Question

A, B, C in order cut a pack of cards replacing them after each cut; on condition that the first who cuts a spade shall win a prize; find their respective chances.

• A. $\frac{15}{37},\frac{12}{37},\frac{10}{37}$
• B. $\frac{16}{37},\frac{11}{37},\frac{10}{37}$
• C. $\frac{16}{37},\frac{12}{37},\frac{9}{37}$
• D. $\frac{14}{37},\frac{13}{37},\frac{10}{37}$

Answer 1

Answer: (C)

$\frac{16}{37},\frac{12}{37},\frac{9}{37}$

Prob.of cutting a spade card $=\frac{13}{52}=\frac{1}{4}.$ and of not cutting it $=1-\frac{1}{4}=\frac{3}{4}$
Let x denote the prob. of cutting a spade card in the long run by A , then $x+\frac{3}{4}x+{\left(\frac{3}{4}\right)}^{2}x=1,$
as it is certain that one of A, B, C will cut a spade card and B and C in their chances cut the spade card under exactly similar conditions.
The spade card under exactly similar conditions.
$\therefore$ their respective probabilities are $\frac{16}{37},\frac{3}{4}\times \frac{16}{37},{\left(\frac{3}{2}\right)}^2\frac{16}{37}$
or respective chances are $\frac{16}{37},\frac{12}{37},\frac{9}{37}.$.