Let $B_0$ be the event that the batch has $0$ defectives, $B_1$ be the event the batch has $1$ defective, and $B_2$ be the event the batch has $2$ defectives.
Let $D_0$ be the event that neither selected component is defective. Problem (a) asks us to find the conditional probabilities $\Pr(B_0|D_0)$, $\Pr(B_1|D_0)$, and $\Pr(B_2|D_0)$. Now the hardest part, identifying precisely what we are after, has been done!
For the calculation, we use the general conditional probability formula
$$\Pr(X|Y)\Pr(Y)=\Pr(X\cap Y).$$
Put $X=B_0$ and $Y=D_0$. We need $\Pr(D_0)$ and $\Pr(B_0\cap D_0)$.
The event $D_0$ can happen in three different ways: (i) Our batch of $10$ is perfect, and we get no defectives in our sample of two; (ii) Our batch of $10$ has $1$ defective, but our sample of two misses them; (iii) Our batch has $2$ defective, but our sample misses them. If it helps, draw a tree that shows the three different paths through which we can end up with no defectives.
For (i), the probability is $(0.5)(1)$. For (ii), the probability that our batch has $1$ defective is $0.35$. Given that it has $1$ defective, the probability that our sample misses it is $\binom{9}{2}/\binom{10}{2}$, which is $8/10$. So the probability of (ii) is $(0.35)(8/10)$. For (iii), the probability our batch has $2$ defective is $0.15$. Given that it has $2$ defective, the probability that our sample misses both is $\binom{8}{2}/\binom{10}{2}$, which is $56/90$. So the probability of (iii) is $(0.15)(56/90)$. We have therefore found that
$$\Pr(D_0)=(0.5)(1)+(0.35)(8/10)+(0.15)(56/90).$$
The probability $\Pr(B_0\cap D_0)$ has been calculated during our calculation of $\Pr(D_0)$. It is $(0.5)(1)$. We conclude that
$$\Pr(B_0|D_0)=\frac{(0.5)(1)}{(0.5)(1)+(0.35)(8/10)+(0.15)(56/90)}.$$
The rest of the calculations for $D_0$ are easy, we have all the information needed. We get
$$\Pr(B_1|D_0)=\frac{(0.35)(8/10)}{(0.5)(1)+(0.35)(8/10)+(0.15)(56/90)}$$
and
$$\Pr(B_2|D_0)=\frac{(0.15)(56/90)}{(0.5)(1)+(0.35)(8/10)+(0.15)(56/90)}.$$
Alternately, we could have used Bayes' Formula directly . I wanted to do it in the above more basic way so that the logic would be clear.
Now unfortunately we have to deal with (b) and (c). But (c) is trivial! For (b), the calculation is as above, a little simpler, because if our sample of two has a defective, it cannot come from a perfect batch.
There is not enough information to solve the second problem. If it gets all its machine pieces from B, which is perfectly possible from the given numbers, then the probability the thing comes from A and is broken is $0$. But it also possible to arrange the buying pattern so that a non-zero proportion is bought from A, and there is a non-zero probability the machine piece is broken and comes from A.
Best Answer
Not quite, $$P=1-(0.95)^{15}-{15\choose 1}((0.95)^{14} \times 0.05)$$