Step 1: Understanding the Question:
The problem provides historical data from 12 games played by A and B, which we use to determine the individual probabilities of each outcome.
A and B are now going to play a series of exactly 3 games.
We must find the probability that the match outcomes represent an alternating sequence of wins for the two players.
Step 2: Key Formula or Approach:
First, calculate the probabilities of a single game: $P(A)$ for A winning, $P(B)$ for B winning.
An alternating win sequence in 3 games can happen in exactly two mutually exclusive ways:
Case 1: A wins, then B wins, then A wins (A-B-A).
Case 2: B wins, then A wins, then B wins (B-A-B).
Because the games are independent, the probability of a sequence is the product of individual probabilities.
Total Probability = $P(A \text{ then } B \text{ then } A) + P(B \text{ then } A \text{ then } B)$.
Step 3: Detailed Explanation:
From the 12 games historically played, A wins 6 times.
The probability that A wins a single game is $P(A) = \frac{6}{12} = \frac{1}{2}$.
B wins 4 times.
The probability that B wins a single game is $P(B) = \frac{4}{12} = \frac{1}{3}$.
(Although there are draws, they interrupt an alternating win sequence, so we don't need to consider them for our specific required sequences.)
We have two possible valid alternating win sequences over 3 games: A-B-A and B-A-B.
Let us calculate the probability for the first sequence, A-B-A.
$P(ABA) = P(A) \times P(B) \times P(A) = \left(\frac{1}{2}\right) \times \left(\frac{1}{3}\right) \times \left(\frac{1}{2}\right) = \frac{1}{12}$.
Let us calculate the probability for the second sequence, B-A-B.
$P(BAB) = P(B) \times P(A) \times P(B) = \left(\frac{1}{3}\right) \times \left(\frac{1}{2}\right) \times \left(\frac{1}{3}\right) = \frac{1}{18}$.
These two sequences are mutually exclusive, so we add their probabilities to find the total probability.
Total Probability = $\frac{1}{12} + \frac{1}{18}$.
The least common multiple (LCM) of 12 and 18 is 36.
We convert both fractions to have a denominator of 36: $\frac{1}{12} = \frac{3}{36}$ and $\frac{1}{18} = \frac{2}{36}$.
Total Probability = $\frac{3}{36} + \frac{2}{36} = \frac{5}{36}$.
Step 4: Final Answer:
The Probability that they will win alternately is 5/36.