Question

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:

• A. \(\frac{5}{72}\)
• B. \(\frac{5}{36}\)
• C. \(\frac{7}{72}\)
• D. \(\frac{7}{36}\)
• E. \(\frac{1}{36}\)

Answer 1

The Correct Option is B

Concept:

• Probability = favourable / total

Step 1: Individual probabilities.
\[ P(A) = \frac{6}{12} = \frac{1}{2}, \quad P(B) = \frac{4}{12} = \frac{1}{3}, \quad P(Draw) = \frac{2}{12} = \frac{1}{6} \]
Step 2: Alternating wins (no draw).
Possible patterns: \[ A \, B \, A \quad \text{and} \quad B \, A \, B \]
Step 3: Compute probability.
\[ P(ABA) = \frac{1}{2} \times \frac{1}{3} \times \frac{1}{2} = \frac{1}{12} \] \[ P(BAB) = \frac{1}{3} \times \frac{1}{2} \times \frac{1}{3} = \frac{1}{18} \] \[ \text{Total} = \frac{1}{12} + \frac{1}{18} = \frac{3+2}{36} = \frac{5}{36} \]
Step 4: Option analysis.

• (A) Incorrect $\times$
• (B) Correct \checkmark
• (C) Incorrect $\times$
• (D) Incorrect $\times$
• (E) Incorrect $\times$

Conclusion:
Thus, the correct answer is
Option (B).