Problem Solving

I was wondering whether there is any elegant way of handling the question below.

John prepared 4 different letters to be sent to 4 different addresses. For each letter, he prepared an envelope with its correct address. If the 4 letters are to be put into the 4 envelopes at random, what is the probability that only 1 letter will be put into the envelope with its correct address?

a) 1/24
b) 1/8
c) 1/4
d) 1/3
e) 3/8

OA: d

Anindya Madhudor wrote:I was wondering whether there is any elegant way of handling the question below.

John prepared 4 different letters to be sent to 4 different addresses. For each letter, he prepared an envelope with its correct address. If the 4 letters are to be put into the 4 envelopes at random, what is the probability that only 1 letter will be put into the envelope with its correct address?

a) 1/24
b) 1/8
c) 1/4
d) 1/3
e) 3/8

OA: d

Suppose there are letters A, B, C, and D which needs to be placed in envelopes 1, 2, 3 and 4 respectively.

There are 4! = 24 total ways you can place the letters.

We need to find the probability for RWWW (i.e. A is the only letter in the correct envelope.)

RWWW means Right wrong wrong wrong.

1/4 of the ways to assign letters are of the form Axxx: ABCD, ABDC, ACBD, ACDB, ADBC, ADCB
2/3 of those have a "wrong" letter in the second envelope (i.e. not B): ACBD, ACDB, ADBC, ADCB
3/4 of these have a "wrong" letter in the third envelope (i.e. not C): ACBD, ACDB, ADBC
2/3 of these have a "wrong" letter in the forth envelope (i.e. not D): ACDB, ADBC
Combined, there is a (1/4)(2/3)(3/4)(2/3) = 1/12 chance of RWWW.

As noted above, the RWWW chance is the same as the chance of WRWW or WWRW or WWWR, so the total chance of 1R3W is 4*1/12 = 1/3.