Question

Five persons entered the lift cabin on the ground floor of an 8 floored house. Suppose that each of them independently and with equal probability can leave the cabin at any floor beginning with the first, then the probability of all 5 persons leaving at different floor is

A. \(\frac{^7P_5}{7^5}\) 

B. \(\frac{7^5}{^7P_5}\) 

C. \(\frac{6}{^6P_5}\) 

D. \(\frac{^5P_5}{5^5}\) 

Answer 1

As building has 8 floors including ground. 

So, they have 7 options to leave the lift. 

∴ total ways in which persons can leave the lift = 7⁵ 

As the number of ways in which all persons leave in a different floor can be given by: 

7× 6× 5× 4× 3 = ⁷P₅ 

∴ P(that all leave lift in different floor) = \(\frac{^7P_5}{7^5}\) 

Our answer matches with option (a) 

∴ Option (a) is the only correct choice