Lock and key

[j_shreyans]

Each of four different locks has a matching key. The keys are randomly reassigned to the locks. What is the probability that exactly two of the keys fit the locks to which they are reassigned?

A)1/8
B)1/6
C)1/4
D)3/8
E)1/2

OAC

[GMATGuruNY]

Each of four different locks has a matching key. The keys are randomly reassigned to the locks. What is the probability that exactly two of the keys fit the locks to which they are reassigned?

a 1/8
b 1/6
c 1/4
d 3/8
e 1/8

One approach:

To determine the probability that an event happens exactly n times:

P(exactly n times) = P(one way) * (total possible ways)

For the problem above, we want to determine the probability that exactly 2 keys are correctly assigned. One way for exactly 2 keys to be correctly assigned is for the first 2 keys to be correct and for the last 2 keys to be wrong:

P(1st key is correct) = 1/4 (out of 4 keys, only 1 correct)
P(2nd key is correct) = 1/3 (out of 3 keys left, only 1 correct)
P(3rd key is wrong) = 1/2 (out of 2 keys left, 1 is wrong)
P(4th key is wrong) = 1/1 (1 key left, and it must be wrong because the correct key was put in the 3rd lock)

Since we want all of these events to happen together, we multiply the fractions. We multiply because the more events we want to happen together, the smaller the probability, and when we multiply fractions, the result keeps getting smaller. Thus, letting C = correct and W = wrong:

P(CCWW) = 1/4 * 1/3 * 1/2 * 1/1 = 1/24.

The result above represents the probability of one particular way that 2 keys could be correct: CCWW. Now we need to multiply by the total number of ways that 2 keys could be correct. Any arrangement of the letters CCWW will give us a good outcome:

Number of ways to arrange CCWW = 4!/2!*2! = 6.

Multiplying the results above, we get:

P(exactly 2 keys correct) = 1/24 * 6 = 1/4.

The correct answer is C.