Search results
Results from the WOW.Com Content Network
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:
As already remarked, most sources in the topic of probability, including many introductory probability textbooks, solve the problem by showing the conditional probabilities that the car is behind door 1 and door 2 are 1 / 3 and 2 / 3 (not 1 / 2 and 1 / 2 ) given that the contestant initially picks door 1 and the ...
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 ≤ π / 2 . To solve such a problem, we first compute the probability that the needle crosses no lines, and then we take its complement.
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 .
Graphs of probability P of not observing independent events each of probability p after n Bernoulli trials vs np for various p.Three examples are shown: Blue curve: Throwing a 6-sided die 6 times gives a 33.5% chance that 6 (or any other given number) never turns up; it can be observed that as n increases, the probability of a 1/n-chance event never appearing after n tries rapidly converges to 0.
Independence is a fundamental notion in probability theory, as in statistics and the theory of stochastic processes.Two events are independent, statistically independent, or stochastically independent [1] if, informally speaking, the occurrence of one does not affect the probability of occurrence of the other or, equivalently, does not affect the odds.
De Morgan's laws represented with Venn diagrams.In each case, the resultant set is the set of all points in any shade of blue. In propositional logic and Boolean algebra, De Morgan's laws, [1] [2] [3] also known as De Morgan's theorem, [4] are a pair of transformation rules that are both valid rules of inference.
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 . The problem can be reframed by describing the boxes as each having one drawer on each of two sides. Each ...