Question

Consider a die with 2 red sides, 2 green sides, and 2 blue sides. Roll the die 5 times, and let X denote the number of times that the die has a red result.

Flip a coin 5 times, and let Y denote the number of times that the coin shows “heads."

a. Find E(X?Y).

b. Find Var?(X?Y).

Answer 1

Here X can take values 0, 1, 2, 3, 4 and 5. And Y can take values 0 ,1 ,2, 3, 4, 5. Since roll is independent from other so X has binomial distribution with parameter n=5 and p=2/ 6 = 1/3. The pdf of X is

$P(X=x)=\binom{5}{x}\left ( \frac{1}{3} \right )^{x}\left ( \frac{2}{3} \right )^{5-x},x=0,1,2,3,4,5$

Following table shows the probabilites of X, rounding to 4 decimal places:

X	P(X=x)
0	0.1317
1	0.3292
2	0.3292
3	0.1646
4	0.0412
5	0.0041

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

Here X has binomial distribution with parameter n=5 and p=0.5. The pdf of Y is

$P(Y=y)=\binom{5}{y}\left ( 0.5 \right )^{y}\left ( 0.5 \right )^{5-y},y=0,1,2,3,4,5$

Following table shows the probabilites of Y, rounding to 4 decimal places:

Y	P(Y=y)
0	0.0313
1	0.1563
2	0.3125
3	0.3125
4	0.1563
5	0.0313

Let W=X-Y. Since X and Y are independent so  

P(W=w)= P(X=x)P(Y=y)

Following table shows the calculations for W:

X	P(X=x)	Y	P(Y=y)	W=X-Y	P(W=w)
0	0.1317	0	0.0313	0	0.00412221
0	0.1317	1	0.1563	-1	0.02058471
0	0.1317	2	0.3125	-2	0.04115625
0	0.1317	3	0.3125	-3	0.04115625
0	0.1317	4	0.1563	-4	0.02058471
0	0.1317	5	0.0313	-5	0.00412221
1	0.3292	0	0.0313	1	0.01030396
1	0.3292	1	0.1563	0	0.05145396
1	0.3292	2	0.3125	-1	0.102875
1	0.3292	3	0.3125	-2	0.102875
1	0.3292	4	0.1563	-3	0.05145396
1	0.3292	5	0.0313	-4	0.01030396
2	0.3292	0	0.0313	2	0.01030396
2	0.3292	1	0.1563	1	0.05145396
2	0.3292	2	0.3125	0	0.102875
2	0.3292	3	0.3125	-1	0.102875
2	0.3292	4	0.1563	-2	0.05145396
2	0.3292	5	0.0313	-3	0.01030396
3	0.1646	0	0.0313	3	0.00515198
3	0.1646	1	0.1563	2	0.02572698
3	0.1646	2	0.3125	1	0.0514375
3	0.1646	3	0.3125	0	0.0514375
3	0.1646	4	0.1563	-1	0.02572698
3	0.1646	5	0.0313	-2	0.00515198
4	0.0412	0	0.0313	4	0.00128956
4	0.0412	1	0.1563	3	0.00643956
4	0.0412	2	0.3125	2	0.012875
4	0.0412	3	0.3125	1	0.012875
4	0.0412	4	0.1563	0	0.00643956
4	0.0412	5	0.0313	-1	0.00128956
5	0.0041	0	0.0313	5	0.00012833
5	0.0041	1	0.1563	4	0.00064083
5	0.0041	2	0.3125	3	0.00128125
5	0.0041	3	0.3125	2	0.00128125
5	0.0041	4	0.1563	1	0.00064083
5	0.0041	5	0.0313	0	0.00012833

Following table shows the pdf of W:

W	P(W=w)
-5	0.00412221
-4	0.03088867
-3	0.10291417
-2	0.20063719
-1	0.25335125
0	0.21645656
1	0.12671125
2	0.05018719
3	0.01287279
4	0.00193039
5	0.00012833

(a-b)

Following table shows the calculations for mean and variance of W:

W	P(W=w)	wP(W=w)	w^2P(W=w)
-5	0.00412221	-0.02061105	0.10305525
-4	0.03088867	-0.12355468	0.49421872
-3	0.10291417	-0.30874251	0.92622753
-2	0.20063719	-0.40127438	0.80254876
-1	0.25335125	-0.25335125	0.25335125
0	0.21645656	0	0
1	0.12671125	0.12671125	0.12671125
2	0.05018719	0.10037438	0.20074876
3	0.01287279	0.03861837	0.11585511
4	0.00193039	0.00772156	0.03088624
5	0.00012833	0.00064165	0.00320825
Total		-0.83346666	3.05681112

So

E(X - Y)E(W)-w P(W-)-0.83346666

$Var(X-Y)=Var(W)=\sum w^{2}P(W=w)-\left [ \sum wP(W=w) \right ]^{2}=2.362144447$