singhsa wrote:There are 4 letters and 4 corresponding envelopes. If we put the 4 letters into the envelopes at random, what is the probability that only one letter was into the exact envelope?

Let's call the envelopes E1, E2, E3 and E4.

P(only E1 gets the correct letter):

P(E1 gets the correct letter) = 1/4 (4 letters total, 1 of them correct)
P(E2 gets the wrong letter) = 2/3 (3 letters left, 2 of them wrong)
P(E3 gets the wrong letter) = 1/2 (2 letters left, 1 of them wrong)
P(E4 gets the wrong letter) = 1/1 (1 letter left, and it must be wrong since we placed the correct letter in either E2 or E3)

Since we need all of these events to happen, we multiply the fractions:

1/4 * 2/3 * 1/2 * 1/1 = 1/12.

Since each envelope has the same probability of getting the correct letter and we have 4 envelopes total, we need to multiply by 4:

4 * 1/12 = 1/3.

GMATGuruNY wrote:

singhsa wrote:There are 4 letters and 4 corresponding envelopes. If we put the 4 letters into the envelopes at random, what is the probability that only one letter was into the exact envelope?

Let's call the envelopes E1, E2, E3 and E4.

P(only E1 gets the correct letter):

P(E1 gets the correct letter) = 1/4 (4 letters total, 1 of them correct)
P(E2 gets the wrong letter) = 2/3 (3 letters left, 2 of them wrong)
P(E3 gets the wrong letter) = 1/2 (2 letters left, 1 of them wrong)
P(E4 gets the wrong letter) = 1/1 (1 letter left, and it must be wrong since we placed the correct letter in either E2 or E3)

Since we need all of these events to happen, we multiply the fractions:

1/4 * 2/3 * 1/2 * 1/1 = 1/12.

Since each envelope has the same probability of getting the correct letter and we have 4 envelopes total, we need to multiply by 4:

4 * 1/12 = 1/3.

Wonderful Explanation!!! Thanks a lot Mitch

I did this question like this -

Total ways to assign letters to envelopes = 4! = 24

Total ways to assign only one correct letter = 4 * 2 (2 ways to assign right envelope to only one letter).

E1 E2 E3 E4
L1 L3 L4 L2
L1 L4 L2 L3

P = 8/24 = 1/3