Question

$2$ players $A,B$ tosses a fair coin in cyclic order $A,A,B,A,A,B\ldots $ till a head appears. If the probability that $A$ gets head first is $p$ , then $\frac{24}{p}$ is equal to

Answer 1

Required probability $=A+\overset{-}{A}A+\overset{-}{A}\overset{-}{A}\overset{-}{B}A+\overset{-}{A}\overset{-}{A}\overset{-}{B}\overset{-}{A}A+\overset{-}{A}\overset{-}{A}\overset{-}{B}\overset{-}{A}\overset{-}{A}\overset{-}{B}A$
$=\frac{1}{2}+\left(\frac{1}{2}\right)^{2}+\left(\frac{1}{2}\right)^{4}+\left(\frac{1}{2}\right)^{5}+\left(\frac{1}{2}\right)^{7}+\left(\frac{1}{2}\right)^{8}+\left(\frac{1}{2}\right)^{10}+\ldots $
$=\left(\frac{1}{2} + \left(\frac{1}{2}\right)^{4} + \left(\frac{1}{2}\right)^{7} + \left(\frac{1}{2}\right)^{10} + \ldots \right)+\left(\left(\frac{1}{2}\right)^{2} + \left(\frac{1}{2}\right)^{5} + \left(\frac{1}{2}\right)^{8} + \ldots \right)$
$=\frac{\frac{1}{2}}{1 - \frac{1}{8}}+\frac{\frac{1}{4}}{1 - \frac{1}{8}}$
$=\frac{4}{7}+\frac{2}{7}=\frac{6}{7}=p$
Hence, $\frac{24}{p}=\frac{24 \times 7}{6}=28$