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Symbolab is an answer engine [1] that provides step-by-step solutions to mathematical problems in a range of subjects. [2] It was originally developed by Israeli start-up company EqsQuest Ltd., under whom it was released for public use in 2011. In 2020, the company was acquired by American educational technology website Course Hero. [3] [4]
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 .
First, you have to understand the problem. [2] After understanding, make a plan. [3] Carry out the plan. [4] Look back on your work. [5] How could it be better? If this technique fails, Pólya advises: [6] "If you cannot solve the proposed problem, try to solve first some related problem. Could you imagine a more accessible related problem?"
For the problem as originally stated, =, = / and = /, which gives = (). This is a vast timespan, even compared to the estimated age of the universe , which is only about 4 × 10 17 s . Furthermore, the length of the rope after such a time is similarly huge, 2.8 × 10 43 429 km, so it is only in a mathematical sense that the ant can ever reach ...
In the simplest case, if one allocates balls into bins (with =) sequentially one by one, and for each ball one chooses random bins at each step and then allocates the ball into the least loaded of the selected bins (ties broken arbitrarily), then with high probability the maximum load is: [8]
In probability theory, the coupon collector's problem refers to mathematical analysis of "collect all coupons and win" contests. It asks the following question: if each box of a given product (e.g., breakfast cereals) contains a coupon, and there are n different types of coupons, what is the probability that more than t boxes need to be bought ...
In the initial problem, the 100 prisoners are successful if the longest cycle of the permutation has a length of at most 50. Their survival probability is therefore equal to the probability that a random permutation of the numbers 1 to 100 contains no cycle of length greater than 50. This probability is determined in the following.
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 ...