Question

Five persons A, B, C, D and E are in the queue of a shop. The probability that A and E are always together is:
a) $\dfrac{1}{4}$
b) $\dfrac{2}{3}$
c) $\dfrac{2}{5}$
d) $\dfrac{3}{5}$

Answer 1

Hint: In the given question, we are provided with the number of people standing in the queue and are asked to evaluate probability considering person A and person B are always together. Since A and E are together, we consider them as 1 person. Further, it could be solved using basic knowledge of probability.

Complete step by step Solution: We are given that there are $5$ persons viz. A, B, C, D, E standing in the queue
Total number of ways they could be arranged = $5!$
Also, people A and E are together
We can consider AE =$1$ person
They can be arranged as AE and EA
The above statement implies, Number of ways A and E can be arranged = $2!$
Now the remaining persons could be arranged in $4!$ ways.
Therefore, the Total number of ways = $4!*2!$
To evaluate the probability of person A and person E always together;
= $\dfrac{4!*2!}{5!}$
= $\dfrac{(4*3*2*1)(2*1)}{(5*4*3*2*1)}$
= $\dfrac{2}{5}$
Thus, the probability that A and E are always together is $\dfrac{2}{5}$

Option c) $\dfrac{2}{5}$ is correct.

Note: Whenever two or more elements are mentioned to be always together, it is expected to consider them as a single element. There is a need to calculate the number of ways these elements could be arranged separately. This key point helps to solve all such types of problems.