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 service of 3 games. The probability that they win alternately, is:

Options

A. `(5)/(12)`
B. `(5)/(36)`
C. `(19)/(27)`
D. `(5)/(27)`

Answer

To solve the problem step by step, we need to calculate the probability that A and B win alternately in a series of 3 games.

### Step 1: Determine the probabilities of A's and B's wins and draws.
- A wins 6 times out of 12 games.
- B wins 4 times out of 12 games.
- There are 2 draws.

**Calculating the probabilities:**
- Probability of A winning (P(A)) = Number of wins by A / Total games = 6/12 = 1/2
- Probability of B winning (P(B)) = Number of wins by B / Total games = 4/12 = 1/3
- Probability of a draw (P(D)) = Number of draws / Total games = 2/12 = 1/6

### Step 2: Identify the winning sequences for alternating wins.
For 3 games, the alternating winning sequences can be:
1. A wins the 1st game, B wins the 2nd game, A wins the 3rd game (ABA).
2. B wins the 1st game, A wins the 2nd game, B wins the 3rd game (BAB).

### Step 3: Calculate the probability for each sequence.
**For the sequence ABA:**
- Probability of A winning the 1st game = P(A) = 1/2
- Probability of B winning the 2nd game = P(B) = 1/3
- Probability of A winning the 3rd game = P(A) = 1/2

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

**For the sequence BAB:**
- Probability of B winning the 1st game = P(B) = 1/3
- Probability of A winning the 2nd game = P(A) = 1/2
- Probability of B winning the 3rd game = P(B) = 1/3

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

### Step 4: Add the probabilities of the two sequences.
Now, we add the probabilities of both sequences to get the total probability of alternating wins:
\[ P(\text{Alternating wins}) = P(ABA) + P(BAB) = \frac{1}{12} + \frac{1}{18} \]

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

Now, add them:
\[ P(\text{Alternating wins}) = \frac{3}{36} + \frac{2}{36} = \frac{5}{36} \]

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

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