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: Understand the setup.
There are three shirts of the same colour but different sizes, one meant for each of the three sons. All three shirts are mixed up in a box in a dark room, so each boy grabs one shirt purely at random. We need the chance that not even one boy ends up with his own shirt.

Step 2: Count the total number of ways to hand out the shirts.
Label the sons 1, 2, 3 and their correct shirts 1, 2, 3. The three shirts can be given to the three boys in \(3! = 3 \times 2 \times 1 = 6\) different ways. Each of these 6 arrangements is equally likely, since the boys pick at random in the dark.

Step 3: List the arrangements where no boy gets his own shirt.
Write each arrangement as the shirt each boy gets, in the order boy 1, boy 2, boy 3:
(1,2,3) fails, since boy 1 gets his own shirt.
(1,3,2) fails, since boy 1 gets his own shirt.
(2,1,3) fails, since boy 3 gets his own shirt.
(2,3,1) works, no boy gets his own shirt.
(3,1,2) works, no boy gets his own shirt.
(3,2,1) fails, since boy 2 gets his own shirt.
So exactly 2 out of the 6 arrangements are complete mismatches, these are called derangements.

Step 4: Compute the probability.
\[ P(\text{no boy gets his own shirt}) = \frac{\text{favourable arrangements}}{\text{total arrangements}} = \frac{2}{6} = \frac{1}{3} \]

Step 5: Check the other options.
\(\frac{1}{2}\) would need 3 out of 6 arrangements to be mismatches, but we listed only 2.
\(\frac{2}{3}\) would need 4 out of 6 arrangements to be complete mismatches, again more than what we found.
\(\frac{1}{4}\) does not match a whole number of arrangements out of 6 at all.

Final Answer:
The probability that none of the boys gets his own shirt is \(\frac{1}{3}\).
\[ \boxed{\frac{1}{3}} \]