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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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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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

Pure Maths

Mechanics Maths

Applied Mathematics

Logic and Functions

Theoretical and Mathematical Physics

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) $Y_{1}, Y_{2}, Y_{3}$