Question

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

Answer 1

Answer: 28

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