Question

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

$p\left(0\right)=1/3;p\left(1\right)=p(-1)=13/55$

$p\left(2\right)=p(-2)=1/11;p\left(3\right)=p(-3)=1/165$

Compute the conditional probability that the gambler wins $i,i=1,2,3,$given that he wins a positive amount.

Short answer

The probabilities are:

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

Step-by-step solution

Step 1:Given information

Let $X$be the winnings of a gambler. Let$p\left(i\right)=P(X=i)$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 $p\left(i\right)=\mathrm{\mathbb{P}}(X=i)$and suppose that $p\left(0\right)=\frac{1}{3},p\left(1\right)=$$p(-1)=\frac{13}{55},p\left(2\right)=p(-2)=\frac{1}{11},p\left(3\right)=p(-3)=\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.

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

Therefore we have:

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

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

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

Therefore, we are done.

Step 3:Final answer

The probabilities are

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