If n letters were put in n envelopes at random, what is the probability that eactlly n-1 of them were in the correct envelopes?
That depends upon how many letters can go into one envelope. If that number is one per envelope, then you cannot have just one in the wrong envelope. The one that is supposed to be in that envelope would also be in a wrong envelope.
1/384
To find the probability that exactly n-1 letters were in the correct envelopes when n letters are put in n envelopes at random, we can apply the principle of derangement.
The principle of derangement states that the number of derangements of n objects (in this case, letters) is the number of arrangements in which none of the objects occupy their original position. Let's denote this number as D(n).
The formula for D(n) is given by:
D(n) = n! * (1/0! - 1/1! + 1/2! - 1/3! + ... + (-1)^n / n!)
Now, let's calculate the probability that exactly n-1 letters were in the correct envelopes.
Firstly, note that there are a total of n! possible arrangements when n letters are put in n envelopes.
Next, for exactly n-1 letters to be in the correct envelopes, there will be only one letter in the incorrect envelope. This means we have (n-1) choices for the misplaced letter.
Therefore, the probability P of exactly n-1 letters being in the correct envelopes is given by:
P = D(n-1) / n!
So, the probability that exactly n-1 letters were in the correct envelopes when n letters are put in n envelopes at random is D(n-1) / n!.
To find the probability that exactly n-1 letters were in the correct envelopes, we need to use the principle of inclusion-exclusion.
Let's assume there are n letters (labeled as letter 1, letter 2, ..., letter n) and n envelopes (labeled as envelope 1, envelope 2, ..., envelope n).
The total number of possible arrangements of putting n letters in n envelopes is n!.
Now, let's consider the favorable cases where exactly n-1 letters are in the correct envelopes. Let's select one letter, say letter 1, and put it in the correct envelope (envelope 1). Now we have (n-1) letters and (n-1) envelopes left.
The remaining (n-1) letters can be arranged in any way among the remaining (n-1) envelopes, which can be done in (n-1)! ways.
However, we also need to consider that the remaining (n-1) letters can be in the correct envelopes in different combinations.
Using the principle of inclusion-exclusion, the probability that exactly n-1 letters were in the correct envelopes can be calculated as follows:
P(exactly n-1 letters in correct envelopes) = (Number of favorable cases)/(Total number of possible cases)
P(exactly n-1 letters in correct envelopes) = [1 * (n-1)! - C(1,1) * (n-1)! + C(2, 1) * (n-1)! - C(3, 1) * (n-1)! + ... + (-1)^(n-1) * C(n-1, 1) * (n-1)!] / n!
Here, C(r, m) represents the combination of selecting m elements from r elements.
Simplifying the above equation, we can find the probability that exactly n-1 letters were in the correct envelopes.
Note: For larger values of n, calculating the exact probability can become computationally intensive. In such cases, approximation methods or simulations can be used to estimate the probability.