The question: An experiment succeeds with a probability of $Z$, which is a random variable uniformly distributed over $[0,1]$. The experiment is conducted twice. Denote $X$ the random variable which equals 1 if the first experiment succeeded, and 0 otherwise. $Y$ is the same, but for the second experiment. What is $P(X=1)$ and $P(X=1, Y=1)$?
It's quite obvious that $P(X=1) = 1/2$, but I'm not sure how to prove it, and I have not the slightest clue on how to calculate $P(X=1, Y=1)$, since clearly $X,Y$ are dependent in one another.
I tried going at it with conditional probability, but it didn't help me much.
Any clues?
Thanks
Yes, the idea is to condition on Z.
When we condition on Z , it becomes a constant and , from your description is the probability that X is 1 : $P(X=1|Z=z) = z$ . Now we use the definition of conditional probability to compute $P(X=1)$ :
$P(X=1) = \int_{0}^{1}P(X=1|Z=z)P(Z=z)dz$ = 0.5
If we condition on Z , X and Y become independent (informal,if we know Z, observing X don't give any aditional information on Y ).
$P(X=1,Y=1|Z=z)=P(X=1|Z=z)P(Y=1|Z=z) = z^{2}$ .
Integrating over z , you find the final answare.
$P(X=1,Y=1) = \int_{0}^{1}P(X=1|Z=z)P(Y=1|Z=z)P(Z=z)dz$ = 1/3