The solution is correct? If I launch 4 dice and at least one even number appears, what is the probability that the sum of the results be even? A: "The sum of the results is even"
B: "At least one even number appears"

A∩B = A

f(x) = (4Cx)*[(1/2)^x]*[(1/2)^(n-x)]

P(A) = f(2)+f(4) = [(4C2)*[(1/2)^4]]+[(1/2)^4]
P(B) = 1-f(0) = 1-[(1/2)^4]

P(A/B) = P(A)/P(B) = (7/16)/(15/16) = 7/15

The solution is correct?
A: You are correct, though I would do the calculation slightly differently: each die can be odd or even with probability $\frac12$, so there are $16$ equally probable results, $8$ of which have even sums.
We should exclude the case where all four dice are odd: odd + odd + odd + odd = even, so the conditional probability is $\dfrac{8-1}{16-1}=\dfrac{7}{15}$ 
A: Your answer is correct, but your work has some errors, as listed below (which work out so as to cancel each other).


*

*$A \cap B$ is not equal to $A$, since $B$ is not a subset of $A$.

*$P(A)$ is equal to $f(0) + f(2) + f(4)$, not $f(2) + f(4)$.

*The formula for $P(A\mid B)$ should be $\displaystyle{\frac{P(A \cap B)}{P(B)}}$, not $\displaystyle{\frac{P(A)}{P(B)}}$.


Here is a corrected version, using your approach ...

$P(A \cap B) = f(2) + f(4) = 7/16$.

$P(B) = 1 - f(0) = 15/16$.

$P(A\mid B) = \displaystyle{\frac{P(A \cap B)}{P(B)}} = \frac{(7/16)}{(15/16)} = 7/15$.
