I approached it the same way. 1/3 chance of getting 1,000 * 1/2 chance of opening one of the other two boxes.thebigkats wrote: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
Game show
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edvhou812
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sanju09
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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!
The 3 keys could be assigned to 3 boxes in 3! ways. Now the contestant could win more than $1000 only if he assigns all keys correctly, which has got only one favorable way. Required probability is [spoiler]1/6
C[/spoiler]
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saurabh2525_gupta
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Hi Mitch,GMATGuruNY wrote: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]
I solved the problem this way,
Let the boxes be
A B C
1 10 1000
let R denote rightly placed key
let W denote wrongly placed key
So the possible combinations can be
RRR
RRW - invalid since when two keys are rightly placed, third has to be right
RWR - invalid since when two keys are rightly placed, third has to be right
RWW
WRR - invalid since when two keys are rightly placed, third has to be right
WRW
WWR
WWW
We are left with 5 possible outcomes and 1 favorable
So the answer should be 1/5
Tell me where am I wrong
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saketk
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We really don't need much of number crunching here.
Total ways=3!
Since, the question says find the probability of person winning more than $1000
Logically we need 2 right combinations to get a sum of 1000, of which one has to be the $1000 box everytime.
Also if 2 out of 3 are arranged correctly then the third will automatically get arranged correctly.
Hence there is only 1 possible case
Answer =1/6
Total ways=3!
Since, the question says find the probability of person winning more than $1000
Logically we need 2 right combinations to get a sum of 1000, of which one has to be the $1000 box everytime.
Also if 2 out of 3 are arranged correctly then the third will automatically get arranged correctly.
Hence there is only 1 possible case
Answer =1/6
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parul9
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I am super bad at probability.
But I got this one.
First I jumped to 1/3, then I thought wait a minute!
There have been multiple game shows on this pattern. The chances of a contestant winning the highest sum are not so fat!
Then I thought harder, and since that did not help, I got to basics.
Probability of an event = no. of time the event can occur/total no. of times
So, Probability of getting 1000$ = no. of times key matches with 1000$ box/no. of key-box combos possible
no. of times key matches with 1000$ box = 1
no. of key-box combos possible = 3! = 6
So, P = 1/6
Answer is C.
n
But I got this one.
First I jumped to 1/3, then I thought wait a minute!
There have been multiple game shows on this pattern. The chances of a contestant winning the highest sum are not so fat!
Then I thought harder, and since that did not help, I got to basics.
Probability of an event = no. of time the event can occur/total no. of times
So, Probability of getting 1000$ = no. of times key matches with 1000$ box/no. of key-box combos possible
no. of times key matches with 1000$ box = 1
no. of key-box combos possible = 3! = 6
So, P = 1/6
Answer is C.
n
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immaculatesahai
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Nice to see an easy question.
So we can picture this in the following way:
3 boxes each with certain amount of money: Box A-($1) Box B- $100) BoxC-($1000)
Winning more than $1000 dollars means, that we need to find out the probability of not only getting box C correct but also box A and B.
Probability of getting all the boxes correct- 3X2X1 (this is permutation and order is important)
=6.
Prob= 1/6.
C wins.
So we can picture this in the following way:
3 boxes each with certain amount of money: Box A-($1) Box B- $100) BoxC-($1000)
Winning more than $1000 dollars means, that we need to find out the probability of not only getting box C correct but also box A and B.
Probability of getting all the boxes correct- 3X2X1 (this is permutation and order is important)
=6.
Prob= 1/6.
C wins.
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ArunangsuSahu
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Total No Of Ways for the arrangement of keys = 3!=6
To get more than $1000 AC or BC has to be prperly keyin. But that implies 3rd one also goes in the right box. It would have been different with 4 boxes
So only One Way
Probability = 1/6
To get more than $1000 AC or BC has to be prperly keyin. But that implies 3rd one also goes in the right box. It would have been different with 4 boxes
So only One Way
Probability = 1/6
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ronnie1985
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The 1000 dollar must open and any one of teh other must also opoen => all the boxes must be opened.
P = (1/3)*(1/2)*(1/1) = 1/6
P = (1/3)*(1/2)*(1/1) = 1/6
Follow your passion, Success as perceived by others shall follow you
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moussaobeid
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GMATGuruNY wrote: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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Ganesh hatwar
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probability = Favorable number of ways / total number of waysDanaJ 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!
= 3/6 = 1/2
Did it manulayy
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SoftwareDrone
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Forgive me, but I am totally lost.
If the contestant is given three keys to three boxes that collectively contain 1101 dollars, and they can open any of the boxes they want, then they are always going to open all three boxes, and the probability of them collecting more than 1000 dollars (i.e. 1101 dollars) is 1.
Right?
If the contestant is given three keys to three boxes that collectively contain 1101 dollars, and they can open any of the boxes they want, then they are always going to open all three boxes, and the probability of them collecting more than 1000 dollars (i.e. 1101 dollars) is 1.
Right?
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NicoleWhite
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This is pretty easy since there is only one favorable outcome. But could someone comment on the correct answer if, say, the question asked the probability of winning at least $1000?
We still have 3! = 6 possible outcomes, but more than one favorable outcome. Please let me know if this would be correct:
Case 1: Get $1000 correct but other two wrong
Case 2: Get $1000 correct and other two correct
= 2/6 = 1/3
Yeah?
We still have 3! = 6 possible outcomes, but more than one favorable outcome. Please let me know if this would be correct:
Case 1: Get $1000 correct but other two wrong
Case 2: Get $1000 correct and other two correct
= 2/6 = 1/3
Yeah?
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surmilsehgal
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Suppose boxes are A B C
& keys are 1 2 3
now to win more than 1000$ he has to get all the keys in the right box.
so our desired event is 1
and total possible ways of arranging the keys is = 3P1 =6 ways
or the ways are as follows:
A-1, B-2, C-3
A-1, B-3, C-2
B-1, A-2, C-3
B-1, A-3, C-2
C-1, A-2, B-3
C-1, A-3, B-2
so the probability is 1/6
& keys are 1 2 3
now to win more than 1000$ he has to get all the keys in the right box.
so our desired event is 1
and total possible ways of arranging the keys is = 3P1 =6 ways
or the ways are as follows:
A-1, B-2, C-3
A-1, B-3, C-2
B-1, A-2, C-3
B-1, A-3, C-2
C-1, A-2, B-3
C-1, A-3, B-2
so the probability is 1/6
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rajeshsinghgmat
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C) 1/6
Let the box containing the money be A,B and C.
Let a,b and c be the respective keys.
Following are the six combinations of Boxes and Keys.
A(a) B(b) C(c)
A(a) B(c) C(b)
A(b) B(a) C(c)
A(b) B(c) C(a)
A(c) B(a) C(b)
A(c) B(b) C(a)
Let the box containing the money be A,B and C.
Let a,b and c be the respective keys.
Following are the six combinations of Boxes and Keys.
A(a) B(b) C(c)
A(a) B(c) C(b)
A(b) B(a) C(c)
A(b) B(c) C(a)
A(c) B(a) C(b)
A(c) B(b) C(a)
