Search results
Results from the WOW.Com Content Network
The Newton–Pepys problem is a probability problem concerning the probability of throwing sixes from a certain number of dice. [ 1 ] In 1693 Samuel Pepys and Isaac Newton corresponded over a problem posed to Pepys by a school teacher named John Smith. [ 2 ]
This is an accepted version of this page This is the latest accepted revision, reviewed on 17 January 2025. Observation that in many real-life datasets, the leading digit is likely to be small For the unrelated adage, see Benford's law of controversy. The distribution of first digits, according to Benford's law. Each bar represents a digit, and the height of the bar is the percentage of ...
The host always reveals a goat and always offers a switch. If and only if he has a choice, he chooses the leftmost goat with probability p (which may depend on the player's initial choice) and the rightmost door with probability q = 1 − p. [38] [34] If the host opens the rightmost ( P=1/3 + q/3 ) door, switching wins with probability 1/(1+q).
The 100 prisoners problem has different renditions in the literature. The following version is by Philippe Flajolet and Robert Sedgewick: [1] The director of a prison offers 100 death row prisoners, who are numbered from 1 to 100, a last chance. A room contains a cupboard with 100 drawers.
The probability distribution of the sum of two or more independent random variables is the convolution of their individual distributions. The term is motivated by the fact that the probability mass function or probability density function of a sum of independent random variables is the convolution of their corresponding probability mass functions or probability density functions respectively.
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 ...
Starting from the rightmost digit, double each digit and add the neighbor. (The "neighbor" is the digit on the right.) If the answer is greater than a single digit, simply carry over the extra digit (which will be a 1 or 2) to the next operation. The remaining digit is one digit of the final result. Example:
A typical example of carry is in the following pencil-and-paper addition: 1 27 + 59 ---- 86 7 + 9 = 16, and the digit 1 is the carry. The opposite is a borrow, as in −1 47 − 19 ---- 28 Here, 7 − 9 = −2, so try (10 − 9) + 7 = 8, and the 10 is got by taking ("borrowing") 1 from the next digit to the left. There are two ways in which ...