A box contains four 40W light bulbs, five 60W bulbs and six 75W bulbs.
If two bulbs are randomly selected from the box of light bulbs, and at least one of them is found to be rated 75W, what is the probability that both of them are 75-W bulbs? $$ P(X=2\mid X\geq1)=\frac{P(X=2)}{P(X\geq1)}=\frac{P(X=2)}{1-P(X=0)} $$
I did not follow this solution Isn't P(A|B) = P(A and B)/P(B)
Why is there no "and" component in the given solution?
My foundation for probability is extremely weak so I apologize if this question is a bit too basic or something. And due to low reputation (extremely new here) I couldn't just ask this in a comment on the solution :)
I wasn't able to follow the analysis in your question. Therefore, I answered your question from scratch.
$A$ is event that both are 75w.
$B$ is event that at least one is 75w.
By Bayes Theorem $\displaystyle P(A|B) = \frac{p(AB)}{p(B)}$.
$p(B)$ can be computed by considering that there may be either one or two 75w bulbs.
$\displaystyle p(B) = \frac{\binom{6}{1}\binom{9}{1}}{\binom{15}{2}} + \frac{\binom{6}{2}}{\binom{15}{2}}.$
$\displaystyle p(AB) = \frac{\binom{6}{2}}{\binom{15}{2}}.$