# Conditional probability and Expectation

• Dec 9th 2009, 05:38 AM
kin
Conditional probability and Expectation
http://us.f6.yahoofs.com/hkblog/_9eQ..._____Dtx9HZmOf
(ii) i have no ideas.......
(iii) P(X=3|Y=6) = P(X=3, Y=6) / P(Y=6)
then ,how to find P(X=3, Y=6)????
• Dec 9th 2009, 05:47 AM
mr fantastic
Quote:

Originally Posted by kin
http://us.f6.yahoofs.com/hkblog/_9eQ..._____Dtx9HZmOf
(ii) i have no ideas.......
(iii) P(X=3|Y=6) = P(X=3, Y=6) / P(Y=6)
then ,how to find P(X=3, Y=6)????

Y ~ Binomial(n = 10, p = 2/5) and X ~ Binomial(n = 5, p = 2/5).

(ii) Note that $Var(Y) = E(Y^2) - (E(Y))^2 \Rightarrow np(1 - p) = E(Y^2) - (np)^2$.

(iii) Try having another think now.
• Dec 9th 2009, 06:47 AM
kin
Quote:

Originally Posted by mr fantastic
Y ~ Binomial(n = 10, p = 2/5) and X ~ Binomial(n = 5, p = 2/5).

(ii) Note that $Var(Y) = E(Y^2) - (E(Y))^2 \Rightarrow np(1 - p) = E(Y^2) - (np)^2$.

(iii) Try having another think now.

(ii) thanks
(iii)
P(X=3,X=6)
= P(Y=6| X=3) * P(X=3)
={ [((3/5)^2) ((2/5)^3) (5C3)]^2 } *[((3/5)^2) ((2/5)^3) (5C3)]
=[((3/5)^2) ((2/5)^3) (5C3)]^3

so,
P(X=3| Y=6)
= P(X=3,X=6)/ P(Y=6)
={[((3/5)^2) ((2/5)^3) (5C3)]^3}/ [((3/5)^4) ((2/5)^6) (10C6)]
=((3/5)^2) ((2/5)^3)((5C3)^3)/ (10C6)
=96/875
AM I RIGHT??