Q. 6.2 - A First Course in Probability

Question

Suppose that $3$ balls are chosen without replacement from an urn consisting of $5$white and$8$ red balls. Let ${\mathrm{X}}_i$ equal $1$ if the $\mathrm{i}$th ball selected is white, and let it equal $0$otherwise. Give the joint probability mass function of

(a) ${\mathrm{X}}_1,{\mathrm{X}}_2$;

(b)${\mathrm{X}}_1,{\mathrm{X}}_2,{\mathrm{X}}_3$ .

Step-by-Step Solution

Answer

a. Probability mass function of ${\mathrm{X}}_1$,${\mathrm{X}}_2$ is $\frac{5}{39}$,$\frac{10}{39}$,$\frac{10}{39}$,$\frac{14}{39}$.

b. Probability mass function of ${\mathrm{X}}_1$,${\mathrm{X}}_2$,${\mathrm{X}}_3$is $\frac{5}{143}$,$\frac{40}{429}$,$\frac{40}{429}$,$\frac{40}{429}$,$\frac{70}{429}$,$\frac{70}{429}$,$\frac{70}{429}$,$\frac{28}{143}$.

Step 1: Calculation for joint probability mass function (part a)

a.

The first ball selected is white, with a probability of$\frac{5}{13}$.

The second picked ball is similarly white, so we now have four white balls in the urn, for a total of twelve balls.

The likelihood is that$\frac{4}{12}$.

Hence

$\mathrm{P}\left({\mathrm{X}}_1=1,{\mathrm{X}}_2=1\right)=\frac{5}{13}\times \frac{4}{12}$

$=\frac{5}{39}$

To get that, use the same strategy.

$\mathrm{P}\left({\mathrm{X}}_1=1,{\mathrm{X}}_2=0\right)=\frac{5}{13}\times \frac{8}{12}$

$=\frac{10}{39}$

$\mathrm{P}\left({\mathrm{X}}_1=0,{\mathrm{X}}_2=1\right)=\frac{8}{13}\times \frac{5}{12}$

$=\frac{10}{39}$

$\mathrm{P}\left({\mathrm{X}}_1=0,{\mathrm{X}}_2=0\right)=\frac{8}{13}·\frac{7}{12}$

$=\frac{14}{39}$

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

b.

From part a,

$\mathrm{P}\left({\mathrm{X}}_1=1,{\mathrm{X}}_2=1,{\mathrm{X}}_3=1\right)=\frac{5}{13}\times \frac{4}{12}\times \frac{3}{11}$

$=\frac{5}{143}$

$\mathrm{P}\left({\mathrm{X}}_1=0,{\mathrm{X}}_2=1,{\mathrm{X}}_3=1\right)=\frac{8}{13}\times \frac{5}{12}\times \frac{4}{11}$

$=\frac{40}{429}$

$\mathrm{P}\left({\mathrm{X}}_1=1,{\mathrm{X}}_2=0,{\mathrm{X}}_3=1\right)=\frac{5}{13}\times \frac{8}{12}\times \frac{4}{11}$

$=\frac{40}{429}$

$\mathrm{P}\left({\mathrm{X}}_1=1,{\mathrm{X}}_2=1,{\mathrm{X}}_3=0\right)=\frac{5}{13}\times \frac{4}{12}\times \frac{8}{11}$

$=\frac{40}{429}$

$\mathrm{P}\left({\mathrm{X}}_1=0,{\mathrm{X}}_2=0,{\mathrm{X}}_3=1\right)=\frac{8}{13}\times \frac{7}{12}\times \frac{5}{11}$

$=\frac{70}{429}$

$\mathrm{P}\left({\mathrm{X}}_1=0,{\mathrm{X}}_2=1,{\mathrm{X}}_3=0\right)=\frac{8}{13}\times \frac{7}{12}\times \frac{5}{11}$

$=\frac{70}{429}$

$\mathrm{P}\left({\mathrm{X}}_1=1,{\mathrm{X}}_2=0,{\mathrm{X}}_3=0\right)=\frac{8}{13}\times \frac{7}{12}\times \frac{5}{11}$

$=\frac{70}{429}$

$\mathrm{P}\left({\mathrm{X}}_1=0,{\mathrm{X}}_2=0,{\mathrm{X}}_3=0\right)=\frac{8}{13}\times \frac{7}{12}\times \frac{6}{11}$

$=\frac{28}{143}$