Question

If 10 letters are to be placed in 10 addressed envelopes, then what is the probability that at least one letter is placed in wrong addressed envelope?

Answer 1

This is a question based on permutations.

The formula to find the permutations of n objects taken 'r' at a time is

nPr = $\frac{n!}{(n-r)!}$

Here there are ten letters and 10 envelopes.

So n= 10 & r = 10

The total number of ways we can place the letters in different envelopes is

10P10

$\frac{10!}{(10-10)!}$

$\frac{10!}{0!}$

But 0! = 1

So total arrangements = 10! = 3628800

Now out of all these arrangements there is one arrangement that all the letters are in the correct envelopes.

So the number of arrangements that at least one letter is in incorrect envelopes = 3628800-1= 3628799

Hence the probability that at least one letter is not in the correct envelope is

= $\frac{3628799}{3628800}$

The probability that at least one letter is placed in wrong addressed envelope is

$\frac{3628799}{3628800}$