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Q.
There are four letters and four envelopes, the letters are placed into the envelopes at random, the probability that all letters are placed in the wrong envelopes is
Probability - Part 2
Solution:
We know from the above given formula that probability that no letter is in right envelope out of $n$ letters and $n$ envelopes is given by
$=\left[\frac{1}{2 !}-\frac{1}{3 !}+\frac{1}{4 !}-\ldots \ldots \ldots .(-1)^{ n } \frac{1}{ n !}\right]$
Since all 4 letters are to be placed in wrong envelopes then required probability
$=\left[\frac{1}{2 !}-\frac{1}{3 !}+\frac{1}{4 !}\right]=\frac{1}{2}-\frac{1}{6}+\frac{1}{24}=\frac{3}{8}$