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. The first object can be given to any of the n persons, and similarly, the subsequent objects can be distributed to any of the n persons independently. Thus, the total number of ways to distribute the objects randomly among the n persons is n^n.
Now, we're interested in the probability that each person gets exactly one object. There are n! (n factorial) ways to distribute the objects such that each person gets one object. This is because there are n choices for the first object, (n - 1) choices for the second object (since it can't go to the person who got the first object), (n - 2) choices for the third object (since it can't go to the first two persons), and so on until the last person, who has only one choice.
Therefore, the probability of each person getting exactly one object is n! / n^n.
However, we want to find the probability that at least one person does not get any objects. This is the complement of the earlier probability. So, the probability that at least one person doesn't get any objects is 1 minus the probability that each person gets exactly one object:
1 - (n! / n^n) = (n^n - n!) / n^n.
This is the final probability that at least one person does not get any objects.
The correct answer is option (D): \(n^n-n!\)
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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.
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Sometimes students get confused between Conditional Probability and Joint Probability. It is essential to know the differences between the two.