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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 ...
The probability that A is the smaller amount is 1/2, and that it is the larger amount is also 1/2. The other envelope may contain either 2A or A/2. If A is the smaller amount, then the other envelope contains 2A. If A is the larger amount, then the other envelope contains A/2.
Banach's match problem is a classic problem in probability attributed to Stefan Banach. Feller [ 1 ] says that the problem was inspired by a humorous reference to Banach's smoking habit in a speech honouring him by Hugo Steinhaus , but that it was not Banach who set the problem or provided an answer.
The Boy or Girl paradox surrounds a set of questions in probability theory, which are also known as The Two Child Problem, [1] Mr. Smith's Children [2] and the Mrs. Smith Problem. The initial formulation of the question dates back to at least 1959, when Martin Gardner featured it in his October 1959 " Mathematical Games column " in Scientific ...
Pages in category "Probability problems" The following 31 pages are in this category, out of 31 total. ... This page was last edited on 1 November 2019, ...
The probability that all prisoners find their numbers is the product of the single probabilities, which is ( 1 / 2 ) 100 ≈ 0.000 000 000 000 000 000 000 000 000 0008, a vanishingly small number. The situation appears hopeless.
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.
The problem can be modelled using a Multinomial distribution, and may involve asking a question such as: What is the expected number of bins with a ball in them? [1] Obviously, it is possible to make the load as small as m/n by putting each ball into the least loaded bin. The interesting case is when the bin is selected at random, or at least ...
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