‘A’,’B’ and ‘C’ in order toss a coin. First one to get a head wins. What are their respective chances of winning?
Answer & explanation
Correct answer: option 3
P(A wins) = P(A wins in Ist attempt) + P(A wins in IInd attempt) + .........
$=\frac{1}{2}+(\frac{1}{2}×\frac{1}{2}×\frac{1}{2})×\frac{1}{2}+(\frac{1}{2})^2×(\frac{1}{2})^2×(\frac{1}{2})^2×\frac{1}{2}+.....=\frac{\frac{1}{2}}{1-\frac{1}{8}}=\frac{4}{7}$
P(B wins) = P (B wins in Ist attempt) + P(B wins in IInd attempt) + ...........
$=\frac{1}{2}×\frac{1}{2}+(\frac{1}{2}×\frac{1}{2}×\frac{1}{2})×\frac{1}{2}×\frac{1}{2}+(\frac{1}{2})^2×(\frac{1}{2})^2×(\frac{1}{2})^2×\frac{1}{2}×\frac{1}{2}+.....$
$=\frac{\frac{1}{2}×\frac{1}{2}}{1-\frac{1}{8}}=\frac{2}{7}$; similarly, P(C wins) = $\frac{1}{7}$