Source: Veritas Prep
On a game show, a contestant is given three keys, each of which opens exactly one of three identical boxes. The first box contains $1, the second $100, and the third $1000. The contestant assigns each key to one of the boxes and wins the amount of money contained in any box that is opened by the key assigned to it. What is the probability that a contestant will win more than $1000?
(A) 1/9
(B) 1/8
(C) 1/6
(D) 1/3
(E) 1/2
Experts: Only Veritas Prep experts, please!
Game show
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IMO DDanaJ wrote:Source: Veritas Prep
On a game show, a contestant is given three keys, each of which opens exactly one of three identical boxes. The first box contains $1, the second $100, and the third $1000. The contestant assigns each key to one of the boxes and wins the amount of money contained in any box that is opened by the key assigned to it. What is the probability that a contestant will win more than $1000?
(A) 1/9
(B) 1/8
(C) 1/6
(D) 1/3
(E) 1/2
Experts: Only Veritas Prep experts, please!
The contestant must get the key correct to the $1000 box in order to win more than $1000.
P($1000) = 1/3
Out of rest two boxes he must have the correct key to at least one to get more than $1000.
P($100) = 1/2 however as this attempt is success the other attempt P($1) is bound to succeed.
P($1) = 1
Thus P(Winning > $1000) = (1/3)*(1/2)*1 = 1/6
So I am not sure about the other case where we consider P($1) = 1/2 and P($100)=1 as both are equally likely.
Thus, 2*1/6 = 1/3
Not sure
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[spoiler]IMO: C[/spoiler]
This is somewhat similar to sibling problems.
The desired outcome is the contestant to win more than $1000.
So the key picked for the box that contains $1000 must match. However, the contestant needs to win more than $1000. So another key picked, either for the $100 box or $1 box, must match. And this also leads to the matching of the final box too (As the first two boxes have already had matched keys).
All three boxes must have their respectively matched keys.
My approach:
Pick the first box (whatever it is, never mind 'bout it), possibility for this box to have its matched key : 1/3 (three keys remain)
Pick the second box, possibility for 2nd box to have its matched key: 1/2 (two keys remain)
The final box, possibility for it to have a matched key is absolute, or 100%, or 1.
Total possibility: 1/3*1/2*1 = 1/6
Correct me if I'm wrong.
This is somewhat similar to sibling problems.
The desired outcome is the contestant to win more than $1000.
So the key picked for the box that contains $1000 must match. However, the contestant needs to win more than $1000. So another key picked, either for the $100 box or $1 box, must match. And this also leads to the matching of the final box too (As the first two boxes have already had matched keys).
All three boxes must have their respectively matched keys.
My approach:
Pick the first box (whatever it is, never mind 'bout it), possibility for this box to have its matched key : 1/3 (three keys remain)
Pick the second box, possibility for 2nd box to have its matched key: 1/2 (two keys remain)
The final box, possibility for it to have a matched key is absolute, or 100%, or 1.
Total possibility: 1/3*1/2*1 = 1/6
Correct me if I'm wrong.
"There is nothing either good or bad - but thinking makes it so" - Shakespeare.
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Here is the official answer and explanation. You will see that it agrees with the above.
Correct answer: (C)
Solution: This tricky probability question is best solved by calculating the total possibilities and then determining which ones are favorable. There are 3! (or 6) ways that the 3 keys could be assigned. We will call the first key A, the second key B, and the third key C. The six unique assignments are:
ABC, ACB, BAC, BCA, CAB, CBA
Therefore, we know that the total number of possible key distributions is 6. How many are favorable? Interestingly, there is only one favorable outcome that will win more than $1000: when all three boxes are assigned the correct keys. To win more than $1000, at least two of the three boxes must open. If two boxes open, the third must as well. Given that there are six unique key distributions, only one will open all three boxes (i.e., ABC and CBA can't both open all three boxes). Thus the probability of winning more than $1000 is 1/6.
Correct answer: (C)
Solution: This tricky probability question is best solved by calculating the total possibilities and then determining which ones are favorable. There are 3! (or 6) ways that the 3 keys could be assigned. We will call the first key A, the second key B, and the third key C. The six unique assignments are:
ABC, ACB, BAC, BCA, CAB, CBA
Therefore, we know that the total number of possible key distributions is 6. How many are favorable? Interestingly, there is only one favorable outcome that will win more than $1000: when all three boxes are assigned the correct keys. To win more than $1000, at least two of the three boxes must open. If two boxes open, the third must as well. Given that there are six unique key distributions, only one will open all three boxes (i.e., ABC and CBA can't both open all three boxes). Thus the probability of winning more than $1000 is 1/6.
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Since I'm not associated with Veritas, I've hidden the explanation below. It will be revealed if you use the quote function to respond.DanaJ wrote:Source: Veritas Prep
On a game show, a contestant is given three keys, each of which opens exactly one of three identical boxes. The first box contains $1, the second $100, and the third $1000. The contestant assigns each key to one of the boxes and wins the amount of money contained in any box that is opened by the key assigned to it. What is the probability that a contestant will win more than $1000?
(A) 1/9
(B) 1/8
(C) 1/6
(D) 1/3
(E) 1/2
Experts: Only Veritas Prep experts, please!
[spoiler]There is only 1 way to win more than $1000: the contestant must assign all of the keys correctly. To win more than $1000, he must assign the $1000 key and at least 1 other key correctly. But it's impossible to assign exactly 2 keys correctly. Once 2 keys have been assigned correctly, the 3rd key must also be assigned correctly, because there will only 1 key and 1 box left, and this 1 remaining key and the 1 remaining box will have to match.
P($1000 box gets the correct key) = 1/3 (out of the 3 keys, only 1 is correct)
P($100 box gets the correct key) = 1/2 (out of the 2 remaining keys, only 1 is correct)
P($1 box gets the correct key) = 1/1 (1 key left, and it must be correct, since the incorrect keys have already been assigned to the other 2 boxes)
Since we need all of these events to happen, we multiply the fractions:
1/3 * 1/2 * 1/1 = 1/6.
The correct answer is C.[/spoiler]
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Hi Guru,GMATGuruNY wrote:
[spoiler]
P($1000 gets the correct key) = 1/3
P($100 box gets the correct key) = 1/2
P($1 box gets the correct key) = 1/1
[/spoiler]
This is the exact point where I was completely confused.
[spoiler]P($1000 gets the correct key) = 1/3
P($100 box gets the correct key) = 1/2
P($1 box gets the correct key) = 1/1
P(Win) = 1/6
What about the other case
P($1000 gets the correct key) = 1/3
P($1 box gets the correct key) = 1/2
P($100 box gets the correct key) = 1/1
P(Win) = 1/6
Thus total 1/3. [/spoiler]
Where am I going wrong? Why shall not we consider the latter?
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My question is -
Why do we have to consider 1000$ in the beginning ?
Why not this - 100$ , 1000$,1$ or 1$,100$,1000$ or 1$,1000$, 100$
I think the contestant can have 1-3 successful shots.
Why do we have to consider 1000$ in the beginning ?
Why not this - 100$ , 1000$,1$ or 1$,100$,1000$ or 1$,1000$, 100$
I think the contestant can have 1-3 successful shots.
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[spoiler]The order of the boxes is irrelevant. To win more than $1000, each of the keys must be assigned correctly. Let's consider the 3 keys that the contestant is placing:
P(first key is put in the correct box) = 1/3
P(second key is put in the correct box) = 1/2
P(third key is put in the correct box) = 1/1
1/3 * 1/2 * 1/1 = 1/6.
Does this help?[/spoiler]
P(first key is put in the correct box) = 1/3
P(second key is put in the correct box) = 1/2
P(third key is put in the correct box) = 1/1
1/3 * 1/2 * 1/1 = 1/6.
Does this help?[/spoiler]
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- shovan85
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Is it because the there are 3 boxes and for getting more than a 1000$ all the boxes are bound to be selected correctly?GMATGuruNY wrote:[spoiler]The order of the boxes is irrelevant. To win more than $1000, each of the keys must be assigned correctly. Let's consider the 3 keys that the contestant is placing:
P(first key is put in the correct box) = 1/3
P(second key is put in the correct box) = 1/2
P(third key is put in the correct box) = 1/1
1/3 * 1/2 * 1/1 = 1/6.
Does this help?[/spoiler]
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[spoiler]Yes. Perhaps it would be easier to think of the problem this way:shovan85 wrote:
Is it because the there are 3 boxes and for getting more than a 1000$ all the boxes are bound to be selected correctly?
3 boxes are placed before the contestant. Each box has a corresponding key.
To win more than 1000 dollars, the contestant must place the 3 keys in the correct order.
There are 3! = 6 ways to arrange the 3 keys. There is only 1 correct arrangement of the 3 keys.
Thus, P(correct arrangement) = 1/6.
Does this help?[/spoiler]
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Great Explanation indeedGMATGuruNY wrote:[spoiler]Yes. Perhaps it would be easier to think of the problem this way:shovan85 wrote:
Is it because the there are 3 boxes and for getting more than a 1000$ all the boxes are bound to be selected correctly?
3 boxes are placed before the contestant. Each box has a corresponding key.
To win more than 1000 dollars, the contestant must place the 3 keys in the correct order.
There are 3! = 6 ways to arrange the 3 keys. There is only 1 correct arrangement of the 3 keys.
Thus, P(correct arrangement) = 1/6.
Does this help?[/spoiler]
Thanks a lot
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Brilliant Approach, 1/3 * 1/2 * 1 = 1/6limestone wrote:[spoiler]IMO: C[/spoiler]
This is somewhat similar to sibling problems.
The desired outcome is the contestant to win more than $1000.
So the key picked for the box that contains $1000 must match. However, the contestant needs to win more than $1000. So another key picked, either for the $100 box or $1 box, must match. And this also leads to the matching of the final box too (As the first two boxes have already had matched keys).
All three boxes must have their respectively matched keys.
My approach:
Pick the first box (whatever it is, never mind 'bout it), possibility for this box to have its matched key : 1/3 (three keys remain)
Pick the second box, possibility for 2nd box to have its matched key: 1/2 (two keys remain)
The final box, possibility for it to have a matched key is absolute, or 100%, or 1.
Total possibility: 1/3*1/2*1 = 1/6
Correct me if I'm wrong.
Great work,
Saurabh Goyal
[email protected]
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Brilliant Approach, 1/3 * 1/2 * 1 = 1/6limestone wrote:[spoiler]IMO: C[/spoiler]
This is somewhat similar to sibling problems.
The desired outcome is the contestant to win more than $1000.
So the key picked for the box that contains $1000 must match. However, the contestant needs to win more than $1000. So another key picked, either for the $100 box or $1 box, must match. And this also leads to the matching of the final box too (As the first two boxes have already had matched keys).
All three boxes must have their respectively matched keys.
My approach:
Pick the first box (whatever it is, never mind 'bout it), possibility for this box to have its matched key : 1/3 (three keys remain)
Pick the second box, possibility for 2nd box to have its matched key: 1/2 (two keys remain)
The final box, possibility for it to have a matched key is absolute, or 100%, or 1.
Total possibility: 1/3*1/2*1 = 1/6
Correct me if I'm wrong.
Great work,
Saurabh Goyal
[email protected]
-------------------------
EveryBody Wants to Win But Nobody wants to prepare for Win.
[email protected]
-------------------------
EveryBody Wants to Win But Nobody wants to prepare for Win.
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Well, i didnt get much into the technical portions because we will be unsure as shovan did in the earlier posts.
Its always to better to approach from a Practical perspective.
We are given three keys namely A B C.
There are three boxes D E F( D is 1$ , E is 100$, F is 1000$)
Let us assume the right combination first
(AD, BE, CF are the only right combinations).
Here are the different scenarios.
ABC can be distributed in 3! ways. that means 6 essential possiblities.
D E F (Boxes)
============
A B C (This combination yields more than 1000$) = Favourable outcome
A C B (This combination yields just 1 $ since A Is properly inserted in D , whereas other two keys mismatch with boxes)
B A C (First two keys mismatch and third one yields 1000$ but its still just 1000$)
B C A (All Three mismatch and yields nothing)
C A B (All Three mismatch and yields nothing)
C B A (B alone matches with its box earning 100$)
So out of this all outcomes we get only the first output in our favourable outcomes.
So the probablility of getting more than 1000$ is 1/6
Its always to better to approach from a Practical perspective.
We are given three keys namely A B C.
There are three boxes D E F( D is 1$ , E is 100$, F is 1000$)
Let us assume the right combination first
(AD, BE, CF are the only right combinations).
Here are the different scenarios.
ABC can be distributed in 3! ways. that means 6 essential possiblities.
D E F (Boxes)
============
A B C (This combination yields more than 1000$) = Favourable outcome
A C B (This combination yields just 1 $ since A Is properly inserted in D , whereas other two keys mismatch with boxes)
B A C (First two keys mismatch and third one yields 1000$ but its still just 1000$)
B C A (All Three mismatch and yields nothing)
C A B (All Three mismatch and yields nothing)
C B A (B alone matches with its box earning 100$)
So out of this all outcomes we get only the first output in our favourable outcomes.
So the probablility of getting more than 1000$ is 1/6
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Hi:
I used simple probability to arrive the answer. Let me know if this approach is wrong and I just got lucky:
We absolutely need $1000 box to be open. Chances of that key getting into the right box is 1/3 (it could go into any box)
Assuming that this is actually in the right box, chances of other keys (2 left) in the right box (either of the two) is also 1/2 (only two possibilities - either both keys in right boxes or both in wrong boxes).
so to arrive >$1000 means that our possibility is 1/3 * 1/2 = 1/6
Certainly I could have gone down the path of listing out the combinations etc. but I got to this answer in few sec through probability
I used simple probability to arrive the answer. Let me know if this approach is wrong and I just got lucky:
We absolutely need $1000 box to be open. Chances of that key getting into the right box is 1/3 (it could go into any box)
Assuming that this is actually in the right box, chances of other keys (2 left) in the right box (either of the two) is also 1/2 (only two possibilities - either both keys in right boxes or both in wrong boxes).
so to arrive >$1000 means that our possibility is 1/3 * 1/2 = 1/6
Certainly I could have gone down the path of listing out the combinations etc. but I got to this answer in few sec through probability