## Binomial Distribution of Discrete Individual Characteristics The Basic Idea

Important characteristics of interest can be inferred from calculations performed on samples drawn from populations of interest. Suppose we have a population of which a certain proportion p of individuals bears a discrete character, and the fraction (1-p) = q lacks the character. Empirically, we know that if one individual is drawn at random from the population, it has a probability p of bearing the character, and a probability q of lacking the character. In general, whenever there are two alternative characters that an individual may possess, the basic frequency distribution of the characters is usually expected to be binomial, and the mathematical treatment of sampling from such a population makes use of the binomial theorem.

The binomial theorem states that the probabilities of a sample of n individuals having 0, 1,2,... r,... n individuals who bear the character are given by successive terms of the expansion of the binomial (q + p)n. The chance that exactly r individuals bear the character is given by q",nq"-1p,{[n(n - 1)]/2}qn-2p2, . . . {n!/[r!(n - r)!]}q"-rpr, . . .pn for r = 0, 1,2,... r,... n individuals, respectively. The probability of r individuals who bear the character in a sample n is given by the general expression

Equation 5.6 is more readily comprehended if it is divided into two parts: the fraction, n!/[r!(n - r)!], equals the number of distinguishable ways of drawing r individuals who have the character [and (n - r) individuals who lack the character] since the joint probability of two independent drawings is the product of their individual probabilities.

Pascal's triangle affords another and a more compact way of presenting the consequences of the binomial concept (Table 5.6). The number of combinations in a given row can be obtained by adding the two numbers on either side of it in the row above. For instance, the probability of drawing two individuals from a set of 5 (n = 5) will be given by

By reading Pascal's triangle across the row for n = 5, we see that 10 is the number of distinguishable ways two individuals who bear the character, and three who do not, can be drawn randomly from a sample of 5: the probability will thus be 10p2q3.

 Sample Expansion of Number of combinations size (n) ( p + q)n n!/[r! (n - r)!]
0 0