TITLE: 5 good and 2 defective bulbs are mixed. What is the probability of finding the two defectives in exactly three attempts?
QUESTION [1 upvotes]: 5 good and 2 defective bulbs are mixed. To find the defective ones, we test them one by one, randomly and without replacement. What is the probability of finding the two defectives in exactly three attempts?
 My work 
I think this exercise can be solved using rule of product,
$P(\text{First good} \cap \text{Second bad} \cap \text{third bad})+P(\text{First bad} \cap \text{Second good} \cap \text{third bad})+P(\text{First bad} \cap \text{Second bad} \cap \text{third good})=P(\text{Finding two defectives in exactly three attempts})=\frac{1}{12}+\frac{1}{21}+\frac{1}{21} $
Is correct this?
Note: Only can be solved using conditional probability.

REPLY [0 votes]: I see the note about only solving with conditional probabilities, but if you wanted a nice way to check your work, your answer should be equivalent to $$\frac{\binom{5}{1} \binom{2}{2}}{\binom{7}{3}}$$
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