probability question please using binomial? We have two machines which produce products of two qualities.The first machine produces 20 products in a hour of the first quality,and  8 products of the second quality.The second machine produces  30 products of the first quality and 9 products of the second quality.From one of the machines we take 2 products,which seem to be of the first quality.whats the probability that the products are from the first machine.?
so,my teacher told us that this is solved using binomial distribution but I dont think so at all.I think it can be solved using P(A/Hi)...
what do you think?
 A: There is a good case for saying you are both right! There are two interoretations of the data. (i) The (first) machine produces $28$ items per hour, exactly $20$ of of which are of the first quality, and $8$ of the second. Similar assumptions about the second. We choose one of the two machines at random, each with probbility $\frac{1}{2}$, and take a sample of $2$ from items produced that hour or (ii) the (first) machine produces $28$ items per hour. With probability $\frac{20}{28}$, an item produced by that machine is of the first quality, and with probability $\frac{8}{28}$ it is of the second quality. Similar assumptions about the second machine. (There are  actually other physically reasonable interpretations, that from the joint productions we take a sample of two. But they are not supported by the wording, which says that from one of the machines we take two items.)
We will use interpretation (ii), which is physically somewhat more reasonable than (i). 
Let $B$ be the event the two items tested are both good, and let $F$ be the event they are from the first machine. We want $\Pr(F|B)$. As usual 
$$\Pr(F|B)=\frac{\Pr(F\cap B)}{\Pr(B)}.$$
We calculate the probabilities on the right.
Under interpretation (ii), we have
$$\Pr(F\cap B)=\frac{1}{2}\binom{28}{2}\left(\frac{20}{28}\right)^2 \left(\frac{8}{28}\right)^0.$$
Similarly, 
$$\Pr(B)=\frac{1}{2}\binom{28}{2}\left(\frac{20}{28}\right)^2 \left(\frac{8}{28}\right)^0+ \frac{1}{2}\binom{39}{2}\left(\frac{30}{39}\right)^2 \left(\frac{9}{39}\right)^0.$$
Remark: Under interpretation (i), we would have
$$\Pr(F\cap B)=\frac{1}{2}\frac{\binom{20}{2}\binom{8}{0}}{\binom{28}{2}},$$
with a similarly constructed expression for $\Pr(B)$. 
