Question

If four letters are placed into 4 addressed envelopes at random, the probability that exactly three letters will go wrong is

• A. $\frac{1}{2}$
• B. $\frac{1}{3}$
• C. $\frac{1}{4}$
• D. $0$

Answer 1

Answer: (B)

$\frac{1}{3}$

Let the Ideal arrangement be A-B-C-D
Consider if A is rightly placed.
Then the number of ways in which the rest of the 3 letters will be wrongly placed is 2.
A-C-D-B
A-D-B-C
Similarly consider B is rightly placed
Then the number of ways in which the rest of the 3 letters will be wrongly placed is 2.
And so on.
Hence for 4 letters, it will be $4\times 2$
Hence the number of ways in which 3 out of 4 letters are wrongly placed =8.
Total number of ways of arranging the 4 letters
$=4!$
$=24$.
Hence the required probability is
$=\frac{4\times 2}{4!}$
$=\frac{8}{24}$
$=\frac{1}{3}$.