Search results
Results From The WOW.Com Content Network
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 ...
We can calculate the probability P as the product of two probabilities: P = P 1 · P 2, where P 1 is the probability that the center of the needle falls close enough to a line for the needle to possibly cross it, and P 2 is the probability that the needle actually crosses the line, given that the center is within reach.
From a permutations perspective, let the event A be the probability of finding a group of 23 people without any repeated birthdays. Where the event B is the probability of finding a group of 23 people with at least two people sharing same birthday, P(B) = 1 − P(A).
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]
Many probability text books and articles in the field of probability theory derive the conditional probability solution through a formal application of Bayes' theorem — among them books by Gill [51] and Henze. [52] Use of the odds form of Bayes' theorem, often called Bayes' rule, makes such a derivation more transparent. [34] [53]
At least one of them is a boy. What is the probability that both children are boys? Gardner initially gave the answers 1 / 2 and 1 / 3 , respectively, but later acknowledged that the second question was ambiguous. [1] Its answer could be 1 / 2 , depending on the procedure by which the information "at least one of them is a ...
One important drawback for applications of the solution of the classical secretary problem is that the number of applicants must be known in advance, which is rarely the case. One way to overcome this problem is to suppose that the number of applicants is a random variable N {\displaystyle N} with a known distribution of P ( N = k ) k = 1 , 2 ...
An alternative method of calculating the odds is to note that the probability of the first ball corresponding to one of the six chosen is 6/49; the probability of the second ball corresponding to one of the remaining five chosen is 5/48; and so on. This yields a final formula of