Problem Solving

swerve wrote:On a game show, there are five boxes, each of which contains a prize. Each box has two locks on it, and to open the box and win the prize a player must have both keys to that box. All ten keys are placed in a bowl and a player will draw four keys. If each lock/key combination is unique, what is the probability that a particular player will win at least one prize by drawing four keys?

A. 2/5
B. 11/21
C. 13/21
D. 2/3
E. 16/21

To win at least one prize, he must open either one box or both boxes.

Let's find the probability he can open both boxes first. To open 2 boxes, he must have all 4 keys to the 2 boxes he has chosen. The number of ways of picking 4 keys from 10 is 10C4 = (10 x 9 x 8 x 7)/(4 x 3 x 2) = 10 x 3 x 7 = 210, and only 4C4 = 1 set of 4 keys are the right keys for the 2 boxes he has chosen; thus, the probability of choosing the right set of keys is 1/210. However, there are 5C2 = (5 x 4)/2 = 10 ways to choose 2 boxes from 5, and each of these 10 selections has 1/210 chance of opening the 2 boxes; therefore, the probability of opening 2 boxes is actually 10 x 1/210 = 1/21.

Of course, he can also open only 1 box. To do that, he must have the 2 keys to that box. Since he has 4 keys, the number of ways the 4 keys contain the 2 keys he needs is 4C2 = (4 x 3)/2 = 6, so the probability he can open 1 box is 6/210. However, it's possible that the 2 keys can open either one of the two boxes he has chosen and there are 10 ways to choose 2 boxes from 5 (as mentioned above), so the probability of opening exactly 1 box is actually 2 x 10 x 6/210 = 120/210 = 12/21.

Therefore, the probability of winning at least one prize is

1/21 + 12/21 = 13/21

Alternate Solution:

We will use the formula:

P(opening at least one box) = 1 - P(opening no boxes)

Thus, we need to calculate the probability that none of the boxes can be opened using the four selected keys.

After the first key is selected, there is a 8/9 probability that the second key is not the matching pair of the first key.

After two keys are selected, there are 2 keys among the remaining 8 keys that form a pair with one of the already selected keys, which means any of the 8 - 2 = 6 keys will not form a pair with one of the previously selected keys. Thus, the probability that the third selected key does not form a pair with the previous two keys is 6/8 = 3/4.

Finally, after three keys are selected, there are 3 keys among the remaining 7 keys that form a pair with one of the already selected keys, which means any of the 7 - 3 = 4 keys will not form a pair with one of the previously selected keys. Thus, the probability that the fourth selected key does not form a pair with the previous three keys is 4/7.

In total, the probability that the four keys do not contain a pair which will open a box is 8/9 x 3/4 x 4/7 = 8/21. Therefore, the probability that the contestant will win at least one prize is 1 - 8/21 = 13/21.

Answer: C