Question (6): If two balls are chosen with replacement from a box containing two
ID: 3355873 • Letter: Q
Question
Question (6): If two balls are chosen with replacement from a box containing two red, two white and two black balls. If we let X and Y denote, respectively, the number of red and white that are chosen, then determine the joint probability distribution of X and Y.
If two balls are chosen with replacement from a box containing two red, two white and two black balls. If we let X and Y denote, respectively, the number of red and white that are chosen, then determine the joint probability distribution of X and Y.
Explanation / Answer
from above probability of choosing 0 red and 0 white balls =P(X=0,Y=0)=P( both balls are black)
=(2/6)*(2/6)=1/9
probability of choosing 1 red and 0 white balls=P(X=1,Y=0) =P(first red and second black+first black and second black)
=(2/6)*(2/6)+(2/6)*(2/6) =2/9
probability of choosing 0 red and 1 white balls=P(X=0,Y=1)=P(first white and second black+first black and second white)
=(2/6)*(2/6)+(2/6)*(2/6) =2/9
probability of 1red and 1 white balls =P(X=1,Y=1)=P(first white and second red+first ed and second white)
=(2/6)*(2/6)+(2/6)*(2/6) =2/9
probability of both red balls=P(X=2,Y=0)=(2/6)*(2/6) =1/9
probability of both white balls=P(X=0,Y=2)=(2/6)*(2/6) =1/9
hence below is joint probability distribution table of X and Y:
please revert for any clarification.
Y X 0 1 2 0 1/9 2/9 1/9 1 2/9 2/9 0 2 1/9 0 0