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Suppose l > t.In this case, integrating the joint probability density function, we obtain: = = (), where m(θ) is the minimum between l / 2 sinθ and t / 2 .. Thus, performing the above integration, we see that, when l > t, the probability that the needle will cross at least one line is
However, if one considers 100 confidence intervals simultaneously, each with 95% coverage probability, the expected number of non-covering intervals is 5. If the intervals are statistically independent from each other, the probability that at least one interval does not contain the population parameter is 99.4%.
Many probability text books and articles in the field of probability theory derive the conditional probability solution through a formal application of Bayes' theorem; among them books by Gill [51] and Henze. [52] Use of the odds form of Bayes' theorem, often called Bayes' rule, makes such a derivation more transparent. [34] [53]
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 ...
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]
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.
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 ...
Its solution is whichever combination of taxis and customers results in the least total cost. Now, suppose that there are four taxis available, but still only three customers. This is an unbalanced assignment problem. One way to solve it is to invent a fourth dummy task, perhaps called "sitting still doing nothing", with a cost of 0 for the ...