Example 1
A game in which a true die is thrown and if a 4, 5, or 6 comes up you win; on a 1, 2, or 3 you lose.
So all the conditions are met. 
where P(x) is the probability that the number of successes equals x. The number of successes, X is a random variable, whereas n and are constants. X is a binomial random variable.
Example 2
A sales rep. calls on 4 clients. The probability of an order from each client is (1/2) and whether she obtains an order from one does not depend on the orders of the others (statistical independence). What is the probability distribution of the number of orders she recieves?
Test the formula

Note: there are a variety of ways for n Bernoulli trials to result in exactly x successes and nx failures. One way is if the first x are successes and the rest are failures.
SSS ... SFFF ... F
another is
SS ... SFFF ... FS.
Because the trials are independent, the probability that the first will occur is
... (1)(1)(1) ... (1) = ^{x}(1)^{nx}
and the probability of the second is
... (1) (1)(1) ... (1) = ^{x}(1)^{nx}
which is the same as the first sequence, so the probability of obtaining any particular sequence of x successes will be the same as the above two sequences.
So, how many ways can n Bernoulli trials give rise to exactly x successes and nx failures? This is just a combinations question.
because each of these sequences are mutually exclusive events we add the probabilities together to get the total probability of x successes in n Bernoulli trials:
n = 6, = 0.2 
n = 6, = 0.3 
n = 6, = 0.7 
n = 6, = 0.8 
n = 5, = 0.4 
n = 10, = 0.4 
n = 20, = 0.4 
< (1/2) skewed to the left, > (1/2) skewed right.
As n gets larger, the binomial probability distribution gets more "bell" shaped.
For larger values of
(i.e.
> 1/2)
switch the designation of success and failure.
recall:
Example 3
= 0.8
so look up:
= 0.2

Example 4
If our sales representative gets no bonus if the number of orders is less than or equal to 3, what is the probability that she won’t get a bonus? P(No Bonus) = P(0) + P(1) + P(2) + P(3) so
= 0.5 Appendix I gives: 0.0625 + 0.2500 + 0.3750 + 0.2500 = 0.9375 but Appendix 12 gives
= 0.5 0.9375 
Recall:
so for our example:
E(X) = (0)P(0) + (1)P(1) + (2)P(2) + (3)P(3) + (4)P(4)
so
E(X) = 0(1/16) + 1(1/4) + 2(3/8) + 3(1/4) +4(1/16) = 2
It can be shown:
Expected Value of a Binomial Random Variable
E(X) = n
Recall:
so for our example:
(X) = [(02)^{2}(1/16) + (12)^{2}(1/4) + (22)^{2}(3/8) + (32)^{2}(1/4) + (42)^{2}(1/16)]^{1/2} = 1
It can be shown:
Standard Deviation of a Binomial Random Variable
Example 5
An oil exploration firm plans to drill 6 holes. They believe that the probability that each hole will yield oil is 0.10. Since they will be drilling in different locations, the outcomes are statistically independent.
solution:
