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How to solve puzzles by graphing the rebounds of a bouncing ball: 1963 Oct: About two new and two old mathematical board games 1963 Nov: A mixed bag of problems 1963 Dec: How to use the odd-even check for tricks and problem-solving 1964 Jan: Presenting the one and only Dr. Matrix, numerologist, in his annual performance 1964 Feb
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 ...
The solvable version of the problem. Here, cups A and C are upside down, and cup B is upright. The three cups problem, also known as the three cup challenge and other variants, is a mathematical puzzle that, in its most common form, cannot be solved. In the beginning position of the problem, one cup is upside-down and the other two are right ...
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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 ...
It can be used to solve a variety of counting problems, such as how many ways there are to put n indistinguishable balls into k distinguishable bins. [4] The solution to this particular problem is given by the binomial coefficient ( n + k − 1 k − 1 ) {\displaystyle {\tbinom {n+k-1}{k-1}}} , which is the number of subsets of size k − 1 ...
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 problem of points, also called the problem of division of the stakes, is a classical problem in probability theory.One of the famous problems that motivated the beginnings of modern probability theory in the 17th century, it led Blaise Pascal to the first explicit reasoning about what today is known as an expected value.
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