Let X be the number of successes in a Bernoulli trial, then X can take 0 or 1 and
P(X =1) = p = "probability of a success"P(X = 0) = 1 - p = q = "probability of failure".
Consider a random experiment of performing n independent Bernoulli trials.Let p be the probability of success, q = 1 - p be the probability of failure.
The probability of x successes and consequently (n - x) failures in n independent trials in a specified order say SSSFFSSFF….FSF is given byP(SSSFFSSFF…FSF)
= P(S) P(S) P(S) P(F) P(F) P(S) P(S) P(F)….P(F) P(S) P(F)= p.p.p.qq.ppq….qpq
= pxqn-xBut x successes can occur in nCx ways.
\ P(X = x) = nCx px qn-x is the probability mass function of exactly x successes.The probability distribution of the number of successes, so obtained is called the binomial distribution.
X P[X = x]0 qn
1 nC1 pqn-12 nC2 p2 qn-2
3 nC3 p3 qn-3. .
. .. .
. .n pn
Note 1:
n and p are called the parameters of the binomial distribution.
Note 2:
If x is a binomial variate with parameters n and p then it is denoted by x = b(n, p).
Note 3:

Example:
5 cards are drawn successively with replacement from well shuffled deck of 52 cards. What is the probability that
i) all the five cards are spadesii) only 3 cards are spades
iii) none is a spade.Suggested answer:
Let X be the random variable for the number of spade cards drawn.
p = probability of drawing a spade card
q = 1 - p
n = 5




Recurrence Relation for the Binomial Distribution
We have
P(X = x + 1) = nCx+1 px+1 qn-x-1

Mean and variance




To find the variance:
We have



= n(n-1) p2 (p+q)n-2 + np
= n2p2 - np2 + np
= n2p2 + np(1-p)
= n2p2+ npqNow V(x) = E(x2) - [E(x)]2
= n2p2 + npq - n2 p2= npq
Example:
If the mean and variance of a binomial distribution are respectively 9 and 6, find the distribution.
Suggested answer:
Mean of a binomial distribution = np = 9
Variance of a binomial distribution = npq = 6



