Mixing
Probability Question : I have three locks and three keys (only One correct key for each lock).
1
$begingroup$
What is the probability that all locks are unlocked by choosing one key and one lock randomly at a time?
Note: After correctly choosing a key to unlock a lock, the key is kept back with the other two keys while the lock is kept aside. So after unlocking first lock, I am left with just 2 locks to choose from but I have all the three keys to choose from.
Please give some explanation to your answer as well.
probability
asked Nov 8 '12 at 6:47
Jatin SehgalJatin Sehgal
61
$endgroup$
|
show 2 more comments
1
$begingroup$
What is the probability that all locks are unlocked by choosing one key and one lock randomly at a time?
Note: After correctly choosing a key to unlock a lock, the key is kept back with the other two keys while the lock is kept aside. So after unlocking first lock, I am left with just 2 locks to choose from but I have all the three keys to choose from.
Please give some explanation to your answer as well.
probability
asked Nov 8 '12 at 6:47
Jatin SehgalJatin Sehgal
61
$endgroup$
2
$begingroup$
Can you at least answer this question: What is the probability that the first lock is unlocked?
$endgroup$
– Brian M. Scott
Nov 8 '12 at 6:50
1
$begingroup$
I am so sorry but I dont know as I am confused over this. I think It should be 1/3.
$endgroup$
– Jatin Sehgal
Nov 8 '12 at 7:03
1
$begingroup$
That’s right. Now when you get to the second lock, has anything really changed? You still have one chance in three of picking the right key, so the probability of unlocking the second lock is ... ?
$endgroup$
– Brian M. Scott
Nov 8 '12 at 7:06
$begingroup$
But in the first chance, I had three locks to choose from to apply my chosen key. In that scenario as well Would you say my answer was correct? Secondly, after successfully unlocking the first lock I ll hv to choose a lock from remaining 2 locks while I still choose a key from all the three keys.......
$endgroup$
– Jatin Sehgal
Nov 8 '12 at 7:11
$begingroup$
The choice of locks doesn’t make any difference. No matter which one you pick, you still have one chance in three of picking the right key.
$endgroup$
– Brian M. Scott
Nov 8 '12 at 7:21
|
show 2 more comments
1
1
1
$begingroup$
What is the probability that all locks are unlocked by choosing one key and one lock randomly at a time?
Note: After correctly choosing a key to unlock a lock, the key is kept back with the other two keys while the lock is kept aside. So after unlocking first lock, I am left with just 2 locks to choose from but I have all the three keys to choose from.
Please give some explanation to your answer as well.
probability
asked Nov 8 '12 at 6:47
Jatin SehgalJatin Sehgal
61
$endgroup$
What is the probability that all locks are unlocked by choosing one key and one lock randomly at a time?
Note: After correctly choosing a key to unlock a lock, the key is kept back with the other two keys while the lock is kept aside. So after unlocking first lock, I am left with just 2 locks to choose from but I have all the three keys to choose from.
Please give some explanation to your answer as well.
probability
probability
asked Nov 8 '12 at 6:47
Jatin SehgalJatin Sehgal
61
asked Nov 8 '12 at 6:47
Jatin SehgalJatin Sehgal
61
asked Nov 8 '12 at 6:47
Jatin SehgalJatin Sehgal
61
asked Nov 8 '12 at 6:47
Jatin SehgalJatin Sehgal
61
asked Nov 8 '12 at 6:47
Jatin SehgalJatin Sehgal
61
61
2
$begingroup$
Can you at least answer this question: What is the probability that the first lock is unlocked?
$endgroup$
– Brian M. Scott
Nov 8 '12 at 6:50
1
$begingroup$
I am so sorry but I dont know as I am confused over this. I think It should be 1/3.
$endgroup$
– Jatin Sehgal
Nov 8 '12 at 7:03
1
$begingroup$
That’s right. Now when you get to the second lock, has anything really changed? You still have one chance in three of picking the right key, so the probability of unlocking the second lock is ... ?
$endgroup$
– Brian M. Scott
Nov 8 '12 at 7:06
$begingroup$
But in the first chance, I had three locks to choose from to apply my chosen key. In that scenario as well Would you say my answer was correct? Secondly, after successfully unlocking the first lock I ll hv to choose a lock from remaining 2 locks while I still choose a key from all the three keys.......
$endgroup$
– Jatin Sehgal
Nov 8 '12 at 7:11
$begingroup$
The choice of locks doesn’t make any difference. No matter which one you pick, you still have one chance in three of picking the right key.
$endgroup$
– Brian M. Scott
Nov 8 '12 at 7:21
|
show 2 more comments
2
$begingroup$
Can you at least answer this question: What is the probability that the first lock is unlocked?
$endgroup$
– Brian M. Scott
Nov 8 '12 at 6:50
1
$begingroup$
I am so sorry but I dont know as I am confused over this. I think It should be 1/3.
$endgroup$
– Jatin Sehgal
Nov 8 '12 at 7:03
1
$begingroup$
That’s right. Now when you get to the second lock, has anything really changed? You still have one chance in three of picking the right key, so the probability of unlocking the second lock is ... ?
$endgroup$
– Brian M. Scott
Nov 8 '12 at 7:06
$begingroup$
But in the first chance, I had three locks to choose from to apply my chosen key. In that scenario as well Would you say my answer was correct? Secondly, after successfully unlocking the first lock I ll hv to choose a lock from remaining 2 locks while I still choose a key from all the three keys.......
$endgroup$
– Jatin Sehgal
Nov 8 '12 at 7:11
$begingroup$
The choice of locks doesn’t make any difference. No matter which one you pick, you still have one chance in three of picking the right key.
$endgroup$
– Brian M. Scott
Nov 8 '12 at 7:21
2
2
$begingroup$
Can you at least answer this question: What is the probability that the first lock is unlocked?
$endgroup$
– Brian M. Scott
Nov 8 '12 at 6:50
$begingroup$
Can you at least answer this question: What is the probability that the first lock is unlocked?
$endgroup$
– Brian M. Scott
Nov 8 '12 at 6:50
1
1
$begingroup$
I am so sorry but I dont know as I am confused over this. I think It should be 1/3.
$endgroup$
– Jatin Sehgal
Nov 8 '12 at 7:03
$begingroup$
I am so sorry but I dont know as I am confused over this. I think It should be 1/3.
$endgroup$
– Jatin Sehgal
Nov 8 '12 at 7:03
1
1
$begingroup$
That’s right. Now when you get to the second lock, has anything really changed? You still have one chance in three of picking the right key, so the probability of unlocking the second lock is ... ?
$endgroup$
– Brian M. Scott
Nov 8 '12 at 7:06
$begingroup$
That’s right. Now when you get to the second lock, has anything really changed? You still have one chance in three of picking the right key, so the probability of unlocking the second lock is ... ?
$endgroup$
– Brian M. Scott
Nov 8 '12 at 7:06
$begingroup$
But in the first chance, I had three locks to choose from to apply my chosen key. In that scenario as well Would you say my answer was correct? Secondly, after successfully unlocking the first lock I ll hv to choose a lock from remaining 2 locks while I still choose a key from all the three keys.......
$endgroup$
– Jatin Sehgal
Nov 8 '12 at 7:11
$begingroup$
But in the first chance, I had three locks to choose from to apply my chosen key. In that scenario as well Would you say my answer was correct? Secondly, after successfully unlocking the first lock I ll hv to choose a lock from remaining 2 locks while I still choose a key from all the three keys.......
$endgroup$
– Jatin Sehgal
Nov 8 '12 at 7:11
$begingroup$
The choice of locks doesn’t make any difference. No matter which one you pick, you still have one chance in three of picking the right key.
$endgroup$
– Brian M. Scott
Nov 8 '12 at 7:21
$begingroup$
The choice of locks doesn’t make any difference. No matter which one you pick, you still have one chance in three of picking the right key.
$endgroup$
– Brian M. Scott
Nov 8 '12 at 7:21
|
show 2 more comments
1 Answer
1
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0
$begingroup$
Possible combinations are 3×2×1=3!=6
In which 1 is correct
P(A)=1÷6=0.166
answered Jan 15 '17 at 6:58
Pankaj JorwalPankaj Jorwal
1
$endgroup$
add a comment |
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1 Answer
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1 Answer
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0
$begingroup$
Possible combinations are 3×2×1=3!=6
In which 1 is correct
P(A)=1÷6=0.166
answered Jan 15 '17 at 6:58
Pankaj JorwalPankaj Jorwal
1
$endgroup$
add a comment |
0
$begingroup$
Possible combinations are 3×2×1=3!=6
In which 1 is correct
P(A)=1÷6=0.166
answered Jan 15 '17 at 6:58
Pankaj JorwalPankaj Jorwal
1
$endgroup$
add a comment |
0
0
0
$begingroup$
Possible combinations are 3×2×1=3!=6
In which 1 is correct
P(A)=1÷6=0.166
answered Jan 15 '17 at 6:58
Pankaj JorwalPankaj Jorwal
1
$endgroup$
Possible combinations are 3×2×1=3!=6
In which 1 is correct
P(A)=1÷6=0.166
answered Jan 15 '17 at 6:58
Pankaj JorwalPankaj Jorwal
1
answered Jan 15 '17 at 6:58
Pankaj JorwalPankaj Jorwal
1
answered Jan 15 '17 at 6:58
Pankaj JorwalPankaj Jorwal
1
answered Jan 15 '17 at 6:58
Pankaj JorwalPankaj Jorwal
1
1
add a comment |
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2
$begingroup$
Can you at least answer this question: What is the probability that the first lock is unlocked?
$endgroup$
– Brian M. Scott
Nov 8 '12 at 6:50
1
$begingroup$
I am so sorry but I dont know as I am confused over this. I think It should be 1/3.
$endgroup$
– Jatin Sehgal
Nov 8 '12 at 7:03
1
$begingroup$
That’s right. Now when you get to the second lock, has anything really changed? You still have one chance in three of picking the right key, so the probability of unlocking the second lock is ... ?
$endgroup$
– Brian M. Scott
Nov 8 '12 at 7:06
$begingroup$
But in the first chance, I had three locks to choose from to apply my chosen key. In that scenario as well Would you say my answer was correct? Secondly, after successfully unlocking the first lock I ll hv to choose a lock from remaining 2 locks while I still choose a key from all the three keys.......
$endgroup$
– Jatin Sehgal
Nov 8 '12 at 7:11
$begingroup$
The choice of locks doesn’t make any difference. No matter which one you pick, you still have one chance in three of picking the right key.
$endgroup$
– Brian M. Scott
Nov 8 '12 at 7:21