Question

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

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

Answer 1

Answer: (C)

$\frac { 2 } { 5 }$

Answer 2

Total number of ways

Favourable number of ways $2.4 !$

Hence required probability $= \frac { 2.4 ! } { 5 ! } = \frac { 2 } { 5 }$.