You type four letters to four different people and address the envelopes. If you then insert the letters into the envelopes randomly, what is the probability that exactly three letters will go into the correct envelopes?
1/2
1/3
1/4
0
Explain your answer.
To solve this problem, we can use the concept of derangements. A derangement is a permutation of a set in which none of the elements appear in their original positions. In this case, we are looking for the probability that exactly three letters go into the correct envelopes, which can be visualized as a derangement of one letter.
Let's start by calculating the total number of possible outcomes. Since there are four letters to be inserted randomly into four envelopes, there are 4! (4 factorial) possible permutations, which is equal to 4 x 3 x 2 x 1 = 24.
Next, let's determine how many derangements are possible. The number of derangements for n objects can be denoted by !n (read as "subfactorial n"). The formula for !n is given by:
!n = n!(1/0! - 1/1! + 1/2! - 1/3! ... + (-1)^n / n!)
For n = 1, the formula simplifies to:
!1 = 1!(1/0! - 1/1!) = 1(1 - 1) = 0
Therefore, there are zero derangements with one letter.
Finally, let's calculate the probability that exactly three letters will go into the correct envelopes. Since there are 0 derangements of one letter out of 24 possible outcomes, the probability is 0/24, which simplifies to 0.
So, the answer to the question is 0.