Online Binomial Distribution Utility  Online Histogram Maker 
In many experiments there are only two outcomes. For instance:
We call such an experiment a Bernoulli trial, and refer to the two outcomes  often arbitrarily  as success and failure. We will be interested in what happens when we perform a sequence of independent Bernoulli trials. For our random variable X we will count the number of successes. For instance:
Q What about the shots on goal example?
A That depends. If the shooter is improving (or getting worse), then the probability of success is changing, and X is not a binomial random variable. For X to be a binomial random variable, we require that the probability of success for each trial be fixed and independent of what happened before.
Binomial Random Variable
A binomial random variable is a random variable that counts the number of successes in a sequence of independent Bernoulli trials with fixed probability of success. Examples include the ones listed above. Now decide which of the following is a binomial random variable. 
Q OK. Now I know what a binomial random variable is. How do we calculate its probability distribution?
A Calculating the probability distribution of X means calculating the following probabilities:
Probability Distribution of Binomial Random Variable
If X is the number of successes in a sequence of n independent Bernoulli trials, then
p = probability of success q = probability of failure = 1p 1. What is the probability of getting heads exactly twice if you flip a fair coin 6 times?
n = number of trials = 6 p = probability of success = .5 q = probability of failure = 1p = 1  .5 = .5 Probability of getting 2 heads = P(X = 2) = C(6, 2) (.5)^{2}(.5)^{62} = 15 .25 .0625 = .2344. 2. What is the probability of getting heads exactly 3 times if you flip a fair coin 6 times?
n = number of trials = 6 p = probability of success = .5 q = probability of failure = 1p = 1  .5 = .5 Probability of getting 3 heads = P(X = 3) = C(6, 3) (.5)^{3}(.5)^{63} = 20 .125 .125 = .3125. 3. You are a telemarketer with a 10% chance of persuading a randomly selected person to switch to your longdistance company. You make 8 calls. What is the probability that exactly one is sucessful? (All answers should be accurate to 4 decimal places.) Using Online Technology

Now that we know how to calculate individual probabilities of the form P(X = x), let us put them all together to obtain a probability distribution:
Example
Your name is Edison and you are an expert penalty goal shooter. your skill has been improved for the past 10 years, and now you are as good as you will ever be. Your success rate has been measured at 80%. Thus, p = .8 and q = .2. You take n = 6 shots on goal, so the possible values of X (the number of successes) are 0, 1, 2, 3, 4, 5, 6. Here is the probability for each value of X:
x  0  1  2  3  4  5  6 
P(X = x)  .000064  .001536  .01536  .08192  .24576  .393216  .262144 
We can use the probability distribution to find the probability that X is in a given range by adding the individual probabilities. (You can also use the Online Binomial Distribution Utility to compute them.) For instance:
Probability of at least 5 successes  =  P(5 X 6)  
=  P(X = 5) + P(X = 6)  
=  .393216 + .262144 = .65536 (or a 65.5% chance)  
Probability of at most 2 successes  =  P(0 X 2)  
=  P(X = 0) + P(X = 1) + P(X = 2)  
=  .000064 + .001536 + .01536 = .01696 (or a 1.7% chance)  
Probability of at least 3 successes  =  P(3 X 6)  
=  1  Probability of at most 2 successes  
Since "at most 2 successes" and "at least 3 successes" are complementary events.  
=  1  .01696 = .98304  
We used the previous answer, and that was a easier than adding P(X = 3), P(X = 4), P(X = 5), and P(X = 6) 
x  0  1  2  3  4  5  6 
P(X = x)  .000064  .001536  .01536  .08192  .24576  .393216  .262144 
Now try some of the exercises in Section 8.2 of Finite Mathematics and Finite Mathematics and Applied Calculus