Question

Let X, Y ~ 10,11 independently. Find P(max(X, Y} > 0.8 1 min(X, Y} = 0.5)

Answer 1

Given that min value out of X, Y is 0.5, probability that the maximum value would be greater than 0.8 is computed here as:

$P(max(X, Y) \geq 0.8 | min(x, Y) = 0.5)$

min(X, Y) = 0.5 if at least one of them is 0.5 and the other is greater than or equal to 0.5

Therefore the required probability here is computed as:

$= P(Y \geq 0.8 | X = 0.5, Y \geq 0.5) + P(X \geq 0.8 | Y = 0.5, X \geq 0.5)$

$= \frac{1 - 0.8}{1 - 0.5} + \frac{1 - 0.8}{1 - 0.5} = 0.8$

Therefore 0.8 is the required probability here.