Question

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

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

Hint:

When two items must always be together, treat them as a block and reduce the number of entities to arrange.

Answer 1

The Correct Option is C

We are asked to find the probability that A and E are always together in the queue of 5 persons. Step 1: Total number of arrangements The total number of ways to arrange 5 persons is: \[ 5! = 120 \] Step 2: Treat A and E as a block If A and E are always together, treat them as a single block. Now, we have 4 entities to arrange: the block (A and E), B, C, and D. The number of ways to arrange these 4 entities is: \[ 4! = 24 \] Step 3: Arranging A and E within the block Within the block, A and E can be arranged in \( 2! = 2 \) ways. Step 4: Probability calculation Therefore, the total number of favorable outcomes is \( 4! \times 2! = 24 \times 2 = 48 \). The probability that A and E are always together is the ratio of favorable outcomes to total outcomes: \[ P(\text{A and E together}) = \frac{48}{120} = \frac{3}{5} \] Thus, the correct answer is \( \frac{3}{5} \).