Defective Components (Binomial Distribution) Components of a certain type are shipped to a supplier in
batches of ten. Suppose that 50% of all batches contain no defective components,
30% contain one defective component, and 20% contain two defective
components. Two components are randomly selected without replacement
from a batch and tested.
(a) (4 marks). If the batch from which the components were selected actually
contains two defectives, what is the probability that neither of
these is selected for testing?
(b) (2 marks). What is the probability that the batch contains two defectives
and neither of these is selected for testing?
(c) (2 marks). What is the probability that neither component selected for
testing is defective?
 A: (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).  
