The probability that a certain door is locked is.6. The key to the door is one of five unidentified keys hanging on a key rack. Two keys are randomly selected before approaching the door. What is the probability that the door may be opened without returning for another key?
2/5 * .6 + .4= ..
99
Well, if we have 5 unidentified keys hanging on a key rack, and 2 keys are randomly selected, we can calculate the probability of the door being opened without returning for another key.
Out of 5 keys, 2 are selected. The probability of selecting the first key that fits the lock is 1/5, as there is only 1 key that will unlock the door out of the 5 available keys.
After the first key is selected, there are 4 keys remaining on the rack, with only 1 key that will unlock the door. So, the probability of selecting the second key that fits the lock is 1/4.
To calculate the overall probability, we need to multiply the individual probabilities together. So, the probability of both keys unlocking the door is (1/5) * (1/4) = 1/20.
Therefore, the probability that the door may be opened without returning for another key is 1/20, or in simpler terms, it's a real "key-nundrum"!
To calculate the probability that the door may be opened without returning for another key, we need to consider the different scenarios.
First, let's determine the probability of selecting the correct key on the first attempt:
Probability of selecting the correct key on the first attempt:
P(key 1 is the correct key) = 1/5
If the first key selected is incorrect, there will be 4 remaining keys on the key rack for the second attempt. The probability of selecting the correct key on the second attempt will depend on whether the first key selected was returned to the rack or not.
If the first key selected is returned to the rack:
Probability of selecting the correct key on the second attempt (assuming first key returned) = 1/4
If the first key selected is not returned to the rack:
Probability of selecting the correct key on the second attempt (assuming first key not returned) = 1/5
Now, let's calculate the overall probability of opening the door without returning for another key by considering all the possible scenarios:
Probability of opening the door on the first attempt:
P(open on 1st attempt) = (P(key 1 is correct)) * (P(key 2 is not selected))
Probability of opening the door on the second attempt (key 1 returned):
P(open on 2nd attempt, key 1 returned) = (P(key 1 is not the correct key)) * (P(key 2 is the correct key, assuming key 1 is returned))
Probability of opening the door on the second attempt (key 1 not returned):
P(open on 2nd attempt, key 1 not returned) = (P(key 1 is not the correct key)) * (P(key 2 is the correct key, assuming key 1 is not returned))
To calculate the overall probability of opening the door without returning for another key, we sum up the probabilities of the different scenarios:
P(open without returning) = P(open on 1st attempt) + P(open on 2nd attempt, key 1 returned) + P(open on 2nd attempt, key 1 not returned)
P(open without returning) = (1/5) * (4/5) + (4/5) * (1/4) + (1/5) * (1/5)
P(open without returning) = 4/25 + 1/5 + 1/25
P(open without returning) = 9/25
Therefore, the probability that the door may be opened without returning for another key is 9/25 or approximately 0.36.
To find the probability that the door may be opened without returning for another key, we need to consider two scenarios:
Scenario 1: The first selected key opens the door.
Scenario 2: The first selected key does not open the door, but the second selected key does.
Let's calculate the probability for each scenario:
Scenario 1: The first selected key opens the door.
In this case, the probability of selecting the correct key on the first attempt is 0.6 (since the probability that the door is locked is 0.6).
Scenario 2: The first selected key does not open the door, but the second selected key does.
In this case, the probability of not selecting the correct key on the first attempt is 0.4 (since the probability that the door is not locked is 0.4). After the first incorrect key is chosen, there are 4 remaining keys on the key rack. The probability of selecting the correct key on the second attempt is 1/4.
To calculate the overall probability, we need to consider both scenarios. Since we are interested in either Scenario 1 or Scenario 2 occurring, we can use the additive rule of probability:
Probability of Scenario 1 + Probability of Scenario 2 = (0.6) + (0.4 * 1/4) = 0.6 + 0.1 = 0.7
Therefore, the probability that the door may be opened without returning for another key is 0.7, or 70%.