A box contains two defective Christmas tree lights that have been inadvertently mixed with eight nondefective lights. If the lights are selected one at a time without replacement and tested until both defective lights are found, what is the probability that both defective lights will be found after exactly three trials? (Round your answer to four decimal places.)

To find the probability of finding both defective lights after exactly three trials, we need to consider the different possible outcomes.

In the first trial, there are 10 lights in total (2 defective + 8 nondefective). The probability of selecting a defective light in the first trial is 2/10 since there are 2 defective lights out of the total 10 lights.

After the first trial, we have 9 lights left (2 defective + 7 nondefective). In the second trial, the probability of selecting a defective light is 1/9 since we already selected one defective light in the first trial, leaving us with 1 defective light out of the remaining 9 lights.

After the second trial, we have 8 lights left (1 defective + 7 nondefective). In the third trial, the probability of selecting the final defective light is 1/8 since we already selected one defective light in the first trial and another defective light in the second trial, leaving us with only 1 defective light out of the remaining 8 lights.

To find the overall probability of these three independent events occurring in sequence, we multiply the individual probabilities:

(2/10) * (1/9) * (1/8) = 1/360

Therefore, the probability of finding both defective lights after exactly three trials is 1/360, or approximately 0.0028 when rounded to four decimal places.