On a game show, a contestant is given three keys, each of

Question

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

(A) 1/9
(B) 1/8
(C) 1/6
(D) 1/3
(E) 1/2

mikemcgarry · Accepted Answer

Hi, there. I'm happy to help with this.

This is a cool question!

Let's call the boxes that contain $1, $100, and $1000, respectively, Box A, Box B, Box C. These are opened, respectively, by Key A, Key B, and Key C. We want to know the probability of winning more than $1000. Notice that if the distribution of keys is: Box A = Key B Box B = Key A Box C = Key C then the contestant wins exactly $1000, not more than $1000. The only configuration that leads to winning more than $1000 is: Box A = Key A Box B = Key B Box C = Key C i.e., getting all three keys correct. That's the only way to be more than $1000. So, really, the question can be rephrased: what is the probability of guessing the order of keys so that each key matches the correct box? Well, for a set of three items, the number of possible permutations is 3! = 3*2*1 = 6. Of those 6 possible permutations, only one of them leads to all three keys being paired with the right box. So, the answer is Probability = 1/6, answer = C Does that make sense? Here's another free practice question involving permutations: https://gmat.magoosh.com/questions/857 The question at that link should be followed by a video solution. Please let me know if you have any other questions. Mike

Bunuel · Answer

Total # of assignment of keys to boxes is 3!=6. Notice that it's impossible to assign only two keys correctly, so to win more than $1000 a contestant must assign all keys to the right boxes. Probability of that is 1/6.

Answer: C.

calreg11 · Answer

thank you guys for the response.

mvictor · Answer

probability of winning 1000 is 1/3. But winning 1000 does not satisfy the condition. The contestant needs to open all boxes correctly so that to win more than 1000. The probability of winning is thus 1/3 * 1/2, since 1/2 is the probability of opening the remaining 2 boxes correctly. This results in 1/6.

TheRzS · Answer

Hi  mikemcgarry

Had the question been,  What is the probability of winning a Prize of 1000 or less?

Would the answer be  1/2 ?

Best
RzS

Cez005 · Answer

For the contestant to win more that $1000, they must unlock the $1000 box and one of the remaining boxes. However, if two boxes are unlocked it is impossible not to unlock the third box. The probability of opening the first two boxes correctly is: (1/3)*(1/2)= 1/6

ankitmani2004 · Answer

It will be 1-1/6 = 5/6.

Thanks.

BrentGMATPrepNow · Answer

There aren't many possible outcomes, so let's start by  listing all possibilities .

Let A, B and C represent the 3 boxes, and let a, b, c, be the keys for those boxes (in that same order). 
 A = $1 prize, B = $100 prize and C = $1000 prize

If the boxes are arranged as A-B-C, then these are all possible arrangements for the keys: 
 1) a - b - c. So, all 3 boxes are opened for a total prize of $1101 
2) a - c - b. Box A is opened for a total prize of $1
3) b - c - a. Zero boxes are opened for a total prize of $0
4) b - a - c. Box C is opened for a total prize of $1000
5) c - b - a. Box B is opened for a total prize of $100
6) c - a - b. Zero boxes are opened for a total prize of $0

Only 1 of the 6 possible outcomes yields a prize greater than $1000

So, P(contestant wins more than $1000) = 1/6

Answer: C

Cheers,
Brent

EMPOWERgmatRichC · Answer

Hi All,

We're told that on a game show, a contestant is given three keys, each 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. We're asked for the probability that a contestant will win more than $1000. This question can be approached in a couple of different ways; 'brute force' works really well here (as Brent has shown). There's also a couple of 'logic shortcuts' that you can use to get to the correct answer without having to do too much math.

To start, with 3 keys, there are 3! = (3)(2)(1) = 6 different ways to assign the keys to the boxes.

From a logic standpoint, you can win 0 boxes, 1 box or 3 boxes -- you CANNOT win just 2 of the boxes (re: if you matched 2 keys and opened 2 boxes, then the last key would have to fit the last box... meaning that you would win all 3 boxes). The only way to win MORE than $1,000 is to win the "$1,000 box" and both of the other 2 boxes (since you can't win just one of them in addition to the $1,000 box).

With 6 possible arrangements of keys - and just 1 way to win all 3 boxes - the probability of winning more than $1,000 is 1/6.

Final Answer:  C

GMAT assassins aren't born, they're made, 
Rich

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