CK-12 Probability and Statistics - Advanced

4.4. The Binomial Probability Distribution http://www.ck12.org

2. A company decides to conduct a survey on customers to see if their new product, a new brand of shampoo,
   will sell well. The company chooses 100 randomly selected customers and ask them to state their preference
   among the new shampoo and two leading shampoos in the market. Letxbe the number of the 100 customers
   who choose the new brand over the other two.

Solution:

1. It is best to review the characteristics of a binomial random variable, in the preceding box. If the first student
   selected in a female, then the probability that the second student is a female is 3/9. Here we have a conditional
   probability: the success of choosing a female student on the second trial depends on the outcome of the first
   trial. Therefore, the trials are not independent andxis not a binomial random variable.


2. In this experiment each customer either states a preference for the new shampoo or does not. The customers’
   preferences are independent of each other and thereforexis a binomial random variable.

Example:

The American Heart Association claims that only 10% of adults over 30 can pass the minimum fitness requirement
that is established by them. Suppose that four adults are randomly selected and given the fitness test. Letxbe the
number of the four who pass the test. What is the probability distribution forx?

Solution:

We first check the characteristics of this experiment. We see that it is a binomial experiment because there are two
outcomes (pass or fail the test) and the outcomes are independent because one student’s success or failure does not
influence any other student’s performance. What are the possible values ofx? They arex= 0 , 1 , 2 , 3 ,4. We will take
each event and study it.

Forx=0 no one passes the fitness test. This is equivalent to

{FFFF}

Fstands for failure on the test. The probability that all will fail the test is:

p(x= 0 ) =p(F F F F) =p(F)p(F)p(F)p(F)

= (. 9 )(. 9 )(. 9 )(. 9 ) = (. 9 )^4 = 0. 6561

Thus, there is a 65.61% chance that all four will fail the fitness test.

Forx=1, only one will pass the test. The list of all possible simple events is:

{SFFF, FSFF, FFSF, FFFS}

Sstands for success on the test. Note that each of these simple events has the same probability,(. 1 )(. 9 )^3. So the
probability of all the simple events is

p(x= 1 ) = 4 [(. 1 )(. 9 )^3 ] = 0. 2916

In other words, there is a chance of 29.16% that only one will pass the test.