91Ó°ÊÓ

$Y_2=0$

Probability mass function $Y_1=1$given that$Y_2=1$

localid="1647199633826" $P\left(Y_1=1âˆ\pounds Y_2=1\right)=\frac{P\left(Y_1=1,Y_2=1\right)}{P\left(Y_2=1\right)}$

Probability mass function $Y_1=1$given that $Y_2=1$

$$\begin{gathered} P\left(Y_1=1âˆ\pounds Y_2=1\right)=\frac{P\left(1,1\right)}{P\left(0,1\right)+P\left(1,1\right)} \\ P\left(Y_1=1âˆ\pounds Y_2=1\right)=\frac{{\displaystyle \frac{1}{26}}}{{\displaystyle \frac{5}{26}}+{\displaystyle \frac{1}{26}}} \\ P\left(Y_1=1âˆ\pounds Y_2=1\right)=\frac{1}{6} \end{gathered}$$

Probability mass function $Y_1=0$given that$Y_2=1$

$$\begin{gathered} P\left(Y_1=0âˆ\pounds Y_2=1\right)=\frac{P\left(0,1\right)}{P\left(0,1\right)+P\left(1,1\right)} \\ P\left(Y_1=0âˆ\pounds Y_2=1\right)=\frac{{\displaystyle \frac{5}{26}}}{{\displaystyle \frac{5}{26}+\frac{1}{26}}} \\ P\left(Y_1=0âˆ\pounds Y_2=1\right)=\frac{5}{6} \end{gathered}$$

Table form of the conditional probability mass function of Y₁ given that $Y_2=1$

$Y_2=0$

Probability mass function $Y_1=1$ given that$Y_2=0$

$P\left(Y_1=1âˆ\pounds Y_2=0\right)=\frac{P\left(Y_1=1,Y_2=0\right)}{P\left(Y_2=0\right)}$

Probability mass function $Y_1=1$given that $Y_2=0$

$$\begin{gathered} P\left(Y_1=1âˆ\pounds Y_2=0\right)=\frac{P\left(1,0\right)}{P\left(1,0\right)+P\left(0,0\right)} \\ P\left(Y_1=1âˆ\pounds Y_2=0\right)=\frac{{\displaystyle \frac{5}{26}}}{{\displaystyle \frac{5}{26}+\frac{15}{26}}} \\ P\left(Y_1=1âˆ\pounds Y_2=0\right)=\frac{5}{20} \\ P\left(Y_1=1âˆ\pounds Y_2=0\right)=\frac{1}{4} \end{gathered}$$

Probability mass function $Y_1=0$given that$Y_2=0$

$$\begin{gathered} P\left(Y_1=0âˆ\pounds Y_2=0\right)=\frac{P\left(0,0\right)}{P\left(1,0\right)+P\left(0,0\right)} \\ P\left(Y_1=0âˆ\pounds Y_2=0\right)=\frac{{\displaystyle \frac{15}{26}}}{{\displaystyle \frac{5}{26}+\frac{15}{26}}} \\ P\left(Y_1=0âˆ\pounds Y_2=0\right)=\frac{15}{20} \\ P\left(Y_1=0âˆ\pounds Y_2=0\right)=\frac{3}{4} \end{gathered}$$