It has been found that if A and B play a game 12 times, A wins 6 times, B wins 4 times and they draw twice. A and B take part in a series of 3 games. The probability that they will win alternately is

Options

A. `(5)/(72)`
B. `(5)/(36)`
C. `(19)/(27)`
D. none of these

Explanation

To solve the problem of finding the probability that A and B win alternately in a series of 3 games, we can follow these steps:

### Step 1: Determine the probabilities of A and B winning
From the previous games played:
- A wins 6 out of 12 games, so the probability of A winning (P(A)) is:
\[ P(A) = \frac{6}{12} = \frac{1}{2} \]
- B wins 4 out of 12 games, so the probability of B winning (P(B)) is:
\[ P(B) = \frac{4}{12} = \frac{1}{3} \]
- The probability of a draw is not needed for this calculation since we are only interested in wins.

### Step 2: Identify the winning combinations for alternating wins
In a series of 3 games, the two possible combinations for A and B to win alternately are:
1. A wins, B wins, A wins (ABA)
2. B wins, A wins, B wins (BAB)

### Step 3: Calculate the probability for each combination
**Combination 1 (ABA):**
- The probability of A winning first: \( P(A) = \frac{1}{2} \)
- The probability of B winning second: \( P(B) = \frac{1}{3} \)
- The probability of A winning third: \( P(A) = \frac{1}{2} \)

Thus, the probability for the combination ABA is:
\[ P(ABA) = P(A) \cdot P(B) \cdot P(A) = \frac{1}{2} \cdot \frac{1}{3} \cdot \frac{1}{2} = \frac{1}{12} \]

**Combination 2 (BAB):**
- The probability of B winning first: \( P(B) = \frac{1}{3} \)
- The probability of A winning second: \( P(A) = \frac{1}{2} \)
- The probability of B winning third: \( P(B) = \frac{1}{3} \)

Thus, the probability for the combination BAB is:
\[ P(BAB) = P(B) \cdot P(A) \cdot P(B) = \frac{1}{3} \cdot \frac{1}{2} \cdot \frac{1}{3} = \frac{1}{18} \]

### Step 4: Calculate the total probability of winning alternately
Now, we add the probabilities of both combinations:
\[ P(\text{alternating wins}) = P(ABA) + P(BAB) = \frac{1}{12} + \frac{1}{18} \]

To add these fractions, we need a common denominator. The least common multiple of 12 and 18 is 36.
- Convert \(\frac{1}{12}\) to have a denominator of 36:
\[ \frac{1}{12} = \frac{3}{36} \]
- Convert \(\frac{1}{18}\) to have a denominator of 36:
\[ \frac{1}{18} = \frac{2}{36} \]

Now, add the two fractions:
\[ P(\text{alternating wins}) = \frac{3}{36} + \frac{2}{36} = \frac{5}{36} \]

### Final Answer
The probability that A and B will win alternately in a series of 3 games is:
\[ \frac{5}{36} \]

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