Prove that the probability that an article chosen at random from $C$ is defective,is $\frac{8}{25}.$ Lot $A$ consists of 3 good and 2 defective articles.Lot $B$ consists of 4 good and 1 defective article.A new lot $C$ is formed by taking 3 articles from $A$ and $2$ from $B$.Prove that the probability that an article chosen at random from $C$ is defective,is $\frac{8}{25}.$

Required probabiliy$=$probability that either $A$ is chosen and then defective item is chosen from it or $B$ is chosen and then defective item is chosen from it.
$=\frac{1}{2}\times \frac{2}{5}+\frac{1}{2}\times \frac{1}{5}=\frac{3}{10}$
but my answer is wrong,means my logic is also wrong.What should be the correct logic to solve this problem.Please help me.Thanks. 
 A: Your logic is correct, however you have the wrong numbers for both the "probability that A is chosen" and "probability that B is chosen". Indeed, you wrote that both these probabilities are $\frac{1}{2}$, but think about it for a second and you'll see why that's slightly off. (It really is nothing more than a case of carefully writing out $\frac{\text{number of favorable outcomes}}{\text{total number of outcomes}}$)
A: Let $G$ denote the event that the item is good.
Let $A$ denote the event that the item came from lot A, and likewise for $B$.
Then $P(G) = P(G | A)P(A) + P(G | B)P(B)$.
Of course $P(A) = \frac{3}{5}, P(B) = \frac{2}{5}$ by the way lot C was formed.
$P(G|A)$ is the probability that an item is good, given that it came from $A$. Well, we could have taken 3 good items, which would have happened with probability ${{3 \choose 3}{2 \choose 0} \over {5 \choose 3}} = \frac{3}{5} \frac{2}{4} \frac{1}{3} = \frac{1}{10}$, and then all of these would have been OK. Or we could have taken 2 good items, with probability ${{3 \choose 2}{2 \choose 1}} \over { 5 \choose 3}$, and then we'd have a $\frac{2}{3}$ chance in that case. Or we could have picked 1 good one, with probability ${{3 \choose 1}{2 \choose 2}} \over { 5 \choose 3}$ and then we'd have had a chance of $\frac{1}{3}$. 
So $$P(G | A) = \frac{1}{10} \cdot 1 + {{{3 \choose 2}{2 \choose 1}} \over { 5 \choose 3}} \cdot \frac{2}{3} + {{{3 \choose 1}{2 \choose 2}} \over { 5 \choose 3}} \cdot \frac{1}{3}$$
We can compute $P(G | B)$ in a similar (somewhat simpler) way. Then compute $P(G)$ by the first formula. 
