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)????

Printable View

- Dec 9th 2009, 04:38 AMkinConditional 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, 04:47 AMmr fantastic
- Dec 9th 2009, 05:47 AMkin
(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??