# Binomial distribution

The prefix bi in binomial experiment refers to the fact that there are two possible outcomes (e.g., head or tail, true or false, working or defective) to each trial in the binomial experiment. A binomial experiment (and the binomial distribution denoted by B(n,p)) is characterized by two parameters:
• n, the number of trials which are performed
• p, the probability of success on a single trial
The outcomes of the trials must be independent of each other, and X or k designates the number of trials which are successes. A BINOMIAL EXPERIMENT COUNTS THE NUMBER OF SUCESSES AMONG n TRIALS.

In order to calculate binomial probabilties, it is necessary to know the number of ways k successes among n trials can occur. For example, the sequences of outcomes SSFSF, SSSFF, and FSSFS all entail three successes (and two failures). Typesetting requires that I use the notation C(n,k) (read as 'the number of combinations of n object taken k at a time' or 'n choose k') for the number of arrangements of k S's and (n-k) F's.

C(n,k) = n!/(k!(n-k)!)

where ! denotes factorial (e.g., 5!= 5 × 4 × 3 × 2 × 1 = 120; 3! = 3 × 2 × 1 = 6; 0! = 1 by definition). C(5,3) = 5!/(3! × 2!) = 120/(6 × 2) = 10.

In order to calculate the probability of exactly k successes note first that the probability of k sucesses (and n-k failures) does not depend on the order of occurrence of the successes and failures:

P(SFSSF)=p(1-p)pp(1-p)=p^3 × (1-p)^2 = p(1-p)(1-p)pp=P(SFFSS) = etc.

The assumption of independence is necessary to get the probabilities of the sequences of outcomes by multiplication.

This reduces the probability of exactly k successes to the number of arrangements of k S's and (n-k) F's times the probability of a given arrangement. The number of arrangements is C(n,k) (which is called the binomial coefficient). The probability of exactly k successes (given n trials) is

P(X=k)=C(n,k)p^k × (1-p)^(n-k)

Example Assume that 25% of fuses are defective, and the fuses in packages of six fuses are independently selected.
• What is the probability that (exactly) two fuses in a package of six are defective? C(6,2)=6!/(2!4!)=720/(2 × 24) = 15; 15 × .25^2 × .75^4 = .2966.
• What is the probability that fewer than two are defective? Fewer than two means 0 or 1. P(X=0)=C(6,0) × .25^0 × .75^6 = 1 × 1 × .1780 = .1780; P(X=1)=C(6,1) × .25^1 × .75^5 = 6 × .25 × .2373 = .3560. P(X=0 or 1) = .1780+.3560 = .5340.

Competency: If 1/5 of the jelly beans in a large bag are licorice, what is the probability that exactly 2 in a handful of 5 jelly beans are licorice? At most 2? More than 2?