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Answers without the blur. Sign up and see all textbooks for free! Q.4.10

Expert-verified Found in: Page 163 ### A First Course in Probability

Book edition 9th
Author(s) Sheldon M. Ross
Pages 432 pages
ISBN 9780321794772 # Let$X$ be the winnings of a gambler. Let $p\left(i\right)=$$P\left(X=i\right)$ and suppose that$p\left(0\right)=1/3;p\left(1\right)=p\left(-1\right)=13/55$$p\left(2\right)=p\left(-2\right)=1/11;p\left(3\right)=p\left(-3\right)=1/165$Compute the conditional probability that the gambler wins $i,i=1,2,3,$ given that he wins a positive amount.

The probabilities are:

$\mathrm{ℙ}\left(X=1\mid Y\right)=\frac{39}{55},\mathrm{ℙ}\left(X=2\mid Y\right)=\frac{3}{11},\mathrm{ℙ}\left(X=3\mid Y\right)=\frac{1}{55}$

See the step by step solution

## Step 1:Given information

Let $X$ be the winnings of a gambler. Let$p\left(i\right)=P\left(X=i\right)$and suppose that

p(0) = 1/3; p(1) = p(−1) = 13/55;

p(2) = p(−2) = 1/11; p(3) = p(−3) = 1/165

## Step 2:Explanation

Let $X$ be a winning of the gambler. Also let us define $p\left(i\right)=\mathrm{ℙ}\left(X=i\right)$ and suppose that $p\left(0\right)=\frac{1}{3},p\left(1\right)=$$p\left(-1\right)=\frac{13}{55},p\left(2\right)=p\left(-2\right)=\frac{1}{11},p\left(3\right)=p\left(-3\right)=\frac{1}{165}$We are to calculate conditional probability of gambler winnning $i=1,2,3$ given that he wins positive amount.

Firstly let us calculate the probability that he won a positive amount.

$\mathrm{ℙ}\left(Y\right)=p\left(1\right)+p\left(2\right)+p\left(3\right)=\frac{13}{55}+\frac{1}{11}+\frac{1}{165}=\frac{55}{165}=\frac{1}{3}$

Therefore we have:

$\mathrm{ℙ}\left(X=1\mid Y\right)=\frac{\mathrm{ℙ}\left(X=1,Y\right)}{\mathrm{ℙ}\left(Y\right)}=\frac{\mathrm{ℙ}\left(X=1\right)}{\mathrm{ℙ}\left(Y\right)}=\frac{\frac{13}{55}}{\frac{1}{3}}=\frac{39}{55}$

$\mathrm{ℙ}\left(X=2\mid Y\right)=\frac{\mathrm{ℙ}\left(X=2,Y\right)}{\mathrm{ℙ}\left(Y\right)}=\frac{\mathrm{ℙ}\left(X=2\right)}{\mathrm{ℙ}\left(Y\right)}=\frac{\frac{1}{11}}{\frac{1}{3}}=\frac{3}{11}$

$\mathrm{ℙ}\left(X=3\mid Y\right)=\frac{\mathrm{ℙ}\left(X=3,Y\right)}{\mathrm{ℙ}\left(Y\right)}=\frac{\mathrm{ℙ}\left(X=3\right)}{\mathrm{ℙ}\left(Y\right)}=\frac{\frac{1}{165}}{\frac{1}{3}}=\frac{1}{55}$

Therefore, we are done.

The probabilities are

$\mathrm{ℙ}\left(X=1\mid Y\right)=\frac{39}{55},\mathrm{ℙ}\left(X=2\mid Y\right)=\frac{3}{11},\mathrm{ℙ}\left(X=3\mid Y\right)=\frac{1}{55}$ ### Want to see more solutions like these? 