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P(A|B) may or may not be equal to P(A), i.e., the unconditional probability or absolute probability of A. If P(A|B) = P(A), then events A and B are said to be independent: in such a case, knowledge about either event does not alter the likelihood of each other. P(A|B) (the conditional probability of A given B) typically differs from P(B|A).
Each value represents the set of shuffles having at least p values m 1, ..., m p in the correct position. Note that the number of shuffles with at least p values correct only depends on p, not on the particular values of . For example, the number of shuffles having the 1st, 3rd, and 17th cards in the correct position is the same as the number ...
P( at least one estimation is bad) = 0.05 ≤ P( A 1 is bad) + P( A 2 is bad) + P( A 3 is bad) + P( A 4 is bad) + P( A 5 is bad) One way is to make each of them equal to 0.05/5 = 0.01, that is 1%. In other words, you have to guarantee each estimate good to 99%( for example, by constructing a 99% confidence interval) to make sure the total ...
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A polynomial P(x) has only finitely many perfect powers for all integers x if P has at least three simple zeros. [18] A generalization of Tijdeman's theorem concerning the number of solutions of y m = x n + k (Tijdeman's theorem answers the case k = 1), and Pillai's conjecture (1931) concerning the number of solutions of Ay m = Bx n + k.
This article is written for those who want to get better at using price to earnings ratios (P/E ratios). To keep it...
A Binomial distributed random variable X ~ B(n, p) can be considered as the sum of n Bernoulli distributed random variables. So the sum of two Binomial distributed random variables X ~ B(n, p) and Y ~ B(m, p) is equivalent to the sum of n + m Bernoulli distributed random variables, which means Z = X + Y ~ B(n + m, p). This can also be proven ...
Assume that there is a counterexample: an integer n ≥ 2 such that there is no prime p with n < p < 2n. If 2 ≤ n < 427, then p can be chosen from among the prime numbers 3, 5, 7, 13, 23, 43, 83, 163, 317, 631 (each being the largest prime less than twice its predecessor) such that n < p < 2n. Therefore, n ≥ 427.