Question

In Problem $6.4$, calculate the conditional probability mass function of $X_1$given that

(a) localid="1647593214168" $X_2=1;$

(b) $X_1=0$

Short answer

Conditional probability mass function of X₁ when$X_2=1is\frac{5}{13},\frac{8}{13}$

Conditional probability mass function of X₁ when$X_2=0is\frac{5}{13},\frac{8}{13}$

Step-by-step solution

Given information (part a)

$X_2=1$

Number of white balls$=5$

number of red balls$=8$

Random variable

role="math" localid="1647197044052" $X_i=\left\{\begin{array}{l}1,\text{if}{i}^{\text{th}}\text{ball selected is white}\\ 0,\text{otherwise}\end{array}\right.$

Here,

role="math" localid="1647197104747" $$\begin{gathered} X_2=1means{2}^{nd}balliswhite \\ {X}_1=0means{1}^{st}ballisred \\ {X}_1=1means{1}^{st}balliswhite \end{gathered}$$

Explanation (part a)

Total number of balls$=5+8=13$

localid="1647197917388" $P\left(X_1=1|X_2=1\right)=P\left(X_1=1\right)$

{Selected ball is replaced in the sum in the urn before the next selection}

localid="1647197966613" $$\begin{gathered} =P\left(whiteball\right) \\ =\frac{N_w}{N_b} \\ =\frac{5}{5+8} \\ =\frac{5}{13} \end{gathered}$$

where N_w= Number of white balls

N_b = Number of balls

localid="1647198047328" $$\begin{gathered} P\left(X_1=0|X_2=1\right)=1-P\left(X_1=1|X_2=1\right) \\ =1-\frac{5}{13} \\ =\frac{8}{13} \end{gathered}$$

Table form of the conditional probability mass function of X₁ given that X₂ $=1$is,

$X_1=x|X_2=1$	$0$	$1$
$P\left(X_1=x|X_2=1\right)$	$\frac{8}{13}$	$\frac{5}{13}$

Given information (part b)

Balls are chosen with replacement from consisting of $5$white and $8$red balls.

$X_2=0$

Random variable:

$X_i=\left\{\begin{array}{l}1,\text{if}{i}^{\text{th}}\text{ball selected is white}\\ 0,\text{otherwise}\end{array}\right.$

Here,

$$\begin{gathered} X_2=1means{2}^{nd}balliswhite \\ {X}_1=0means{1}^{st}ballisred \\ {X}_1=1means{1}^{st}balliswhite \end{gathered}$$

Explanation (part b)

Total number of balls $=5+8=13$

$P\left(X_1=1|X_2=0\right)=P\left(X_1=1\right)$

{Selected ball is replaced in the sum in the urn before the next selection}

$$\begin{gathered} =P\left(whiteball\right) \\ =\frac{N_w}{N_b} \\ =\frac{5}{5+8} \\ =\frac{5}{13} \end{gathered}$$

where N_w= Number of white balls

N_b = Number of balls

localid="1647198073976" $$\begin{gathered} P\left(X_1=0|X_2=0\right)=1-P\left(X_1=1|X_2=1\right) \\ =1-\frac{5}{13} \\ =\frac{8}{13} \end{gathered}$$

Table form of the conditional probability mass function of X₁ given that localid="1647527988521" $X_2=0$is,

$X_1=x|X_2=0$	$0$	1
$P\left(X_1=x|X_2=0\right)$	$\frac{8}{13}$	$\frac{5}{13}$