Q. 6.3 - A First Course in Probability

Question

In Problem $6.2$, suppose that the white balls are numbered, and let ${\mathrm{Y}}_{\mathrm{i}}$equal $1$ if the $\mathrm{i}$th white ball is selected and$0$ otherwise. Find the joint probability mass function of

(a)${\mathrm{Y}}_1,{\mathrm{Y}}_2$

(b) $\mathrm{Y\_}\left\{1\right\}, \mathrm{Y\_}\left\{2\right\}, \mathrm{Y\_}\left\{3\right\}$

Step-by-Step Solution

Answer

a. Joint probability mass function of ${\mathrm{Y}}_1$,${\mathrm{Y}}_2$is$\frac{1}{26}$,$\frac{5}{26}$,$\frac{5}{26}$,$\frac{15}{26}$.

b. Joint probability mass function of $\mathrm{Y\_}\left\{1\right\}$,$\mathrm{Y\_}\left\{2\right\}$,$\mathrm{Y\_}\left\{3\right\}$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 1: Calculation for joint probability mass function (part a)

a.

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

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

Hence

$\mathrm{P}\left({\mathrm{Y}}_1=1,{\mathrm{Y}}_2=1\right)=\frac{\left(\begin{array}{c}11\\ 1\end{array}\right)}{\left(\begin{array}{c}13\\ 3\end{array}\right)}$

$=\frac{1}{26}$

To get that, use the same strategy.

$\mathrm{P}\left({\mathrm{Y}}_1=1,{\mathrm{Y}}_2=0\right)=\frac{\left(\begin{array}{c}11\\ 2\end{array}\right)}{\left(\begin{array}{c}13\\ 3\end{array}\right)}$

$=\frac{5}{26}$

$\mathrm{P}\left({\mathrm{Y}}_1=0,{\mathrm{Y}}_2=1\right)=\frac{\left(\begin{array}{c}11\\ 2\end{array}\right)}{\left(\begin{array}{c}13\\ 3\end{array}\right)}$

$=\frac{5}{26}$

$\mathrm{P}\left({\mathrm{Y}}_1=0,{\mathrm{Y}}_2=0\right)=\frac{\left(\begin{array}{c}11\\ 2\end{array}\right)}{\left(\begin{array}{c}13\\ 3\end{array}\right)}$

$=\frac{15}{26}$

Step 2: Calculation for joint probability mass function (part b)

b.

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

$\mathrm{P}\left({\mathrm{Y}}_1=1,{\mathrm{Y}}_2=1,{\mathrm{Y}}_3=1\right)=\frac{\left(\begin{array}{c}10\\ 0\end{array}\right)}{\left(\begin{array}{c}13\\ 3\end{array}\right)}$

$=\frac{1}{286}$

$\mathrm{P}\left({\mathrm{Y}}_1=0,{\mathrm{Y}}_2=1,{\mathrm{Y}}_3=1\right)=\frac{\left(\begin{array}{c}10\\ 1\end{array}\right)}{\left(\begin{array}{c}13\\ 3\end{array}\right)}$

$=\frac{5}{143}$

$\mathrm{P}\left({\mathrm{Y}}_1=1,{\mathrm{Y}}_2=0,{\mathrm{Y}}_3=1\right)=\frac{\left(\begin{array}{c}10\\ 1\end{array}\right)}{\left(\begin{array}{c}13\\ 3\end{array}\right)}$

$=\frac{5}{143}$

$\mathrm{P}\left({\mathrm{Y}}_1=1,{\mathrm{Y}}_2=1,{\mathrm{Y}}_3=0\right)=\frac{\left(\begin{array}{c}10\\ 1\end{array}\right)}{\left(\begin{array}{c}13\\ 3\end{array}\right)}$

$=\frac{5}{143}$

$\mathrm{P}\left({\mathrm{Y}}_1=0,{\mathrm{Y}}_2=0,{\mathrm{Y}}_3=1\right)=\frac{\left(\begin{array}{c}10\\ 2\end{array}\right)}{\left(\begin{array}{c}13\\ 3\end{array}\right)}$

$=\frac{45}{286}$

$\mathrm{P}\left({\mathrm{Y}}_1=0,{\mathrm{Y}}_2=1,{\mathrm{Y}}_3=0\right)=\frac{\left(\begin{array}{c}10\\ 2\end{array}\right)}{\left(\begin{array}{c}13\\ 3\end{array}\right)}$

$=\frac{45}{286}$

$\mathrm{P}\left({\mathrm{Y}}_1=1,{\mathrm{Y}}_2=0,{\mathrm{Y}}_3=0\right)=\frac{\left(\begin{array}{c}10\\ 2\end{array}\right)}{\left(\begin{array}{c}13\\ 3\end{array}\right)}$

$=\frac{45}{286}$

$\mathrm{P}\left({\mathrm{Y}}_1=0,{\mathrm{Y}}_2=0,{\mathrm{Y}}_3=0\right)=\frac{\left(\begin{array}{c}10\\ 3\end{array}\right)}{\left(\begin{array}{c}13\\ 3\end{array}\right)}$

$=\frac{60}{143}$