There are two boxes, each containing two components. Each component is defective with probability 1/4, independent of all other components. The probability that exactly one box contains exactly one defective component equals?
[Math] exactly one box contains exactly one defective component
probabilityprobability theorystatistics
Related Solutions
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.
(a) Imagine that two components are selected, one at a time, from a group of $10$ that contains $2$ bad. The probability the first item selected is good is $\frac{8}{10}$. Given that the first item was good, the probability the second item is good is $\frac{7}{9}$. So our probability is $\frac{8}{10}\cdot\frac{7}{9}$.
Another way: There are $\binom{10}{2}$ equally likely ways to select $2$ items. There are $\binom{8}{2}$ ways to select $2$ good. For the probability, divide.
(b) The probability a batch with $2$ bad is selected is $0.2$. Given this has happened, the probability both items tested are good is the answer to (a)$. Multiply.
(c) This will require some calculation. We get that the tested items are both good in $3$ different ways: (i) The batch of $10$ chosen is all good or (ii) the batch chosen has $1$ bad, but it is not one of the tested items or (ii) the batch chosen has $2$ bad, neither of which is tested. We find the probabilities of (i), (ii), and (iii), and add up.
In (b), we found the probability of (iii).
If the items are all good, then for sure the $2$ tested items will be good. Thus the probability of (i) is $(0.5)(1)$.
To find the probability of (ii), use the method that we used to solve (a) and (b).
Best Answer
P(1 faulty in box) = C(2,1) (1/4)(3/4) = 3 / 8
the chances of a box having other than 1 fault is 5/8
The P(1 exactly one fault in exactly one box) = C(2,1)(3/8)(5/8) = 15/32