Question 1194851: A box contains 24 bulbs, 3 of which are defective. If two bulbs are selected and tested, find the probability that both are defective.
Found 2 solutions by Boreal, greenestamps:
Answer by Boreal(15235)   (Show Source):
You can put this solution on YOUR website! bulb 1 has a 3/24 chance of being defective. Once chosen, the probability of the second one's being defective is 2/23. That product is 6/552 or 1/92.
Answer by greenestamps(13198)  (Show Source):
You can put this solution on YOUR website!
The solution from @Boreal uses one of two primary methods for solving problems like this -- by imagining selecting one bulb at a time and finding the probability that each is defective, then multiplying those probabilities.
In this problem, that is the much easier method. However, you should know the other primary method, which will be required when the problem is more complex.
The probability that both bulbs selected are defective is the number of ways 2 of the 3 defective bulbs could be selected, divided by the total number of ways 2 of the 24 bulbs could be selected.
# of ways to select 2 of the 3 defective bulbs: 3 choose 2 = C(3,2) = 3
# of ways to select 2 of the 24 bulbs: 24 choose 2 = C(24,2) = (24*23)/2 = 23*12=276
P(selecting two defective bulbs) = 3/276 = 1/92
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