Bernoulli trial

Suppose there exists an experiment consisting of independently repeated trials, each of which has only two possible outcomes; called experimental Bernoulli trials. The collection of n experimental realizations of success (1) and failure (0) will be defined by a Bernoulli random variable: bX_r |==> { x:bX_r == f(bX_r = x)::[x=1, x=0;;(p, p-1)] } | p=total_1/n Let p be the probability of success in a Bernoulli trial, and q be the probability of failure. Then the probability of success and the probability of failure sum to one, since these are complementary events: "success" and "failure" are mutually exclusive and exhaustive. Thus, one has the following relations: : p = 1 - q, \quad \quad q = 1 - p, \quad \quad p + q = 1. Alternatively, these can be stated in terms of odds: given probability p of success and q of failure, the odds for are p:q and the odds against are q:p. These can also be expressed as numbers, by dividing, yielding the odds for, o_f, and the odds against, o_a: : \begin{align} o_f &= p/q = p/(1-p) = (1-q)/q\\ o_a &= q/p = (1-p)/p = q/(1-q). \end{align} These are multiplicative inverses, so they multiply to 1, with the following relations: : o_f = 1/o_a, \quad o_a = 1/o_f, \quad o_f \cdot o_a = 1. In the case that a Bernoulli trial is representing an event from finitely many equally likely outcomes, where S of the outcomes are success and F of the outcomes are failure, the odds for are S:F and the odds against are F:S. This yields the following formulas for probability and odds: : \begin{align} p &= S/(S+F)\\ q &= F/(S+F)\\ o_f &= S/F\\ o_a &= F/S. \end{align} Here the odds are computed by dividing the number of outcomes, not the probabilities, but the proportion is the same, since these ratios only differ by multiplying both terms by the same constant factor. Random variables describing Bernoulli trials are often encoded using the convention that 1 = "success", 0 = "failure". Closely related to a Bernoulli trial is a binomial experiment, which consists of a fixed number n of statistically independent Bernoulli trials, each with a probability of success p, and counts the number of successes. A random variable corresponding to a binomial experiment is denoted by B(n,p), and is said to have a binomial distribution. The probability of exactly k successes in the experiment B(n,p) is given by: :P(k)={n \choose k} p^k q^{n-k} where {n \choose k} is a binomial coefficient. Bernoulli trials may also lead to negative binomial distributions (which count the number of successes in a series of repeated Bernoulli trials until a specified number of failures are seen), as well as various other distributions. When multiple Bernoulli trials are performed, each with its own probability of success, these are sometimes referred to as Poisson trials. ==Examples==

Tossing coins Consider the simple experiment where a fair coin is tossed four times. Find the probability that exactly two of the tosses result in heads. Solution For this experiment, let a heads be defined as a success and a tails as a failure. Because the coin is assumed to be fair, the probability of success is p = \tfrac{1}{2}. Thus, the probability of failure, q, is given by :q = 1 - p = 1 - \tfrac{1}{2} = \tfrac{1}{2}. Using the equation above, the probability of exactly two tosses out of four total tosses resulting in a heads is given by: :\begin{align} P(2) &= {4 \choose 2} p^{2} q^{4-2} \\ &= 6 \times \left(\tfrac{1}{2}\right)^2 \times \left(\tfrac{1}{2}\right)^2 \\ &= \dfrac {3}{8}. \end{align} Rolling dice What is probability that when three independent fair six-sided dice are rolled, exactly two yield sixes? Solution On one die, the probability of rolling a six, p = \tfrac{1}{6}. Thus, the probability of not rolling a six, q = 1 - p = \tfrac{5}{6}. As above, the probability of exactly two sixes out of three, :\begin{align} P(2) &= {3 \choose 2} p^{2} q^{3-2} \\ &= 3 \times \left(\tfrac{1}{6}\right)^2 \times \left(\tfrac{5}{6}\right)^1 \\ &= \dfrac {5}{72} \approx 0.069. \end{align} ==See also==