The General Social Survey asked a sample of adults how many siblings (brothers and sisters) they had (X) and also how many children they had (Y). We show results for those who had no more than 4 children and no more than 4 siblings. Assume that the joint probability mass function is given in the following contingency table: y x 0 1 2 3 4 0 0.03 0.01 0.02 0.01 0.01 1 0.09 0.05 0.08 0.03 0.01 2 0.09 0.05 0.07 0.03 0.02 3 0.06 0.04 0.07 0.04 0.02 4 0.04 0.04 0.04 0.03 0.02
The General Social Survey asked a sample of adults how many siblings (brothers and sisters) they had (X) and also how many children they had (Y). We show results for those who had no more than 4 children and no more than 4 siblings. Assume that the joint
y | |||||
---|---|---|---|---|---|
x | 0 | 1 | 2 | 3 | 4 |
0 | 0.03 | 0.01 | 0.02 | 0.01 | 0.01 |
1 | 0.09 | 0.05 | 0.08 | 0.03 | 0.01 |
2 | 0.09 | 0.05 | 0.07 | 0.03 | 0.02 |
3 | 0.06 | 0.04 | 0.07 | 0.04 | 0.02 |
4 | 0.04 | 0.04 | 0.04 | 0.03 | 0.02 |
Find the conditional probability mass function pY|X(y|4). (Round the final answer to four decimal places.)
pY|X(0|4) =
pY|X(1|4) =
pY|X(2|4) =
pY|X(3|4) =
pY|X(4|4) =
Find the conditional probability mass function pX|Y (x|3). (Round the final answer to four decimal places.)
pX|Y(0|3) =
pX|Y(1|3) =
pX|Y(2|3) =
pX|Y(3|3) =
pX|Y(4|3) =
Find the conditional expectation E(X|Y = 3). (Round the final answer to four decimal places.)
E(X|Y = 3) =
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