Problem Solving

AAPL wrote:Veritas Prep

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. 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

OA C

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

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

AAPL wrote:Veritas Prep

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. 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

OA C

Let's call the boxes A, B, and C, and their correct keys are 1, 2, and 3, respectively. Since there are 3! = 6 ways to assign the 3 keys to the 3 boxes, let's analyze each of these 6 possible outcomes and calculate the amount of money that will be won for each:

Combo 1: A-1, B-2, C-3 --> In this combo, we see that he will get $1101.
Combo 2: A-1, B-3, C-2 --> In this combo, we see that he will get $1.
Combo 3: A-2, B-1, C-3 --> In this combo, we see that he will get $1000.
Combo 4: A-2, B-3, C-1--> In this combo, we see that he will get $0.
Combo 5: A-3, B-1, C-2 --> In this combo, we see that he will get $0.
Combo 6: A-3, B-2, C-1 --> In this combo, we see that he will get $100.

We see that of these 6 possible outcomes, only the first one gets him more than $1000. Therefore, the probability is 1/6.

Alternate Solution:

We should notice that the only way the contestant can win more than $1000 is if all keys are assigned to the correct boxes. The reason for this is that as soon as the key for the $1000 box and one of the other boxes are assigned correctly, the only remaining key must belong to the only remaining box.

There are 3! = 6 ways to assign the keys to the boxes and in only one of these assignments the contestant is winning more than $1000; therefore the probability is 1/6.

Answer: C