Question

Suppose that 3 balls are chosen without replacement from an urn consisting of 5 white and 8 red balls. Let Xi equal 1 if the ith ball selected is white, and let it equal 0 otherwise. Give the joint probability mass function of (a) X1,X2; (b) X1,X2,X3.

Answer 1

Answer with Step-by-step explanation:

We are given that

Total number of chosen balls =3

Total number of white balls=5

Total number of red balls=8

$X_i=1$ for ith selected

white ball

$X_i=0$ for ith red ball

a.$P(X_1=1,X_2=1)=\frac{5}{13}\cdot\frac{4}{12}=\frac{5}{39}$

$P(X_1=1,X_2=1)=\frac{5}{39}$

$P(X_1=1,X_2=0)=\frac{5}{13}\cdot \frac{8}{12}=\frac{10}{39}$

$P(X_1=1,X_2=0)=\frac{10}{39}$

$P(X_1=0,X_2=1)=\frac{8}{13}\times \frac{5}{12}=\frac{10}{39}$

$P(X_1=0,X_2=1)=\frac{10}{39}$

$P(X_1=0,X_2=0)=\frac{8}{13}\times \frac{7}{12}=\frac{14}{39}$

$P(X_1=0,X_2=0)=\frac{14}{39}$

b.$P(X_1=1,X_2=1,X_3=1)=\frac{5}{13}\times \frac{4}{12}\times \frac{3}{11}$

$P(X_1=1,X_2=1,X_3=1)=\frac{5}{143}$

$P(X_1=0,X_2=1,X_3=1)=\frac{8}{13}\times \frac{5}{12}\times \frac{4}{11}$

$P(X_1=0,X_2=1,X_3=1)=\frac{40}{429}$

$P(X_1=1,X_2=0,X_3=1)=\frac{5}{13}\times \frac{8}{12}\times \frac{4}{11}$

$P(X_1=1,X_2=0,X_3=1)=\frac{40}{429}$

$P(X_1=1,X_2=1,X_3=0)=\frac{5}{13}\times \frac{4}{12}\times \frac{8}{11}$

$P(X_1=1,X_2=1,X_3=0=\frac{40}{429}$

$P(X_1=0,X_2=0,X_3=1)=\frac{8}{13}\times \frac{7}{12}\times \frac{5}{11}$

$P(X_1=0,X_2=0,X_3=1)=\frac{70}{429}$

$P(X_1=0,X_2=1,X_3=0)=\frac{8}{13}\times \frac{5}{12}\times \frac{7}{11}$

$P(X_1=0,X_2=1,X_3=0)=\frac{70}{429}$

$P(X_1=0,X_2=0,X_3=0)=\frac{8}{13}\times \frac{7}{12}\times \frac{6}{11}$

$P(X_1=0,X_2=0,X_3=0)=\frac{28}{143}$

$P(X_1=1,X_2=0,X_3=0)=\frac{5}{13}\times \frac{8}{12}\times \frac{7}{11}$

$P(X_1=1,X_2=0,X_3=0)=\frac{70}{429}$