n objects are distributed at random among n persons. The number of ways in which this can be done so that at least one of them will not get any object is
- \(n!-n\)
- \(n^n-n\)
- \(n^n-n^2\)
- \(n^n-n!\)
The Correct Option is D
Solution and Explanation
The initial step involves distributing the n objects among n persons.
Each object can be given to any of the n persons independently, so the total number of ways to distribute the objects is:
Total ways = $n^n$
We now consider the case where each person gets exactly one object. This is a perfect one-to-one assignment or permutation, which can happen in:
Favorable ways = $n!$
So, the probability that each person gets exactly one object is:
$\dfrac{n!}{n^n}$
But we are interested in the probability that at least one person does not get any object, which is the complement of the above case:
Required probability = $1 - \dfrac{n!}{n^n} = \dfrac{n^n - n!}{n^n}$
So, the final answer is: $n^n - n!$
The correct option is (D)
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Concepts Used:
Conditional Probability
Conditional Probability is defined as the occurrence of any event which determines the probability of happening of the other events. Let us imagine a situation, a company allows two days’ holidays in a week apart from Sunday. If Saturday is considered as a holiday, then what would be the probability of Tuesday being considered a holiday as well? To find this out, we use the term Conditional Probability.
Let’s discuss certain theorems of Conditional Probability:
- Let us consider a random experiment where the sample space S is considered as space and two events namely A and B happen there. Then, the formula would be:
P(S | B) = P(B | B) = 1.
Proof of the same: P(S | B) = P(S ∩ B) ⁄ P(B) = P(B) ⁄ P(B) = 1.
[S ∩ B indicates the outcomes common in S and B equals the outcomes in B].
- Now let us consider any two events namely A and B happening in a sample space ‘s’, then, P(A ∩ B) = P(A).
P(B | A), P(A) >0 or, P(A ∩ B) = P(B).P(A | B), P(B) > 0.
This theorem is named as the Multiplication Theorem of Probability.
Proof of the same: As we all know that P(B | A) = P(B ∩ A) / P(A), P(A) ≠ 0.
We can also say that P(B|A) = P(A ∩ B) ⁄ P(A) (as A ∩ B = B ∩ A).
So, P(A ∩ B) = P(A). P(B | A).
Similarly, P(A ∩ B) = P(B). P(A | B).
The interesting information regarding the Multiplication Theorem is that it can further be extended to more than two events and not just limited to the two events. So, one can also use this theorem to find out the conditional probability in terms of A, B, or C.
Read More: Types of Sets
Sometimes students get confused between Conditional Probability and Joint Probability. It is essential to know the differences between the two.