Practicing Success

Target Exam

CUET

Subject

-- Mathematics - Section B1

Chapter

Probability

Question:

‘A’,’B’ and ‘C’ in order toss a coin. First one to get a head wins. What are their respective chances of winning?

Options:

$\frac{1}{3}$

$\frac{2}{3}$

$\frac{4}{7}$

$\frac{3}{7}$

Correct Answer:

$\frac{4}{7}$

Explanation:

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) = $\frac{1}{7}$