Solve the given problem involving conditional probability

[Delta2]

Ehm, not so cheerful here, check your post #18 again, P(W) is the addition of those two.

 

[chwala]

Ehm, not so cheerful here, check your post #18 again, P(W) is the addition of those two.

Yes, i am aware that $P(W)= \dfrac{4n-3}{n(n+1)}$ is the addition given by

$P(X=B,Y=W) = \dfrac{3}{n} × \dfrac{3}{n+1}= \dfrac{9}{n(n+1)}$

and$P(X=W,Y=W) = \dfrac{n-3}{n} × \dfrac{4}{n+1}= \dfrac{4(n-3)}{n(n+1)}$

My post $30$ is correct.

 

[Delta2]

yes ok right, it is $4n-3$ there i thought you had $4(n-3)$.