Q. 6.3 In Problem 6.2, suppose that the... [FREE SOLUTION]

91影视

A First Course in Probability

Sheldon M. Ross

$Math Studyset 91影视 Explanations$ Math

9th Edition 路 432 pages

ISBN: 9780321794772

Chapter 6: Q. 6.3 (page 271)

In Problem $6.2$ , suppose that the white balls are numbered, and let $Y_{i}$ equal $1$ if the $i$ th white ball is selected and $0$ otherwise. Find the joint probability mass function of
(a) $Y_{1}, Y_{2}$
(b) $Y_{1}, Y_{2}, Y_{3}$

Short Answer

Expert verified

a. Joint probability mass function of $Y_{1}$ , $Y_{2}$ is $\frac{1}{26}$ , $\frac{5}{26}$ , $\frac{5}{26}$ , $\frac{15}{26}$ .

b. Joint probability mass function of $Y_{1}$ , $Y_{2}$ , $Y_{3}$ is $\frac{1}{286}$ , $\frac{5}{143}$ , $\frac{5}{143}$ , $\frac{5}{143}$ , $\frac{45}{286}$ , $\frac{45}{286}$ , $\frac{45}{286}$ , $\frac{60}{143}$ .

Step by step solution

Calculation for joint probability mass function (part a)

There are $(\begin{matrix} 13 \\ 3 \end{matrix})$ ways to select three balls from a total of thirteen.

There are $(\begin{matrix} 11 \\ 1 \end{matrix})$ ways to do this if the first two white balls have been chosen.

Hence

$P (Y_{1} = 1, Y_{2} = 1) = \frac{(\begin{matrix} 11 \\ 1 \end{matrix})}{(\begin{matrix} 13 \\ 3 \end{matrix})}$

$= \frac{1}{26}$

To get that, use the same strategy.

$P (Y_{1} = 1, Y_{2} = 0) = \frac{(\begin{matrix} 11 \\ 2 \end{matrix})}{(\begin{matrix} 13 \\ 3 \end{matrix})}$

$= \frac{5}{26}$

$P (Y_{1} = 0, Y_{2} = 1) = \frac{(\begin{matrix} 11 \\ 2 \end{matrix})}{(\begin{matrix} 13 \\ 3 \end{matrix})}$

$= \frac{5}{26}$

$P (Y_{1} = 0, Y_{2} = 0) = \frac{(\begin{matrix} 11 \\ 2 \end{matrix})}{(\begin{matrix} 13 \\ 3 \end{matrix})}$

$= \frac{15}{26}$

Calculation for joint probability mass function (part b)

Use the same concept as in part one (a). Consider all potential scenarios and utilize a combinatoric argument to arrive at that conclusion.

$P (Y_{1} = 1, Y_{2} = 1, Y_{3} = 1) = \frac{(\begin{matrix} 10 \\ 0 \end{matrix})}{(\begin{matrix} 13 \\ 3 \end{matrix})}$

$= \frac{1}{286}$

$P (Y_{1} = 0, Y_{2} = 1, Y_{3} = 1) = \frac{(\begin{matrix} 10 \\ 1 \end{matrix})}{(\begin{matrix} 13 \\ 3 \end{matrix})}$

$= \frac{5}{143}$

$P (Y_{1} = 1, Y_{2} = 0, Y_{3} = 1) = \frac{(\begin{matrix} 10 \\ 1 \end{matrix})}{(\begin{matrix} 13 \\ 3 \end{matrix})}$

$= \frac{5}{143}$

$P (Y_{1} = 1, Y_{2} = 1, Y_{3} = 0) = \frac{(\begin{matrix} 10 \\ 1 \end{matrix})}{(\begin{matrix} 13 \\ 3 \end{matrix})}$

$= \frac{5}{143}$

$P (Y_{1} = 0, Y_{2} = 0, Y_{3} = 1) = \frac{(\begin{matrix} 10 \\ 2 \end{matrix})}{(\begin{matrix} 13 \\ 3 \end{matrix})}$

$= \frac{45}{286}$

$P (Y_{1} = 0, Y_{2} = 1, Y_{3} = 0) = \frac{(\begin{matrix} 10 \\ 2 \end{matrix})}{(\begin{matrix} 13 \\ 3 \end{matrix})}$

$= \frac{45}{286}$

$P (Y_{1} = 1, Y_{2} = 0, Y_{3} = 0) = \frac{(\begin{matrix} 10 \\ 2 \end{matrix})}{(\begin{matrix} 13 \\ 3 \end{matrix})}$

$= \frac{45}{286}$

$P (Y_{1} = 0, Y_{2} = 0, Y_{3} = 0) = \frac{(\begin{matrix} 10 \\ 3 \end{matrix})}{(\begin{matrix} 13 \\ 3 \end{matrix})}$

$= \frac{60}{143}$

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91影视

In Problem $6.2$ , suppose that the white balls are numbered, and let $Y_{i}$ equal $1$ if the $i$ th white ball is selected and $0$ otherwise. Find the joint probability mass function of
(a) $Y_{1}, Y_{2}$
(b) $Y_{1}, Y_{2}, Y_{3}$

Short Answer

Step by step solution

Calculation for joint probability mass function (part a)

Calculation for joint probability mass function (part b)

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91影视

In Problem 6.2, suppose that the white balls are numbered, and let Yiequal 1if the ith white ball is selected and0 otherwise. Find the joint probability mass function of(a)Y1,Y2(b)Y_{1},Y_{2},Y_{3}

Short Answer

Step by step solution

Calculation for joint probability mass function (part a)

Calculation for joint probability mass function (part b)

One App. One Place for Learning.

Most popular questions from this chapter

Recommended explanations on Math Textbooks

Mechanics Maths

Statistics

Applied Mathematics

Pure Maths

Geometry

Decision Maths

Study anywhere. Anytime. Across all devices.

In Problem $6.2$ , suppose that the white balls are numbered, and let $Y_{i}$ equal $1$ if the $i$ th white ball is selected and $0$ otherwise. Find the joint probability mass function of
(a) $Y_{1}, Y_{2}$
(b) $Y_{1}, Y_{2}, Y_{3}$