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In company A, each of the p employees was selected in all of the n

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Aabhash777
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In company A, each of the p employees was selected in all of the n [#permalink] New post   25 Sep 2022, 03:15
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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]
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C
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ankitgoswami
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In company A, each of the p employees was selected in all of the n [#permalink] New post   25 Sep 2022, 14:26
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Aabhash777 wrote:
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]


p employees can select one of n management schools in \(n^p\) ways

Similarly, in only one way out of \(n^p\) ways, all p employees can select one school

No of ways of selecting that one school = n

So prob = \(\frac{1* n}{n^p }\)
= \(\frac{1}{n^{p-1}}\)

Ans is C
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Re: In company A, each of the p employees was selected in all of the n [#permalink] New post   27 Sep 2022, 09:27
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Aabhash777 wrote:
I did not get why is it n^p and not p^n?

ankitgoswami wrote:
Aabhash777 wrote:
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]


p employees can select one of n management schools in \(n^p\) ways

Similarly, in only one way out of \(n^p\) ways, all p employees can select one school

No of ways of selecting that one school = n

So prob = \(\frac{1* n}{n^p }\)
= \(\frac{1}{n^{p-1}}\)

Ans is C


One employees can select one of n management schools in n ways

Two employees can select one of n management schools in n x n ways
.
.
.
P employees can select one of n management schools in n x n x n x n x n x n x n x.....(p times) ways

=\(n^p\) ways

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Re: In company A, each of the p employees was selected in all of the n [#permalink] New post   26 Sep 2022, 21:58
I did not get why is it n^p and not p^n?

ankitgoswami wrote:
Aabhash777 wrote:
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]


p employees can select one of n management schools in \(n^p\) ways

Similarly, in only one way out of \(n^p\) ways, all p employees can select one school

No of ways of selecting that one school = n

So prob = \(\frac{1* n}{n^p }\)
= \(\frac{1}{n^{p-1}}\)

Ans is C
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Re: In company A, each of the p employees was selected in all of the n [#permalink] New post   27 Sep 2022, 05:25
Aabhash777 wrote:
I did not get why is it n^p and not p^n?

ankitgoswami wrote:
Aabhash777 wrote:
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]


p employees can select one of n management schools in \(n^p\) ways

Similarly, in only one way out of \(n^p\) ways, all p employees can select one school

No of ways of selecting that one school = n

So prob = \(\frac{1* n}{n^p }\)
= \(\frac{1}{n^{p-1}}\)

Ans is C



Using some actual numbers can help.

2 people, 3 schools.

People are assigned schools, schools aren't assigned people.

If schools were assigned to people, you could have a situation where 1 person is assigned to more than 1 school or no school at all.

So each of the 2 people above has 3 choices of school and for each selection 1 makes, the other has 3 choices

3*3= 3^2 , equivalent to n^p

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Re: In company A, each of the p employees was selected in all of the n [#permalink] New post   27 Sep 2022, 07:46
Given: In company A, each of the p employees was selected in all of the n management schools in the State of Michigan.

Asked: If each of the p employees chose one of those schools, what is the probability that all the p employees chose the same school?

Favorable ways = n
Total ways = n^p

Probability that all the p employees chose the same school = n/n^p = 1/n^{p-1}

IMO C
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Re: In company A, each of the p employees was selected in all of the n [#permalink]
27 Sep 2022, 07:46
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