Solve the given problem involving conditional probability

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The discussion revolves around solving a conditional probability problem involving two boxes with white and black balls. The probabilities for drawing a white ball from box Y depend on the number of balls in box X, with calculations showing that for n=4, the required probability is 13/20, while for n=5, it is 17/30. Participants emphasize the importance of considering both scenarios: drawing a black or a white ball from box X, and using a probability tree to visualize outcomes. The final expressions for the probabilities are derived through combining the outcomes from both scenarios, leading to a clearer understanding of the problem. The conversation highlights the complexity of conditional probability and the necessity of careful calculations.
  • #31
Ehm, not so cheerful here, check your post #18 again, P(W) is the addition of those two.
 
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  • #32
Delta2 said:
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.
 
  • #33
yes ok right, it is ##4n-3## there i thought you had ##4(n-3)##.
 

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