91Ó°ÊÓ

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

Let $X$be a winning of the gambler. Also let us define $p\left(i\right)=\mathrm{â„™}(X=i)$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.

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

Therefore we have:

$\mathrm{â„™}(X=1âˆ\pounds Y)=\frac{\mathrm{â„™}(X=1,Y)}{\mathrm{â„™}\left(Y\right)}=\frac{\mathrm{â„™}(X=1)}{\mathrm{â„™}\left(Y\right)}=\frac{\frac{13}{55}}{\frac{1}{3}}=\frac{39}{55}$

$\mathrm{â„™}(X=2âˆ\pounds Y)=\frac{\mathrm{â„™}(X=2,Y)}{\mathrm{â„™}\left(Y\right)}=\frac{\mathrm{â„™}(X=2)}{\mathrm{â„™}\left(Y\right)}=\frac{\frac{1}{11}}{\frac{1}{3}}=\frac{3}{11}$

$\mathrm{â„™}(X=3âˆ\pounds Y)=\frac{\mathrm{â„™}(X=3,Y)}{\mathrm{â„™}\left(Y\right)}=\frac{\mathrm{â„™}(X=3)}{\mathrm{â„™}\left(Y\right)}=\frac{\frac{1}{165}}{\frac{1}{3}}=\frac{1}{55}$

Therefore, we are done.

The probabilities are

$\mathrm{â„™}(X=1âˆ\pounds Y)=\frac{39}{55},\mathrm{â„™}(X=2âˆ\pounds Y)=\frac{3}{11},\mathrm{â„™}(X=3âˆ\pounds Y)=\frac{1}{55}$