Question

1. A box contains 3 white and 2 black balls. The white balls are labelled by 1, 2, and 3, and the black balls by 4 and 5. A ball is randomly picked from the box. Let ? be the number shown on the picked ball, and ? = 1 if the picked ball is black; ? = 0 otherwise. Find

a. ?(? = 1);

b. ?(? = 4, ? = 1);

c. ?(??);

d. ???(?|? = 4).

Answer 1

The Sample space of this experiment is {1,2,3,4,5}

The possible outcomes and the corresponding values of X and Y are given below.

outcome X Y Probability

1 1 0 1/5

2 2 0 1/5

3 3 0 1/5

4 4 1 1/5

5 5 1 1/5

(a) P(Y = 1) = P(outcome = 4 or 5) = 2/5

(of 5 equally likely out comes 2 are favourable to this event)

(b) P(X=4,Y=1) = P(outcome = 4) = 1/5

(c) E(XY) = (1)(0)(1/5) +(2)(0)(1/5)+(3)(0)(1/5)+(4)(1)(1/5)+5(1)(1/5) =0+0+0+4/5+5/5=9/5 = 1.8

(d) Conditional distribution Y|X=4

When X =4, Y=1 with probability 1.

E(Y|X=4) = (1)(1) =1

E(Y2|X=4) = (12)(1) =(1)(1)=1

Var(Y|X=4) = E(Y2|X=4) -(E(Y|X=4)) 2 = 1-12 = 1-1=0

Also when X=4, Y is constant 1, so Var(Y|X=4).