Section 5-3Binomial

Probability Distributions

Binomial Probability Distribution

A binomial probability distribution results from a procedure that meets all the following requirements:

1. The procedure has a fixed number of trials.

2. The trials must be independent. (The outcome of any individual trial doesn’t affect the probabilities in the other trials.)

3. Each trial must have all outcomes classified into two categories (commonly referred to as success and failure).

4. The probability of a success remains the same in all trials.

Notation for Binomial Probability Distributions

S and F (success and failure) denote the two possible categories of all outcomes; p and q will denote the probabilities of S and F, respectively, so

P(S) = p (p = probability of success)

P(F) = 1 – p = q (q = probability of failure)

Notation (continued)

n denotes the fixed number of trials.

x denotes a specific number of successes in n trials, so x can be any whole number between 0 and n, inclusive.

p denotes the probability of success in one of the n trials.

q denotes the probability of failure in one of the n trials.

P(x) denotes the probability of getting exactly x successes among the n trials.

Example 1: Determine whether or not the given procedure results in a binomial distribution. If it does not binomial, identify at least one requirement that is not satisfied. Treating 863 subjects with Lipitor (Atorvastatin) and recording whether there is a “yes” response when they are asked if they experienced a headache (based on data from Pfizer, Inc.)

Yes, all four requirements are met.

Example 2: Determine whether or not the given procedure results in a binomial distribution. If it does not binomial, identify at least one requirement that is not satisfied. Treating 152 couples with YSORT gender selection method developed by the Genetics & IVF Institute and recording the ages of the parents.

No, there are more that two possible outcomes for the ages of the parents.

Important Hints

Be sure that x and p both refer to the same category being called a success.

When sampling without replacement, consider events to be independent if n < 0.05N.

Methods for Finding Probabilities

We will now discuss three methods for finding the probabilities corresponding to the random variable x in a binomial distribution.

Method 1: Using the Binomial Probability Formula

for x = 0, 1, 2, . . ., n

where

n = number of trials

x = number of successes among n trials

p = probability of success in any one trial

q = probability of failure in any one trial (q = 1 – p)

!

( )! !

x n xnP x p q

n x x

Binomial Probability Formula

Number of outcomes with exactly x successes among n trials

The probability of x successes among n trials

for any one particular order

!

( )! !

x n xnP x p q

n x x

Example 3: Assume that a procedure yields a binomial distribution with a trial repeated n times. Use the binomial probability formula to find the probability of x successes given the probability p of success on a single trial.

a) n = 12, x = 10, p = 3/4

b) n = 20, x = 4, p = 0.15

!

! !x n xn

P x p qn x x

10 212!( 10) 0.75 0.25

2!10!

0.232

P x

4 1620!( 4) 0.15 0.85

16!4!

0.182

P x

Minitab Table results can be used to find binomial probabilities.

Method 2: Using Minitab Table Technology

Minitab

Example 4: Refer to the Minitab display. (When blood donors were randomly selected, 45% of them had blood that is Group O (based on data from the Greater New York Blood Program).) The display shows the probabilities obtained by entering the values of n = 5 and p = 0.45.

a) Find the probability that at least 1 of the 5 donorshas Group O blood.

b) If at least 1 Group O donor is needed, is it reasonable to expect that at least 1 will be obtained?

Yes, it is reasonable to expect that at least one group O donors will be obtained since 0.950 > 0.05.

x P(x)

0 0.050328

1 0.205889

2 0.336909

3 0.275653

4 0.112767

5 0.018453

( 1) 1 ( 0)

1 0.050328

0.950

P x P x

We will be doing binomial probabilities using the calculator:  *Keyword “exact”…find probability with binompdf(number of trials, probability of success, number of successes) *Keywords “less than or equal to” …find probability with binomcdf(number of trials, probability of success, number of successes)

**If it says “at least” or “greater than or equal to” use 1 – binomcdf with 1 less number for successWhat you are doing here is 1 minus what you DO NOT WANT and that will give you the probability that you do want.

Method 3: Using TI 83/84 Technology

Example 5: Use the TI 83/84 calculator to calculate the following probabilities.a) Calculate the probability of tossing a coin 20 times and getting exactly 9 heads.

n = 20, p = 0.5, x = 9 binompdf(20, 0.5, 9) = 0.160    b) Calculate the probability of tossing a coin 20 times and getting less than 6 heads. n = 20, p = 0.5, x < 6 ≤ 5 binomcdf(20, 0.5, 5) = 0.021    c) Calculate the probability of tossing a coin 32 times and getting at least 14 heads. n = 32, p = 0.5, x ≤ 13 1 – binomcdf(32, 0.5, 13) = 0.811

Example 5 continued:  d) About 1% of people are allergic to bee stings. What is the probability that exactly 1 person in a class of 25 is allergic to bee stings? n = 25, p = 0.01, x = 1 binompdf(25, 0.01, 1) = 0.196    e) Refer to (d), what is the probability that 4 or more of them are allergic to bee stings?

n = 25, p = 0.01, x ≤ 3 1 – binomcdf(25, 0.01, 3) = 0.00011

Example 6: Use the TI 83/84 calculator to calculate the following probabilities.

a) At a college, 53% of students receive financial aid. In a random group of 9 students, what is the probability that exactly 5 of them receive financial aid? n = 9, p = 0.53, x = 5 binompdf(9, 0.53, 5) = 0.257     b) Now using (a), what is the probability that fewer than 3 students in the class receive financial aid?

n = 9, p = 0.53, x ≤ 2 binomcdf(9, 0.53, 2) = 0.064

Example 7: The brand name of Mrs. Fields (cookies) has a 90% recognition rate (based on data from Franchise Advantage). If Mrs. Fields herself wants to verify that rate by beginning with a small sample of 10 randomly selected consumers, find the probability that exactly 9 of the 10 consumers recognize her brand name. Also find the probability that the number who recognize her brand name is not 9.

n = 10, p = 0.90, x = 9 binompdf(10, 0.90, 9) = 0.387

1 – P(it is 9) = 1 – 0.387 = 0.613