Problem Solving

Tanya prepared four different letters and four corresponding envelopes, each envelope with one correct address on it. If the letters were randomly placed in the envelopes, what is the prob that only 1 letter would be put into its correct envelope?

Answer is 1/3...but how?

Do the following trick. Let's represent the relation between letters and envelops in this way

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

If I want to know the possible combinations I have to permute the envelops, for instance another combination would be

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

So the total possible combinations are the permutations of the envelops 4!

Now we want the combinations that only have one match correct.

Let's say that the first match is correct

L1 ---------- E1

now we have to generate the other combinations but without a match

L2 ---------- E2
L3 ---------- E3
L4 ---------- E4

For L2 we have two possible envelops E3 and E4.
if you pick E3, for E3 you have to pick E4 since you cannot pick E4 for L4.
if you pick E4, for E3 you have to pick E2 since you cannot pick E3 for L3.

So given the choice E3 or E4 the others are automatic. You only have two options.

Since we assume that the first match is correct but this can happen in any of the other
we have 4 x 2 = 8 combinations in which only one letter is correctly addressed.

Probability = desire outcomes/ total = 8/4! = 1/3

total number of ways in which the letters can be kept in 4 envelops
=4 x3 x 2 x1= 24

one letter should be placed in one right envelop so it has 1 place
the next letter can be placed in only 2 places i.e. 3-1 for the right envelop
The next letter can be kept 2 places
and the same for the last letter
so the required probability is 1 X 2 x 2 x 2 / 24
=1/3

The way I think is when they say "only one" letter in correct envelop makes it real easy.

If you want only one letter in correct envelop that means all other 3 letters in wrong envelop. Find out probability of that.

For other 3 letters, since one of the remaining 3 envelop in correct they can put in one of other two envelopes only. So there are only only two ways the second/third/forth letter can be put in wrong out of three....so 1/3

total number of ways in which the letters can be kept in 4 envelops
=4 x3 x 2 x1= 24

one letter should be placed in one right envelop so it has 1 place
the next letter can be placed in only 2 places i.e. 3-1 for the right envelop
The next letter can be kept 2 places
and the same for the last letter
so the required probability is 1 X 2 x 2 x 2 / 24
=1/3