# Lecture 7: The Binomial Distribution

### Bernoulli Trials

Bernoulli Trials
3 conditions
1. Each trial results in one of two possible outcomes, which are termed either "success" or "failure".
2. The probability of success remains the same from one trial to the next.
3. The outcomes of the trials are independent of one another.

 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. 2 possible out comes: win, lose. The probability of winning remains constant at 0.5 from one trial to the next. The chances of winning are not increased from previous trials b/c the dice have no memory.  So all the conditions are met.

The Binomial Distribution
Suppose that n Bernoulli trials occur and that the probability of success on each trial equals . Then the number of successes occuring in these n trials has a binomial probability distribution. i.e.

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?

• 3 conditions met.
• n = 4
= 0.5

Test the formula

 P(0): must get no orders from all 4 so P(0) = (1/2)4 b/c of statistical independence. P(1): 4 mutually exclusive ways to get one order, each way has a probability of (1/16) b/c (1/2) on one order and then no order on three is (1/2) 3 , so (1/2) 4 = (1/16). Because these outcomes are mutually exclusive, we add the probability of these 4 ways so 4 * (1/2) 4 = (4/16) = (1/4). P(2): Six ways to get two orders out of four people. So as above, because of mutual exclusivity, we add the individual probabilities to get 6 * (1/2) 4 = (6/16) = (3/8).

### Derivation of the Formula

Note: there are a variety of ways for n Bernoulli trials to result in exactly x successes and n-x 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-)n-x

and the probability of the second is

... (1-) (1-)(1-) ... (1-) = x(1-)n-x

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 n-x 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:

Family of Distributions
The binomial probability distribution is not a single distribution but a family of distributions, each of which depends on the values of n and .

 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.

### Tables

The Binomial Distribution
Appendix Table I gives values of P(x) for n = 1 to 20, and = 0.05 to 0.50.

For larger values of  (i.e. > 1/2) switch the designation of success and failure.

recall:

 Example 3 = 0.8 x = 4 n = 10 so look up: = 0.2 x = 6 n = 10 the value in the table is 0.0055.

The Cumulative Binomial Distribution (Appendix Table 12)
Use the cumulative binomial distribution to find the probability that x is less than or equal to some amount.

 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 n = 4 Appendix I gives: 0.0625 + 0.2500 + 0.3750 + 0.2500 = 0.9375 but Appendix 12 gives = 0.5 n = 4 k = 3 (x <= 3) 0.9375

### Mean

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

### Standard Deviation

Recall:

so for our example:

(X) = [(0-2)2(1/16) + (1-2)2(1/4) + (2-2)2(3/8) + (3-2)2(1/4) + (4-2)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.
 (a) If the firm needs two or more holes to produce oil to stay in business, what is the probability that they will stay in business? (b) Give the expected value and standard deviation of the number of holes that will yield oil.

solution:

 (a) 1 - P(# holes is 0 or 1) so n = 6 = 0.1 (1 - ) = 0.9  P(0 or 1) = P(0) + P(1) = (6! / 0!6!)(0.1)0(0.9)6 + (6! / 1!5!)(0.1)(0.9)5 = 0.5314 + 0.3543 = 0.8857 so 1 - 0.8857 = 0.1143 = P(firm will stay in business) (b) E(X) = 6(0.1) = 0.6  (X) = [6(0.1)(0.9)]1/2 = (0.54)1/2 = 0.73