Binomial Trials and Binomial Distribution
Consider a random experiment whose outcomes can be
classified as success or failure. It means that experiment results in only two
outcomes E1(success) or E2 (failure). Further assume
that experiment can be repeated several times, probability of success or
failure in any trial are p and q (p + q = 1) and don’t vary from trial to trial
and finally different trials are independent. Such a experiment is called Binomial
experiment and trials are said to be binomial trials. For instance tossing of a
fair coin several times, each time outcome would be either a success (say
occurrence of head) or failure (say occurrence of tail).
A probability distribution representing the binomial
trials is said to binomial distribution.
Let us consider a Binomial experiment which has been
repeated ‘n’ times. Let the probability of success and failure in any trial be
p and q respectively and we are interested in the probability of occurrence of
exactly ‘r’ successes in these n trials. Now number of ways of choosing ‘r’
success in ‘n’ trials = nCr. Probability of ‘r’ successes
and (n-r) failures is pr×qn-r. Thus probability
of having exactly r successes = nCr×pr×qn-r.
Let ‘X’ be random variable representing the number of
successes, then
P(X = r) = nCr×pr×qn-r (r = 0, 1, 2, L , n)
Remarks:
q Probability of
utmost ‘r’ successes in n trials = 
q Probability of
atleast ‘r’ successes in n trials = 
q Probability of
having 1st success at the rth trial = p×qr-1
Example.1
A die is thrown 7 times. What is the chance that an odd
number turns up .
(i) exactly 4 times
(ii) at least 4 times?
Solution
Probability of success = 
&⇒ p = 
and
(i) For exactly 4 successes required probability
= 7C4. 
(ii) For atleast 4 successes required probability
= 7C4 
= 
.
Example.2
A
and B play a series of games which cannot be drawn and p, q are their
respective chances of winning a single game. What is the chance that A wins m
games before B wins n games.
Solution:
For this to happen, A must win atleast m out of the
first m + n - 1 games, \ The required probability
= m + n - 1Cmpmqn
- 1 + m + n - 1Cm + 1 pm + 1 qn
- 2 + .... + m + n - 1 Cm + n – 1pm + n -1.
