Question

Five persons A, B, C, D and E are in queue at a shop. The probability that A and B are always together is.

• A. \( \frac{2}{5} \)
• B. \( \frac{1}{4} \)
• C. \( \frac{3}{5} \)
• D. \( \frac{2}{3} \)

Hint:

When two objects must be together, treat them as a single unit and then calculate the arrangements.

Answer 1

The Correct Option is A

There are 5 people, and we want to find the probability that A and B are always together. 1. The total number of ways to arrange 5 people in the queue is \( 5! \): \[ 5! = 120 \] 2. Now, if A and B are always together, treat them as one unit. So, we have 4 units to arrange (A-B, C, D, E). The number of ways to arrange these 4 units is \( 4! \): \[ 4! = 24 \] 3. Within the A-B unit, A and B can be arranged in \( 2! = 2 \) ways. 4. Therefore, the number of favorable outcomes is \( 4! \times 2! = 24 \times 2 = 48 \). 5. The probability is the ratio of favorable outcomes to total outcomes: \[ \text{Probability} = \frac{48}{120} = \frac{2}{5} \] Thus, the correct answer is (A) \( \frac{2}{5} \).