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. $\frac{19}{27}$
• D. none of these

Answer 1

Answer: (B)

$\frac{5}{36}$

$P\left(Awin\right)=\frac{6}{12}=\frac{1}{2}P\left(Bwin\right)=\frac{4}{12}=\frac{1}{3}$
The alternative win in 3 games can be in two possible ways as follows
ABA or BAB
Let's call this event to be E
$P\left(E\right)=P\left(A\right)\times P\left(B\right)\times P\left(A\right)+P\left(B\right)\times P\left(A\right)\times P\left(B\right)=\frac{1}{2}\times \frac{1}{3}\times \frac{1}{2}+\frac{1}{3}\times \frac{1}{2}\times \frac{1}{3}=\frac{1}{12}+\frac{1}{18}=\frac{5}{36}$