Binomial Distribution
Consider a batch of N light bulbs. Each bulb may be defective (S) or nondefective (F). The experiment involves selecting a light bulb and checking whether it is S or F. This experiment is called a Bernoulli Experiment since it has only two outcomes Success and Failure. Suppose it is known that there are M defective light bulbs in the batch. If we represent success by 1 and failure by 0, then
P (Success) = P (X = 1) = M/N = p (say)
P (Failure) = P (X = 0) = 1  p = q (say)
X is said to be a random variable with Bernoulli distribution.
(Notice that a Bernoulli experiment can always be replicated by a (biased) coin with Head = 1, Tail = 0, P(1) = p)
Suppose the Bernoulli experiment is repeated n times under the same condition. That is, after the light bulb is tested, it is put back into the batch. This way, the probabilities p and q remain unchanged. (This type of sampling is called Sampling with Replacement.)
Let X = Number of successes in n trials.
Then, P(X = x) = 

px q^{n  x}, x = 0, 1, 2, ..., n where 


We sum up the Bernoulli Process as follows:
1. Each trial has only two possible outcomes.
In our example, the two possible outcomes are whether a bulb is defective or nondefective.
2. The probability of the outcome of any trial remains fixed over time.
In our example, the probability of the bulb being defective or nondefective remains fixed throughout.
3. The trials are statistically independent.
In our example, the outcome of the bulb being defective or nondefective does not affect the outcome of any other bulb being so.
Example
Find the probability of getting exactly three heads in 4 tosses of a biased coin, where
P(H) = 3/4 and P(T) = 1/4
P(X = 3)=


(0.75)^{3} (0.25) = 4 x (0.75)^{3} x (0.25) 
=

0.421875 

It can be shown for the Binomial Distribution
m = E(x) = np
s^{2} = V(X) = npq
