P(H)=3/4=0.75
if head appears then X=1 and urn 1 is chosen
P(label 1 from urn 1)=0.30
P(label 2 from urn 1)=0.40
P(label 3 from urn )=0.30
and P(label 1 from urn 1 and head)=0.75*0.30=0.225
similarly.....we will do other calculations
-----------------------------
joint distribution table obtained is
|
|
X=0 |
X=1 |
|
| Y=1 |
0.05 |
0.225 |
0.275 |
| Y=2 |
0.125 |
0.3 |
0.425 |
| Y=3 |
0.075 |
0.225 |
0.3 |
|
total |
0.25 |
0.75 |
1 |
|
X |
0 |
1 |
total |
|
P(X=x) |
0.25 |
0.75 |
1 |
|
|
|
|
|
E(x)=X*P(X) |
0 |
0.75 |
0.75 |
E(X)=0.75
----------------------------------------
|
Y |
P(Y) |
E(Y)=Y*P(Y) |
| 1 |
0.275 |
0.275 |
| 2 |
0.425 |
0.85 |
| 3 |
0.3 |
0.9 |
|
total |
1 |
2.025 |
E(Y)=2.025
---------------------------------------
Y|X
|
|
X=0 |
X=1 |
P(Y|X=0) |
P(Y|X=1) |
|
Y*P(Y|X=0) |
Y*P(Y|X=1) |
| Y=1 |
0.05 |
0.225 |
0.2 |
0.3 |
|
0 |
0 |
| Y=2 |
0.125 |
0.3 |
0.5 |
0.4 |
|
0.5 |
0.4 |
| Y=3 |
0.075 |
0.225 |
0.3 |
0.3 |
|
0.6 |
0.6 |
|
total |
0.25 |
0.75 |
1 |
1 |
|
1.1 |
1 |
so, E(Y | X)=?Y*P(Y|X=0) +? Y*P(Y|X=1) = 1.1+1=2.1
---------------------------------------------------
X|Y
|
|
X=0 |
X=1 |
|
| Y=1 |
0.05 |
0.225 |
0.275 |
| Y=2 |
0.125 |
0.3 |
0.425 |
| Y=3 |
0.075 |
0.225 |
0.3 |
|
|
|
|
|
|
|
|
|
P(X|Y=1) |
0.1818182 |
0.818182 |
1 |
|
P(X|Y=2) |
0.2941176 |
0.705882 |
1 |
|
P(X|Y=3) |
0.25 |
0.75 |
1 |
|
|
|
|
|
X*P(X|Y=1) |
0 |
0.818182 |
0.818182 |
|
X*P(X|Y=2) |
0 |
0.705882 |
0.705882 |
|
X*P(X|Y=3) |
0 |
0.75 |
1.524064 |
E(X | Y) = ?X*P(X|Y=1)+?X*P(X|Y=2)+?X*P(X|Y=3) =
0.818181818+0.705882353+1.524064171 =
3.0481
