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 win alternately, is

• A. $\frac{5}{12}$
• B. $\frac{5}{36}$
• C. $\frac{19}{27}$
• D. $\frac{5}{27}$

Answer 1

The correct option is B

$\frac{5}{36}$

Explanation for the correct answer:

Probability to win alternatively:

Probability of $A$ winning $=\frac{6}{12}=\frac{1}{2}$

Probability of $B$ winning $=\frac{4}{12}=\frac{1}{3}$

Probability of draw $=\frac{2}{12}=\frac{1}{6}$

They can win alternatively $2$ ways $ABA$ or $BAB.$

Probability they win alternatively $=P\left(ABA\right)+P\left(BAB\right)$

$$\begin{gathered} =\left(P\right(A)\times P(B)\times P(A\left)\right)+\left(P\right(B)\times P(A)\times P(B\left)\right) \\ =\left(\frac{1}{2}\right)\times \left(\frac{1}{3}\right)\times \left(\frac{1}{2}\right)+\left(\frac{1}{3}\right)\times \left(\frac{1}{2}\right)\times \left(\frac{1}{3}\right) \\ =\frac{1}{12}+\frac{1}{18} \\ =\frac{3+2}{36} \\ =\frac{5}{36} \end{gathered}$$

Therefore, the probability that they win alternately, is $\frac{5}{36}$.

Hence, the correct answer is option (B).