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

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!

GMATGuruNY wrote:

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!

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.

[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]

GMATGuruNY wrote:

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!

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.

[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]

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!