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Many mathematical problems have been stated but not yet solved. These problems come from many areas of mathematics, such as theoretical physics, computer science, algebra, analysis, combinatorics, algebraic, differential, discrete and Euclidean geometries, graph theory, group theory, model theory, number theory, set theory, Ramsey theory, dynamical systems, and partial differential equations.
What is the probability of winning the car by switching given the player has picked door 1 and the host has opened door 3? The answer to the first question is 2 / 3 , as is shown correctly by the "simple" solutions. But the answer to the second question is now different: the conditional probability the car is behind door 1 or door 2 ...
Probability distribution of the length of the longest cycle of a random permutation of the numbers 1 to 100. The green area corresponds to the survival probability of the prisoners. In the initial problem, the 100 prisoners are successful if the longest cycle of the permutation has a length of at most 50.
There are many longstanding unsolved problems in mathematics for which a solution has still not yet been found. The notable unsolved problems in statistics are generally of a different flavor; according to John Tukey, [1] "difficulties in identifying problems have delayed statistics far more than difficulties in solving problems."
The Dirac delta function, although not strictly a probability distribution, is a limiting form of many continuous probability functions. It represents a discrete probability distribution concentrated at 0 — a degenerate distribution — it is a Distribution (mathematics) in the generalized function sense; but the notation treats it as if it ...
Comparing p(n) = probability of a birthday match with q(n) = probability of matching your birthday. In the birthday problem, neither of the two people is chosen in advance. By contrast, the probability q(n) that at least one other person in a room of n other people has the same birthday as a particular person (for example, you) is given by
Bertrand's box paradox: the three equally probable outcomes after the first gold coin draw. The probability of drawing another gold coin from the same box is 0 in (a), and 1 in (b) and (c). Thus, the overall probability of drawing a gold coin in the second draw is 0 / 3 + 1 / 3 + 1 / 3 = 2 / 3 .
Pages in category "Probability problems" The following 31 pages are in this category, out of 31 total. This list may not reflect recent changes. B. Balls into bins ...