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TS EAMCET · Maths · Probability

5 persons entered a lift cabin in the cellar of a 7 -floor building apart from cellar. If each of them independently and with equal probability can leave the cabin at any floor out of the 7 floors beginning with the first, then the probability of all the 5 persons leaving the cabii at different floors is

  1. A \(\frac{360}{2401}\)
  2. B \(\frac{5}{54}\)
  3. C \(\frac{51}{71}\)
  4. D \(\frac{5}{18}\)
Verified Solution

Answer & Solution

Correct Answer

(A) \(\frac{360}{2401}\)

Step-by-step Solution

Detailed explanation

Total floors \(=7\) Total number of ways to exit \(=7^5\) If 5 persons leave at 5 different floors then number of ways \(=\) arranging 5 people in 7 ways \(={ }^7 \mathrm{P}_5\) Required probability \(=\frac{{ }^7 \mathrm{P}_5}{7^5}\) \[ =\frac{7 !}{2 ! .7^5}=\frac{360}{2401} \]