If n objects are distributed at random among n persons, the probability that at least one of them will not get anything is

1 - \frac{(n - 1)!}{n^{n - 1}}

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\frac{(n - 1)!}{n^{n}}

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1 - \frac{(n - 1)!}{n^{n}}

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None of these

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Solution

The correct option is B $1 - \frac{(n - 1)!}{n^{n - 1}}$
The first object can be given to any of the $n$ persons.
But the second, third and other objects, too, can go to any of the $n$ persons,
Therefore the total number of ways of distributing the $n$ objects randomly among $n$ persons is $n^{n}$ .
There are $^{n} P_{n} = n!$ ways in which each person gets exactly one objects, so the probability of this happening is
$\frac{n!}{n^{n}} = \frac{(n - 1)!}{n^{n - 1}}$
Hence the probability that at least one person does not get any objects is
$1 - \frac{(n - 1)!}{n^{n - 1}}$

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Similar questions

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