Question

Two persons A and B throw a (fair) die(six-faced cube with faces numbered from $1$ to $6$) alternatively, starting with A. The first person to get an outcome different from the previous one by the opponent wins. The probability that B wins is$?$

• A. $5/6$
• B. $6/7$
• C. $7/8$
• D. $8/9$

Answer 1

The correct option is B $6/7$

$A$ rolls a dice, then probability of any face is $\frac{6}{6}=1$

If $B$ has to win, his dice must show different face from that of $A$

So, probability of getting different face is $\frac{5}{6}$

If $B$ has to win then the process is continues like $=\frac{6}{6}$( A win+( A not win) (B not win)(A win) +(A not win)(B not win) (A not win)(Bwin).......)

$\therefore P=(\frac{6}{6}\times \frac{5}{6}+\frac{6}{6}\times \frac{1}{6}\times \frac{1}{6}\times \frac{5}{6}+\frac{6}{6}\times \frac{1}{6}\times \frac{1}{6}\times \frac{1}{6}\times \frac{5}{6}.......)$

$P=\frac{5}{6}+\frac{5}{6^3}+\frac{5}{6^5}+....$

This forms a G.P with first term $=\frac{5}{6}$ and common ratio $=\frac{1}{36}$

$\therefore P=\frac{\frac{5}{6}}{1-\frac{1}{36}}=\frac{6}{7}$

Hence, option $\left(B\right)$ is correct.