91Ó°ÊÓ

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

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

$=\frac{5}{39}$

To get that, use the same strategy.

localid="1649431448658" $\mathrm{P}\left({\mathrm{X}}_1=1,{\mathrm{X}}_2=0\right)=\frac{5}{13}×\frac{8}{12}$

$=\frac{10}{39}$

localid="1649431473375" $\mathrm{P}\left({\mathrm{X}}_1=0,{\mathrm{X}}_2=1\right)=\frac{8}{13}×\frac{5}{12}$

$=\frac{10}{39}$

localid="1649431502228" $\mathrm{P}\left({\mathrm{X}}_1=0,{\mathrm{X}}_2=0\right)=\frac{8}{13}·\frac{7}{12}$

$=\frac{14}{39}$

b.

From part a,

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

$=\frac{5}{143}$

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

$=\frac{40}{429}$

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

localid="1649431919032" $=\frac{40}{429}$

localid="1649431941084" $\mathrm{P}\left({\mathrm{X}}_1=1,{\mathrm{X}}_2=1,{\mathrm{X}}_3=0\right)=\frac{5}{13}×\frac{4}{12}×\frac{8}{11}$

localid="1649432000877" $=\frac{40}{429}$

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

role="math" $=\frac{70}{429}$

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

role="math" localid="1649432156487" $=\frac{70}{429}$

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

$=\frac{70}{429}$

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

$=\frac{28}{143}$