Probability of choosing two bulbs with the same rating given that one has a specific rating I am trying to teach myself statistics, and working through Jay DeVore's excellent text of "Probability and Statistics for Engineering and the Sciences". The problem is as follows:
We have box of the following:


*

*{A} - 40W light bulbs, N(A) = 4 

*{B} - 65W light bulbs, N(B) = 5 

*{C} - 75W light bulbs, N(C) = 6


We select 2 bulbs at random. Given that at least one of these is rated 75W, what is the probability that they are both rated 75W?
Using a probability tree, I was able figure out the answer by dividing the probability of both bulbs being 75W (6/15)*(5/14) by all the cases that included a 75-W bulb including the case where both were 75 W.
However, how should I do this using combinatorics?
Thanks
 A: Imagine if you will that each and every bulb has a unique id label.
I.e. you have 40watt bulbs: $A_1, A_2, A_3,A_4$, you have 65watt bulbs $B_1,\dots, B_5$, you have 75watt bulbs $C_1,\dots, C_6$.
Notice that every pair of two bulbs is equally likely to be drawn.  For convenience, let us assume order doesn't matter.  (The same arguments will apply for if we had let order matter and it won't affect the answer either way).  We then set as our sample space $S = \{\text{all ways of drawing two bulbs}\}$.
As we are interested in a conditional probability, we restrict our attention to a subset of the sample space, what I will call $F$, where $F = \{\text{all ways of drawing two bulbs where at least one of them is 75watt}\}$
Let us count how large $F$ is.
There are two cases to consider: either exactly one bulb is 75watt, or both bulbs are 75watt.  There are $\binom{6}{1}\binom{5+4}{1}$ number of outcomes for the first case, and there are $\binom{6}{2}$ number of outcomes for the second case.  Thus, $|B| = \binom{6}{1}\binom{9}{1} + \binom{6}{2}= 6\cdot 9 + 15 = 69$.
Let $E = \{\text{all ways of drawing two 75watt bulbs}\}$.  Notice, we calculated $E$ above, as it was exactly the second case to consider in calculating $F$.
We have then $Pr(E|F) := \frac{|E\cap F|}{|F|} = \frac{|E|}{|F|} = \frac{15}{69}=\frac{5}{23}$
note: this version of the definition of probability only works in an equiprobable setting like this one

If you were to have let order matter, it might be clearer why this gives the same answer as you calculated with the tree diagram.
When order matters, you have that $|F|$ splits into three cases:


*

*first is 75watt, second is 75watt ($6\cdot 5$)

*first is 75watt, second is not ($6\cdot 9$)

*first is not, second is 75watt ($9\cdot 6$)
and that $|E|$ corresponds to the first case above with a total of $6\cdot 5$ outcomes.
You have then that $Pr(E|F) = \frac{6\cdot 5}{6\cdot 5 + 6\cdot 9 + 9\cdot 6} = \frac{30}{138} = \frac{5}{23}$
A: Combine the 40w and 65w bulbs into one group (Low) of nine bulbs.
The other group (High) has the six 75w bulbs. Let $X$ be the
number of H bulbs you get in two draws. The random variable
can take values 0, 1, and 2. $P(X = i) = {6 \choose i}{9 \choose {2-i}}/{15 \choose 2},$ for $i = 0,1,2.$ Find all three and check
that they add to 1.
You want the conditional probability 
$$P(X = 2|X \ge 1) = \frac{P(X = 2, X\ge 1)}{P(X \ge 1)} = 
\frac{P(X = 2)}{P(X \ge 1)}.$$
If you don't already know, you will soon see that $X$ has a
hypergeometric distribution.
Some kinds of software will compute the distribution. Here is
output from R. You can use the results to check yours.
 i = 0:2; pdf = dhyper(i, 6, 9, 2)
 cbind(i, pdf)
       i        pdf
 ##    0  0.3428571
 ##    1  0.5142857
 ##    2  0.1428571

A: Let $A$=Event that both bulbs are 75W and $B=$Event that at least one bulb is 75W. Using Bayes Theorem one can write,
$$P(A|B)=\frac{P(B|A)P(A)}{P(B)} = \frac{P(A)}{P(B)}$$
since $P(B|A)=1$ ie if you know both bulbs are 75W then you can be certain at least one bulb is 75W. $P(A)$ is straightforward, assuming drawing the bulbs are independent of one another then $$P(A)=\frac{6}{15}\times\frac{5}{14}=\frac{30}{210}$$ Now consider $B^c$=Event that neither bulb is 75W, then (again assuming independence) $$P(B^c)=\frac{9}{15}\times \frac{8}{14} = \frac{72}{210}$$ And therefore,
$$P(A|B)=\frac{\frac{30}{210}}{1-\frac{72}{210}}=\frac{5}{23}$$
