Question

A mother bought three shirts of the same colour but of different sizes, one each for her three sons. All three shirts were kept together in a box in a dark room. Each of the three boys took one shirt at random from the box. What is the probability that none of the boys ends up with his own shirt?

• A. \(\frac{1}{2}\)
• B. \(\frac{1}{3}\)
• C. \(\frac{2}{3}\)
• D. \(\frac{1}{4}\)

Hint:

Count the total 3! = 6 ways to hand out the shirts, then count the arrangements with no match, either directly or with the derangement formula, to get 2 out of 6.

Answer 1

The Correct Option is B

Step 1: Recognise this as a derangement problem.
A derangement is an arrangement where nothing goes to its own correct place. Here we want a derangement of 3 shirts among 3 boys, so none of them ends up with the shirt actually meant for him.

Step 2: Use the derangement formula.
For $n$ items, the number of derangements is
$!n = n!\left(1 - \frac{1}{1!} + \frac{1}{2!} - \frac{1}{3!} + \cdots \pm \frac{1}{n!}\right)$
For $n = 3$:
$!3 = 3!\left(1 - 1 + \frac{1}{2} - \frac{1}{6}\right) = 6\left(\frac{1}{2} - \frac{1}{6}\right) = 6 \times \frac{1}{3} = 2$
So there are 2 ways to give out the shirts so that no boy gets his own.

Step 3: Compare with the total number of ways.
The shirts can be given to the 3 boys in $3! = 6$ total ways, all equally likely since the boys pick blindly in the dark.

Step 4: Find the probability.
$P(\text{derangement}) = \dfrac{!3}{3!} = \dfrac{2}{6} = \dfrac{1}{3}$
This matches the well known result that the chance of a complete derangement of 3 items is always $\frac{1}{3}$.
\[ \boxed{\frac{1}{3}} \]