Question of counting and probability

[Alexsandro]

Could someone help me with this question ?

There are two locks on the door and the keys are among the six different ones you carry in your pocket. In a hurry you dropped one somewhere.

a) What is the probability that you can still open the door ?

b) What is the probability that the first two keys you you try will open the door ?

The answers are:
a)2/3
b)4!/6!

 

[Tide]

a) There is one chance in three that one of the two keys to the locks was lost so the probablility that neither of them was lost is 2/3.

b) There is only one way in which the first two keys you select are the "good" keys (the order doesn't matter!) but 4! ways to arrange the remaining 4 keys (including the one designated as lost.) Compare this with 6! ways of arranging all 6 keys and you have the probability of opening the lock on the first two tries as 4!/6!

 

[Alexsandro]

doubt

a) There is one chance in three that one of the two keys to the locks was lost so the probablility that neither of them was lost is 2/3.

b) There is only one way in which the first two keys you select are the "good" keys (the order doesn't matter!) but 4! ways to arrange the remaining 4 keys (including the one designated as lost.) Compare this with 6! ways of arranging all 6 keys and you have the probability of opening the lock on the first two tries as 4!/6!

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Thanks for help.
But, I don't understand why "There is one chance in three that one of the two keys to the locks was lost". Could you explain better ?

 

[NateTG]

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Thanks for help.
But, I don't understand why "There is one chance in three that one of the two keys to the locks was lost". Could you explain better ?

There are two keys that you care about in order to open the door. Let's call them key A and key B. The chance that key A is lost is 1 in 6. The chance that key B is lost is also 1 in 6. Since these are exclusive events (both never happen at the same time) we add the probabilities: 1/6+1/6=1/3. So you have a 1/3 chance of not being able to open the door. That's a 2/3 chance of being able to open the door.

 

[NateTG]

b) There is only one way in which the first two keys you select are the "good" keys (the order doesn't matter!) but 4! ways to arrange the remaining 4 keys (including the one designated as lost.) Compare this with 6! ways of arranging all 6 keys and you have the probability of opening the lock on the first two tries as 4!/6!

This is somewhat incorrect.

The correct answer to the question, as it is posted here is 2 in 30:
Since there is a missing key, there is a chance to open the door only 2 out of 3 times.
Of those 2 out of 3 times, you need to pick the correct two keys out of five. Since one of these depends on the other, multiply the probabilities, and come up with 2 in 30=1/15.

Now, assuming that the key was not dropped, the chance that the first two keys open the door is 1 in 15. Clearly you must pick a key that opens a lock as the first key. Since there are 2 in 6 keys that open one of the locks, the chance for that to happen is 1/3. If you've picked one of the two keys, you need to chose the correct key for the second which is one in 5. Once again, one of these depends on the other, so the total probability is their product: 1/15.

The chance that the first key picked fits into the first lock is 1 in 6. If the first key fits into the first lock, then the chance that the second key fits into the second lock is 1 in 5. Once again, one of these depends on the other, so the total probability is their product: 1/30. So the only way to come up with 6!/4! is to assume that the keys must work in the order that they are chosen.

 

[Tide]

Nate,

The way I analyzed (b) was as follows: There are 6 keys and I asked how many ways are there of arranging them such that the "correct" keys are the first ones chosen. Imagine 6 placeholders with the first two placeholders dedicated to the "correct" keys. One of the remaining positions corresponds to the lost key and it doesn't matter which place it is. There is exactly one way for the two keys to be in the proper positions (again, order doesn't matter) but 4! ways of arranging the remaining keys. Since there are 6! ways of arranging all 6 keys the probability is 4!/6! = 1/30.

Using your approach, there is a 2/3 probability of the two correct keys being in the collection of 5 remaining keys. Given that, then again there is exactly 1 way for the first two keys to be the correct ones (order doesn't matter!) but 3! ways of arranging the remaining 3 keys while the total number of arrangements of 5 keys is 5!. So, the desired probability is 2/3 X 3! / 5! = 1/30.

 

[NateTG]

The question (as posted) is:

b) What is the probability that the first two keys you you try will open the door ?

So, let's say that one of the arrangements you consider legal is ABCDEF so A and B open the door, but then B and A will also open the door, even if they don't do so in that order.

If the question were "What is the chance that you will open the door if you try two keys at random?" then it indeed would be $\frac{4!}{6!}=\frac{1}{30}$ (assuming at least one of the two keys only opens one of the locks).