Question

4. Suppose urn 11 is filled with 60% green balls and 40% red balls, and urn T is filled with 40% green balls and 60% red balls. Someone will flip a coin and then select a ball from urn H or T depending on whether the coin lands heads or tails, respectively. Let X be 1 or 0 if the coin lands heads or tails, and let Y be 1 or 0 if the ball is green or red (a) Write out the joint distribution of X and Y in a table. b) Find EY]. What is the probability that the ball is green? c) Find VarlYX- and Var YX- and Var Y]. Thinking of variances as measur- ing uncertainty, explain why one of these variances is larger than the others d) Suppose you see that the ball is green. What is the probability that the coin turned up heads?

Answer 1

(a)

When a fair coin has been flipped the possible outcomes are head and tail with equal probability. So we have

P(X=1) = 0.5, P(X=0) = 0.5

When head came, urn H selected so we have

P(Y=1 | X=1) = 0.60, P(Y=0 | X=1) = 0.40

When tail came, urn T selected so we have

P(Y=1 | X=0) = 0.40, P(Y=0 | X=0) = 0.60

Now the joint probabilities are

P(Y=1 and X=1) = P(Y=1 | X=1)P(X=1) = 0.60 *0.50 = 0.30

P(Y=0 and X=1) = P(Y=0 | X=1)P(X=1) = 0.40 *0.50 = 0.20

P(Y=1 and X=0) = P(Y=1 | X=0)P(X=0) = 0.40 *0.50 = 0.20

P(Y=0 and X=0) = P(Y=0 | X=0)P(X=0) = 0.60 *0.50 = 0.30

Following table shows the joint pdf:

		X
		0	1	P(Y=y)
Y	0	0.3	0.2	0.5
	1	0.2	0.3	0.5
	P(X=x)	0.5	0.5	1

(b)

Following table shows the calculations for E(Y):

Y	P(Y=y)	yP(Y=y)
0	0.5	0
1	0.5	0.5
Total		0.5

So,

The probability that ball is green is

P(Y=1) = 0.5

(c)

Following table shows the calculations for Var(Y):

So,

---------------

Following table shows the conditional pdf of Y given X=0 and calculations for Var(Y|X=0):

So,

Following table shows the conditional pdf of Y given X=1 and calculations for Var(Y|X=1):

So,

(d)