To solve the problem step by step, we need to calculate the probabilities involved in the game where A and B throw a pair of dice alternately. A wins if he gets a total of 6, and B wins if she gets a total of 7. A starts the game, and we want to find the probability of A winning in the third throw.
### Step 1: Calculate the probability of A winning (P(A))
A can win by rolling a total of 6. The combinations to achieve this with two dice are:
- (1, 5)
- (2, 4)
- (3, 3)
- (4, 2)
- (5, 1)
Thus, there are 5 successful outcomes for A.
The total number of outcomes when rolling two dice is 36 (6 faces on the first die multiplied by 6 faces on the second die).
So, the probability of A winning is:
\[ P(A) = \frac{5}{36} \]
### Step 2: Calculate the probability of A losing (P(A'))
Since A can either win or lose, we can find the probability of A losing:
\[ P(A') = 1 - P(A) = 1 - \frac{5}{36} = \frac{31}{36} \]
### Step 3: Calculate the probability of B winning (P(B))
B can win by rolling a total of 7. The combinations to achieve this are:
- (1, 6)
- (2, 5)
- (3, 4)
- (4, 3)
- (5, 2)
- (6, 1)
Thus, there are 6 successful outcomes for B.
So, the probability of B winning is:
\[ P(B) = \frac{6}{36} = \frac{1}{6} \]
### Step 4: Calculate the probability of B losing (P(B'))
Similarly, the probability of B losing is:
\[ P(B') = 1 - P(B) = 1 - \frac{1}{6} = \frac{5}{6} \]
### Step 5: Determine the sequence of events for A to win in the third throw
For A to win in the third throw, the following sequence of events must occur:
1. A must lose on the first throw.
2. B must lose on the second throw.
3. A must win on the third throw.
The probabilities for these events are:
- Probability that A loses on the first throw: \( P(A') = \frac{31}{36} \)
- Probability that B loses on the second throw: \( P(B') = \frac{5}{6} \)
- Probability that A wins on the third throw: \( P(A) = \frac{5}{36} \)
### Step 6: Calculate the overall probability of A winning in the third throw
Now, we multiply these probabilities together:
\[
P(\text{A wins in third throw}) = P(A') \times P(B') \times P(A) = \left(\frac{31}{36}\right) \times \left(\frac{5}{6}\right) \times \left(\frac{5}{36}\right)
\]
Calculating this:
\[
= \frac{31 \times 5 \times 5}{36 \times 6 \times 36} = \frac{775}{7776}
\]
### Final Answer
Thus, the probability of A winning the game in the third throw is:
\[
\frac{775}{7776}
\]
