Question

In Problem $6.3$, calculate the conditional probability mass function of $Y_1$given that

(a) localid="1647528969986" $Y_2=1;$

(b) localid="1647528979412" $Y_2=0.$

Short answer

Conditional probability mass function of Y₁ when $Y_2=1$is $\frac{1}{6},\frac{5}{6}$.

Conditional probability mass function of Y₁ whenlocalid="1647529059438" $Y_2=0is\frac{1}{4},\frac{3}{4}.$

Step-by-step solution

Given information (part a)

$Y_2=0$

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

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

Explanation (part a)

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

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

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

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

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

$Y_1/Y_2$	$0$	$1$
$0$	$\left(\frac{10}{13}\right)\left(\frac{9}{12}\right)=\frac{15}{26}$	$\left(\frac{10}{13}\right)\left(\frac{3}{12}\right)=\frac{5}{26}$
1	$\left(\frac{3}{12}\right)\left(\frac{10}{13}\right)=\frac{5}{26}$	$\left(\frac{3}{13}\right)\left(\frac{2}{12}\right)=\frac{1}{26}$

Given information (part b) 

$Y_2=0$

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

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

Explanation (part b)

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

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

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

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