A secretary addresses 50 different letters and envelopes to 50 different people, but the letters are randomly mixed before being put into envelopes. What is the probability that at least one letter gets into the correct envelope?
.632
To find the probability that at least one letter gets into the correct envelope, we can use the principle of inclusion-exclusion.
Let's consider the opposite event, where no letter gets into the correct envelope. This is also known as a derangement. We need to count the number of derangements of the 50 letters.
The total number of possible arrangements of 50 letters is 50!.
To find the number of derangements, we can use the derangement formula: D(n) = n!(1 - 1/1! + 1/2! - 1/3! + ... + (-1)^n/n!), where n is the number of elements to be deranged.
Using this formula for n = 50, we get:
D(50) = 50!(1 - 1/1! + 1/2! - 1/3! + ... + (-1)^50/50!)
Now, we can find the probability of no letter getting into the correct envelope by dividing the number of derangements by the total number of possible arrangements:
P(no letter in correct envelope) = D(50) / 50!
Finally, we can find the probability that at least one letter gets into the correct envelope by subtracting the probability of no letter in the correct envelope from 1:
P(at least one letter in correct envelope) = 1 - P(no letter in correct envelope)
Calculating the values for P(no letter in correct envelope) and P(at least one letter in correct envelope) require a lot of computations due to the large factorials involved. However, using approximations, we can estimate that the probability of at least one letter getting into the correct envelope is about 0.63212 or approximately 63.2%.
Note: This approximation assumes a large number of possibilities and can be further improved by using more precise methods like recursion or dynamic programming for calculating the derangement formula.