10.2 I FUNDAMENTALS OF BUSINESS MATHEMATICS AND STATISTICSTheoretical Distribution
If a single coin is tossed, the outcomes are two: head or tail Probability of head is^12 and tail is^12. Thus
()qp+=+=n ()^112211.
If two coins are thrown, the outcomes are four :
HH TH HT TT
PP qp pq qq
p² 2pq q²
Thus for two coins (p+q)^2 =p^2 +2pq+q^2. This binomial expansion is called binomial distribution.
Thus when coins A and B are tossed, the outcomes are : A and B fall with heads up, A head up and B tails
up, A tail up and B head up and A and B fall with tail up.
Probability of 2 heads = p × p = p^2
Probability of 1 head and 1 tail = (p×q) + (q×p) = (pq+pq) = 2pq
Probability of 2 tails = q x q = q^2. The sum is p^2 + 2pq + q^2 as the expansion of (p+q)^2.
If three coins are tossed, the following are the outcomes (p+q)^3 = p^3 +3p^2 q+3q^2 p+q^3 (p for head and q for
tail). The outcomes are :
HHH HHT HTH THH HTT THT TTH TTT
p^3 p^2 qp^2 qqp^2 pq^2 pq^2 q^2 pq^3
= p^3 + 3p^2 q + 3q^2 p + q^3 = (p+q)^3When p =^12 and q =^12 , the probability of outcome113
⎝⎠⎜⎜⎜⎛⎞ 22 +=+++=⎟⎟⎟⎟ 18 3 8 3 8 18 1.
Thus a simple rule to find out the probabilities of 3H, 2H, 1H, 0H is as follows:3 heads = p^3 ==() (^21318)
2 heads =^33 pq^2 =× =()^11222 ()^38
2 head = ()()
2 112
22
33 3
pq =× = 80 head = q^3 ==()^12381
Thus in term of binomial expansion it is
= (p + q)n
10.2.1. OBTAINING BINOMIAL COEFFICIENT
For n trials the binomial probability distribution consists of (n +1) terms, the successive binomial coefficient
being nCo, nC 1 , nC 2 , .... nCn–1, nCn.
To find the terms of the expansion, we use the expansion of (p+q)n. Since nCo = nCn=1, the first and last
coefficient will always be one. Binomial coefficient will be symmetric form. The values of the binomial
coefficient for different values of n can be obtained easily form Pascal’s triangle given below :