Question

1. 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

0 beacause there are total 10 letters and only 10 envelopes

Answer 2

Answer:

$\textbf{Hence, The probability that at least one letter is placed}\\\textbf{in wrong addressed envelope is = }\bf\frac{3628799}{3628800}$

Step-by-step explanation:

This is a question based on permutations.

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

$_{r}^{n}\textrm{P}=\frac{n!}{(n-r)!}$

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

$_{10}^{10}\textrm{P}=\frac{10!}{(0)!}$

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

$\textbf{Hence, The probability that at least one letter is placed}\\\textbf{in wrong addressed envelope is = }\bf\frac{3628799}{3628800}$