Question

A father purchases dresses for his three daughters. The dresses are of the same colour but of different sizes. The dress is kept in a dark room. What is the probability that all the three will not choose their own dress?

• A. \( \frac{2}{3} \)
• B. \( \frac{1}{3} \)
• C. \( \frac{1}{6} \)
• D. \( \frac{1}{9} \)
• E.

Hint:

The number of derangements of \( n \) objects is the number of ways to arrange the objects such that no object appears in its original position. The formula for the number of derangements is: \[ !n = n! \left(1 - \frac{1}{1!} + \frac{1}{2!} - \cdots + \frac{(-1)^n}{n!}\right) \]

Answer 1

The Correct Option is B

The total number of ways in which the three daughters can select their dresses is given by: \[ 3! = 6 \] Step 1: To determine the number of favorable cases where none of the daughters picks her own dress, we apply the concept of derangements. The number of derangements for 3 objects is: \[ !3 = 2 \] Step 2: The probability that none of the daughters selects her own dress is calculated as: \[ \frac{!3}{3!} = \frac{2}{6} = \frac{1}{3} \] Thus, the probability that all three daughters will not pick their own dress is \( \boxed{\frac{1}{3}} \).