Q79 / 160 · TS EAMCET 2023 (13 May Shift 2) · Mathematics
Two players A and B alternatively toss 3 . coins simultaneously. The player who gets 2 heads and 1 tail first, wins the game. If game continues until someone wins and if $A$ begins the game, the probability that $B$ wins the game is
- A$\frac{24}{39}$
- B$\frac{4}{7}$
- Correct answerC$\frac{15}{39}$
- D$\frac{3}{7}$
Solution
Let probability of $A$ to win $=P(A)$
and Probability of $B$ to win $=P(B)$
If Probability of $A$ to loss $=P(A)$
and Probability of $B$ to loss $=P\left(B^{\prime}\right)$
Also $P(2$ head 1 tail $)=\frac{3}{8}$
$$
\begin{aligned}
& P(B \text { won })=P\left(A^{\prime}\right) \cdot P(B)+P\left(A^{\prime}\right) \cdot P\left(B^{\prime}\right) P\left(A^{\prime}\right) P(B) \\
& +P\left(A^{\prime}\right) P\left(B^{\prime}\right) P\left(A^{\prime}\right) P\left(B^{\prime}\right) \cdot P\left(A^{\prime}\right) P(B)+\ldots . . \\
& =\frac{5}{8} \cdot \frac{3}{8}+\left(\frac{5}{8}\right)^3 \cdot \frac{3}{8}+\left(\frac{5}{8}\right)^5 \cdot \frac{3}{8}+\ldots \\
& =\frac{3}{8} \cdot \frac{5}{8}\left[1+\left(\frac{5}{8}\right)^2+\left(\frac{5}{8}\right)^4+\ldots . .\right. \\
& =\frac{5}{64} \frac{1}{1-\left(\frac{5}{8}\right)^2}=\frac{15}{64} \times \frac{64}{39}=\frac{15}{39} .
\end{aligned}
$$
Hence, the correct answer is option: 'C'.