Chapter 6: Q. 6.36 (page 273)

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

Expert verified

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" $Y_{2} = 0 i s \frac{1}{4}, \frac{3}{4} .$

Step by step solution

Given information (part a)

$Y_{2} = 0$

Probability mass function $Y_{1} = 1$ given that $Y_{2} = 1$

localid="1647199633826" $P (Y_{1} = 1 ∣ Y_{2} = 1) = \frac{P (Y_{1} = 1, Y_{2} = 1)}{P (Y_{2} = 1)}$

Explanation (part a)

Probability mass function $Y_{1} = 1$ given that $Y_{2} = 1$

$P (Y_{1} = 1 ∣ Y_{2} = 1) = \frac{P (1, 1)}{P (0, 1) + P (1, 1)} P (Y_{1} = 1 ∣ Y_{2} = 1) = \frac{\frac{1}{26}}{\frac{5}{26} + \frac{1}{26}} P (Y_{1} = 1 ∣ Y_{2} = 1) = \frac{1}{6}$

Probability mass function $Y_{1} = 0$ given that $Y_{2} = 1$

$P (Y_{1} = 0 ∣ Y_{2} = 1) = \frac{P (0, 1)}{P (0, 1) + P (1, 1)} P (Y_{1} = 0 ∣ Y_{2} = 1) = \frac{\frac{5}{26}}{\frac{5}{26} + \frac{1}{26}} P (Y_{1} = 0 ∣ Y_{2} = 1) = \frac{5}{6}$

Table form of the conditional probability mass function of Y₁ given that $Y_{2} = 1$

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

Given information (part b)

$Y_{2} = 0$

Probability mass function $Y_{1} = 1$ given that $Y_{2} = 0$

$P (Y_{1} = 1 ∣ Y_{2} = 0) = \frac{P (Y_{1} = 1, Y_{2} = 0)}{P (Y_{2} = 0)}$

Explanation (part b)

Probability mass function $Y_{1} = 1$ given that $Y_{2} = 0$

$P (Y_{1} = 1 ∣ Y_{2} = 0) = \frac{P (1, 0)}{P (1, 0) + P (0, 0)} P (Y_{1} = 1 ∣ Y_{2} = 0) = \frac{\frac{5}{26}}{\frac{5}{26} + \frac{15}{26}} P (Y_{1} = 1 ∣ Y_{2} = 0) = \frac{5}{20} P (Y_{1} = 1 ∣ Y_{2} = 0) = \frac{1}{4}$

Probability mass function $Y_{1} = 0$ given that $Y_{2} = 0$

$P (Y_{1} = 0 ∣ Y_{2} = 0) = \frac{P (0, 0)}{P (1, 0) + P (0, 0)} P (Y_{1} = 0 ∣ Y_{2} = 0) = \frac{\frac{15}{26}}{\frac{5}{26} + \frac{15}{26}} P (Y_{1} = 0 ∣ Y_{2} = 0) = \frac{15}{20} P (Y_{1} = 0 ∣ Y_{2} = 0) = \frac{3}{4}$

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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

Step by step solution

Given information (part a)

Explanation (part a)

Given information (part b)

Explanation (part b)

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In Problem 6.3, calculate the conditional probability mass function of Y1given that(a) localid="1647528969986" Y2=1;(b) localid="1647528979412" Y2=0.

Short Answer

Step by step solution

Given information (part a)

Explanation (part a)

Given information (part b)

Explanation (part b)

One App. One Place for Learning.

Most popular questions from this chapter

Recommended explanations on Math Textbooks

Applied Mathematics

Geometry

Statistics

Calculus

Theoretical and Mathematical Physics

Decision Maths

Study anywhere. Anytime. Across all devices.

In Problem $6.3$ , calculate the conditional probability mass function of $Y_{1}$ given that
(a) localid="1647528969986" $Y_{2} = 1;$
(b) localid="1647528979412" $Y_{2} = 0 .$