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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."
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.
Since this is the short needle case, l < a, l < b. Let (x,y) mark the coordinates of the needle's midpoint and let φ mark the angle formed by the needle and the x-axis. Similar to the examples described above, we consider x, y, φ to be independent uniform random variables over the ranges 0 ≤ x ≤ a, 0 ≤ y ≤ b, − π / 2 ≤ φ ...
Among the simple solutions, the "combined doors solution" comes closest to a conditional solution, as we saw in the discussion of methods using the concept of odds and Bayes' theorem. It is based on the deeply rooted intuition that revealing information that is already known does not affect probabilities .
Sleeping Beauty problem: A probability problem that can be correctly answered as one half or one third depending on how the question is approached. Three Prisoners problem , also known as the Three Prisoners paradox: [ 3 ] A variation of the Monty Hall problem .
Goldbach’s Conjecture. One of the greatest unsolved mysteries in math is also very easy to write. Goldbach’s Conjecture is, “Every even number (greater than two) is the sum of two primes ...
P(B | A) is the proportion of outcomes with property B out of outcomes with property A, and P(A | B) is the proportion of those with A out of those with B (the posterior). The role of Bayes' theorem can be shown with tree diagrams. The two diagrams partition the same outcomes by A and B in opposite orders, to obtain the inverse probabilities ...
Then considering the case with p = a and q = b, the last vote counted is either for the first candidate with probability a/(a + b), or for the second with probability b/(a + b). So the probability of the first being ahead throughout the count to the penultimate vote counted (and also after the final vote) is: