Q. 6.35 - A First Course in Probability

Question

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

(a) $X_2=1;$ 

(b) $X_1=0$ 

Step-by-Step Solution

Answer

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

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

Step 1: Given information (part a)

$X_2=1$

Number of white balls$=5$

number of red balls$=8$

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,

$$\begin{gathered} X_2=1 means 2^{nd} ball is white \\ {X}_1=0 means 1^{st} ball is red \\ {X}_1=1 means 1^{st} ball is white \end{gathered}$$

Step 2: Explanation (part a)

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

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

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

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

where N_w = Number of white balls

N_b = Number of balls

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

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

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

Step 1: 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,

$$\begin{gathered} X_2=1 means 2^{nd} ball is white \\ {X}_1=0 means 1^{st} ball is red \\ {X}_1=1 means 1^{st} ball is white \end{gathered}$$

Step 2: 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}

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

where N_w = Number of white balls

N_b = Number of balls

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

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

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