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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 isA. \(\frac{^7P_5}{7^5}\) B. \(\frac{7^5}{^7P_5}\) C. \(\frac{6}{^6P_5}\) D. \(\frac{^5P_5}{5^5}\) |
Answer» 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 |
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