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$\mathrm{A, B, C \: and \: D}$ cut a pack of 52 cards successively in the order given. If the person who cuts, a spade first receives Rs 350 , then the expectation of $\mathrm{A}$ is
Option: 1
$129$

Option: 2
$\frac{64}{175}$

Option: 3
$350$

Option: 4
$128$

Answers (1)

best_answer

Let $\mathrm{E}$ be the event of any one cutting a spade in one cut and let $\mathrm{S}$ be the sample space then

$\mathrm{ n(E)={ }^{13} C_1 }$

and $\mathrm{\quad n(S)={ }^{52} C_1 }$

$\mathrm{\therefore P(E)=p=\frac{n(E)}{n(S)}=\frac{{ }^{13} C_1}{{ }^{52} C_1}=\frac{13}{52}=\frac{1}{4} }$

$\mathrm{ \Rightarrow \quad P(E)=p=\frac{1}{4} }$

$\mathrm{ \therefore P(\bar{E})=q=1-p=\frac{3}{4} }$

The probability of $\mathrm{ A }$ winning (when $\mathrm{ A }$ starts the game)

$\mathrm{ =p+q q q p+q q q q q q p+\ldots \infty }$

$\mathrm{ \Rightarrow \quad p+q^3 p+q^6 p+\ldots \infty }$

$\mathrm{=\frac{p}{1-q^3} }$ (sum of infinite GP)

$\mathrm{ =\frac{\frac{1}{4}}{1-\left(\frac{3}{4}\right)^3} }$

$\mathrm{ =\frac{64}{175} }$

$\mathrm{ \therefore }$ Expectation of $\mathrm{A=\mathrm{Rs} 350 \times \text { probability } }$
$\mathrm{=\mathrm{Rs}\: 350 \times \frac{64}{175} }$

$\mathrm{ =\mathrm{Rs} \: 128 \text {. } }$

Hence option 4 is correct

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Rishabh

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