Conditional probability question with multiple draws

The question states that we have the following number of light bulbs:

• 4x 40W bulbs
• 5x 60W bulbs
• 6x 75W bulbs

Say that we draw two bulbs from this total of 15 bulbs. Given that one of the two is a 75W bulb, what are the odds that both are 75W bulbs?

I thought that the answer would simply be 5/14 (0.357), IE the odds of drawing a 75W bulb knowing that one of them has already been removed from the pool, but my textbook says the answer is 0.217 and I have no clue why. Can someone explain?

Solution 1:

$5/14$ would be true if the question stated that given the first drawn bulb was $75$ W, what was the probability that the second bulb would be $75$ W too.

In this case someone tells you that one of the two drawn bulbs is $75$ W. If $A$ is the event of both bulbs being $75$ W and $B$ is the event of at least one of the bulbs being $75$ W -

$ \displaystyle P(A \cap B) = \frac{6}{15} \cdot \frac{5}{14} = \frac{1}{7}$

$ \displaystyle P(B) = 1 - \frac{9}{15} \cdot \frac{8}{14} = \frac{23}{35}$

(I am subtracting the probability of both bulbs not being $75$ W from $1$, which gives me the probability that at least one of the bulbs is $75$ W)

Applying Bayes' Theorem, $ \displaystyle P(A |B) = \frac{5}{23}$