Question

A and B throw a pair of dice. A wins if he throws 6 before B throws 7 and B wins if he throws 7 before A throws 6. If A begins, his chance of winning is :

• A. $0.4$
• B. $0.1$
• C. $0.5$
• D. $0.7$

Answer 1

The correct option is B $0.1$

The sum $6$ can be get in $5$ ways

i.e., $\left[\right(1,5),(2,4),(3,3),(4,2),(5,1\left)\right]$

$\bullet$ The probability of $A$ throwing $6$ is $\frac{5}{36}$

$\bullet$ The probability of $A$ not throwing $=1-\frac{5}{36}$

$=\frac{31}{36}$

Similarly the probability of $B$ throwing $7$ is $\frac{6}{36}$

$\bullet$ probability of $B$ not throwing $=1-\frac{6}{36}$

$=\frac{30}{36}=\frac{5}{6}$

Now, $A$ can win if the throw & $6$ in the first, third, fifth, throw the chance of $A$ winning $=0.1$

$\left[B\right]$ option.