Probability of getting a working product from three groups of products... The first group of products has one third of defect products and two thirds of working products. The rest two groups have all working products. Find the probability that the random product is working and from the first group.
edit: What I've done:
A - a product is taken from the first group, $p(A) = 0.3333 = 33.33\%$
B - working product form the first group, $p(B) = 0.6666 = 66.66\%$
X - a product is taken from the first group and is working, $p(X) = 0.3333 \cdot 0.6666 = 0.2221 = 22.21\%$
I don't know if I'm any right.
 A: Consider the following events:
$A\equiv$"the product is picked from the first group"
$C\equiv$"the product we pick is working".
What we are looking for, is the probability that both $A$ and $C$ occur, that is, we want to compute $\Pr(A\cap C)$, the probability of the event $A\cap C$.
In your attempt at a solution, $A\cap C$ is the event you label $B$. Now, if we look at the question statement, $\frac23$ is the proportion of working product in the first group. Thus, $\frac23$ can be seen as the probability of picking a working product given that we have picked a product from the first group (if this distinction confuses you don't hesitate to write a comment and I'll further clarify, it is a rather important point), i.e.
$\Pr(C|A)\equiv$"Probability that $C$ occurs given that we know $A$ occurred".
I agree with your derivation of $\Pr(A)=\frac13$. Now, knowing that for arbitrary events $E,F$, the conditional probability $\Pr(E|F)$ is defined as
$$\Pr(E|F)=\frac{\Pr(E\cap F)}{\Pr(F)},$$
you should now have all you need to get $\Pr(A\cap C)$.
