed{\frac{5}{324}}
\]