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)=(5)/(13)\cdot(4)/(12)=(5)/(39)`

`P(X_1=1,X_2=1)=(5)/(39)`

`P(X_1=1,X_2=0)=(5)/(13)\cdot (8)/(12)=(10)/(39)`

`P(X_1=1,X_2=0)=(10)/(39)`

`P(X_1=0,X_2=1)=(8)/(13)* (5)/(12)=(10)/(39)`

`P(X_1=0,X_2=1)=(10)/(39)`

`P(X_1=0,X_2=0)=(8)/(13)* (7)/(12)=(14)/(39)`

`P(X_1=0,X_2=0)=(14)/(39)`

b.
`P(X_1=1,X_2=1,X_3=1)=(5)/(13)* (4)/(12)* (3)/(11)`

`P(X_1=1,X_2=1,X_3=1)=(5)/(143)`

`P(X_1=0,X_2=1,X_3=1)=(8)/(13)* (5)/(12)* (4)/(11)`

`P(X_1=0,X_2=1,X_3=1)=(40)/(429)`

`P(X_1=1,X_2=0,X_3=1)=(5)/(13)* (8)/(12)* (4)/(11)`

`P(X_1=1,X_2=0,X_3=1)=(40)/(429)`

`P(X_1=1,X_2=1,X_3=0)=(5)/(13)* (4)/(12)* (8)/(11)`

`P(X_1=1,X_2=1,X_3=0=(40)/(429)`

`P(X_1=0,X_2=0,X_3=1)=(8)/(13)* (7)/(12)* (5)/(11)`

`P(X_1=0,X_2=0,X_3=1)=(70)/(429)`

`P(X_1=0,X_2=1,X_3=0)=(8)/(13)* (5)/(12)* (7)/(11)`

`P(X_1=0,X_2=1,X_3=0)=(70)/(429)`

`P(X_1=0,X_2=0,X_3=0)=(8)/(13)* (7)/(12)* (6)/(11)`

`P(X_1=0,X_2=0,X_3=0)=(28)/(143)`

`P(X_1=1,X_2=0,X_3=0)=(5)/(13)* (8)/(12)* (7)/(11)`

`P(X_1=1,X_2=0,X_3=0)=(70)/(429)`