SOLUTION: a box contains 15 keys. Three of them open the main door. Two keys are selected randomly without replacement. What is the probablity that the first key does not open the main dorr
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-> SOLUTION: a box contains 15 keys. Three of them open the main door. Two keys are selected randomly without replacement. What is the probablity that the first key does not open the main dorr
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Question 251395: a box contains 15 keys. Three of them open the main door. Two keys are selected randomly without replacement. What is the probablity that the first key does not open the main dorr and the second key does open the main door? Answer by solver91311(24713) (Show Source):
The probability that the first key selected will not open the door is the number of keys in the bag that won't open the door (15 - 3 = 12) divided by the number of keys in the bag altogether, so 12/15 = 4/5 = 80%.
The probability that the second key selected WILL open the door is the number of keys in the bag that will open the door, which is still 3 because the first key we picked didn't open the door, divided by the number of keys in the the bag altogether, which is now 14 because we didn't replace the first key selected, so: 3/14.
If you are calculating the probability that this happens OR that happens (either thing happening is a success), then you add the probabilities. On the other hand, if you are calculating the probability that this happens AND that happens (both have to happen to be a success), then you multiply the probabilities.
