Question

6 marbles in a bag, 2 are white the rest are red. The marbles are removed from the bag one at a time. Let the random variable Y represent the draw from the bag in which the last white marble has been drawn. Determine the probability distribution of Y.

Answer 1

below is probability distribution of Y :

P(Y=2) =P(First 2 are white) =(2/6)*(1/5)=1/15

P(Y=3) =P(one of 1st 2 are white and 3rd is white) =((₂C₁)*(₄C₁)/(₆C₂))*(1/4)=(2*4/15)*(1/4)=2/15

P(Y=4)=P(one of 1st 3 are white and 4th is white) =((₂C₁)*(₄C₂)/(₆C₃))*(1/3)=(2*6/20)*(1/3)=1/5

P(Y=5)=P(one of 1st 4 are white and 5th is white) =((₂C₁)*(₄C₃)/(₆C₄))*(1/2)=(2*4/15)*(1/2)=4/15

P(Y=6)=P(one of 1st 5 are white and 6th is white) =((₂C₁)*(₄C₄)/(₆C₅))*(1/1)=(2*1/6)*(1/1)=1/3