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04 Feb 2025, 01:30
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C
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In company A, each of the p employees was selected in all of the n management schools in the State of Michigan. If each of the p employees chose one of those schools, what is the probability that all the p employees chose the same school?
A. n/p
B. p/n
C. 1/n^(p - 1)
D. 1/n^p
E. 1/p^n
A. n/p
B. p/n
C. 1/n^(p - 1)
D. 1/n^p
E. 1/p^n
Updated on: 04 Feb 2025, 18:12
Originally posted by tgsankar10 on 04 Feb 2025, 04:43.
Last edited by tgsankar10 on 04 Feb 2025, 18:12, edited 1 time in total.
Last edited by tgsankar10 on 04 Feb 2025, 18:12, edited 1 time in total.
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Each of the \(p\) employees was selected in all the \(n\) schools. Hence each employee has \(n\) choices to make.
The probability of one school being chosen by one employee is \(\frac{1}{n}\)
All the employees can select the same school in only 1 way. Hence the probability of one school being chosen by all \(p\) employees is \(\frac{1}{n^p}\)
Since there are n schools, the probability that the same school being chosen by all is \(\frac{n}{n^p}=\frac{1}{n^{(p-1)}}\)
Answer: C
The probability of one school being chosen by one employee is \(\frac{1}{n}\)
All the employees can select the same school in only 1 way. Hence the probability of one school being chosen by all \(p\) employees is \(\frac{1}{n^p}\)
Since there are n schools, the probability that the same school being chosen by all is \(\frac{n}{n^p}=\frac{1}{n^{(p-1)}}\)
Answer: C
04 Feb 2025, 12:48
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Favourable outcome = There are n possibilities to chose the same school
Total outcomes = Every employee has n options, n^p
Probability = n/(n^p) = 1/(n^(p-1))
Total outcomes = Every employee has n options, n^p
Probability = n/(n^p) = 1/(n^(p-1))
Bunuel
In company A, each of the p employees was selected in all of the n management schools in the State of Michigan. If each of the p employees chose one of those schools, what is the probability that all the p employees chose the same school?
A. n/p
B. p/n
C. 1/n^(p - 1)
D. 1/n^p
E. 1/p^n
A. n/p
B. p/n
C. 1/n^(p - 1)
D. 1/n^p
E. 1/p^n
