A box contains two defective Christmas tree lights that have been inadvertently mixed with eight non-defective 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?© BrainMass Inc. brainmass.com October 10, 2019, 5:10 am ad1c9bdddf
There are total 2 + 8 = 10 light bulbs in the box.
to pick 3 out of 10, the total outcomes are: 10C3 = 10 * 9 * 8 / 3! = ...
This solution calculates the probability that both defective lights will be found after exactly three trials.