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Then, with probability at least /, there is a unique set in that has the minimum weight among all sets of . It is remarkable that the lemma assumes nothing about the nature of the family F {\displaystyle {\mathcal {F}}} : for instance F {\displaystyle {\mathcal {F}}} may include all 2 n − 1 {\displaystyle 2^{n}-1} nonempty subsets.
In the simplest case, if one allocates balls into bins (with =) sequentially one by one, and for each ball one chooses random bins at each step and then allocates the ball into the least loaded of the selected bins (ties broken arbitrarily), then with high probability the maximum load is: [8]
From a permutations perspective, let the event A be the probability of finding a group of 23 people without any repeated birthdays. Where the event B is the probability of finding a group of 23 people with at least two people sharing same birthday, P(B) = 1 − P(A).
In probability theory, the coupon collector's problem refers to mathematical analysis of "collect all coupons and win" contests. It asks the following question: if each box of a given product (e.g., breakfast cereals) contains a coupon, and there are n different types of coupons, what is the probability that more than t boxes need to be bought ...
The solution to this particular problem is given by the binomial coefficient (+), which is the number of subsets of size k − 1 that can be formed from a set of size n + k − 1. If, for example, there are two balls and three bins, then the number of ways of placing the balls is ( 2 + 3 − 1 3 − 1 ) = ( 4 2 ) = 6 {\displaystyle {\tbinom {2 ...
At least one of them is a boy. What is the probability that both children are boys? Gardner initially gave the answers 1 / 2 and 1 / 3 , respectively, but later acknowledged that the second question was ambiguous. [1] Its answer could be 1 / 2 , depending on the procedure by which the information "at least one of them is a ...
An alternative method of calculating the odds is to note that the probability of the first ball corresponding to one of the six chosen is 6/49; the probability of the second ball corresponding to one of the remaining five chosen is 5/48; and so on. This yields a final formula of
Randomly assigning each variable to be true with probability 1/2 gives an expected 2-approximation. More precisely, if each clause has at least k variables, then this yields a (1 − 2 −k)-approximation. [8] This algorithm can be derandomized using the method of conditional probabilities. [9]