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.
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?