5.3 – Binomial Probability Distributions: Objectives: 1. Understand the Binomial Distribution 2. Calculate probabilities using the binomial distribution formula. 3. Find probabilities using a table and excel spreadsheets.

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. 3. Each trial must have all outcomes classified into two categories. 4. The probability of a success remains the same in all trials. When selecting a sample for a statistical analysis, we usually sample without replacement. Recall that sampling without replacement results in dependent events, which violates the second requirement in the above definition. However, we can assume independence by applying the 5% guideline learned previously. If a sample size is no more than 5% of the size of the population, treat the selections as being independent (even if the selections are made without replacement, so they are technically dependent).

For each of the following examples, determine whether the given procedure results in a binomial distribution. If not, state the reason why.

Example: Choosing 8 marbles from a box of 40 marbles (20 purple, 12 red, and 8 green) one at a time with replacement, keeping track of the colors of the marbles chosen. Solution: Not binomial: there are more than two outcomes for each trial.

Example: Choosing 10 marbles from a box of 40 marbles (20 purple, 12 red, and 8 green) one at a time with replacement, keeping track of the number of red marbles chosen. Solution: Procedure results in a binomial distribution.

Example: Choosing 5 people (without replacement) from a group of 58 people, of which 15 are women, keeping track of the number of men chosen. Solution: Not binomial: the trials are not independent.

Example: Rolling a single die 47 times, keeping track of the "fives" rolled. Solution: Procedure results in a binomial distribution

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. P(S) = p

(p = probability of success)

P(F) = 1 – p = q

(q = probability of failure)

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.

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. Example: If a die is rolled 20 times, what is the probability of rolling exactly 8 five’s? a. Does this procedure result in a binomial distribution? Explain why. b. If the procedure does result is a binomial distribution, identify the values of i. ii. iii. iv.

n x p q

Solution: a. The procedure is a binomial distribution because all four requirements are met. There are a fixed number of trials, the trials are independent, each outcome is either a success or failure and each favorable outcome has the same probability. b. The values are as follows: i. ii. iii. iv.

n = 20 x=8 p = 1/6 q = 5/6

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

P( x) =

n! ⋅ p x ⋅ q n− x ( n − x )! x!

Example: 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. n = 30, x = 12, p = 0.20 Solution: Using the binomial probability formula yields:

n! ⋅ p x ⋅ q n− x (n − x)! x! 30! = ⋅ (0.20)12 ⋅ (0.80)18 (18)!12! = 0.006

P ( x) =

Example: 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. n = 64, x = 3, p = 0.04 Solution: Using the binomial probability formula yields:

n! ⋅ p x ⋅ q n− x (n − x)! x! 64! = ⋅ (0.04) 3 ⋅ (0.96) 61 (61)!3! = 0.221

P ( x) =

Example: A tennis player makes a successful first serve 48% of the time. If she serves 9 times, what is the probability that she gets exactly 3 first serves in? Assume that each serve is independent of the others. Solution: Because this problem meets the requirements of a binomial probability distribution, we can use the binomial probability formula. n! P ( x) = ⋅ p x ⋅ q n− x (n − x)! x! 9! = ⋅ (0.48) 3 ⋅ (0.52) 6 (6)!3! = 0.184

Justification of the Formula: The binomial probability formula can be broken down into two component parts. The first part is essentially ssentially the combinations formula which determines the number of outcomes with exactly x successes among n trials and the second part determines the probability probability of x successes among n trials for any one particular order.

Example: Multiple-choice choice questions on a test each have 4 possible answers, one of which is correct. Assume that you guess the answers to 4 such questions. a. Use the multiplication rule to find the probability that the first two guesses are wrong and the third and fourth guesses are correct. That is, find P(WWCC), where C denotes a correct answer and W denotes a wrong answer. b. Make a complete list of the different different possible arrangements of 2 wrong answers and 2 correct answers, and then find the probability for each entry in the list. c. Based on the preceding results, what is the probability of getting exactly 2 correct answers when 4 guesses are made?

Solution: a. First, let’s identify n, x, p, and q. If P(S) = the probability of guessing correctly on a single question, then P (F) = the probability of guessing incorrectly. Then we have the following: n = 4, x = 2, p = 0.25, q = 0.75 Therefore, the probability of P (WWCC) is: P(WWCC ) = (0.75)(0.75)(0.25)(0.25) = (0.75) 2 (0.25) 2 = (0.563)(0.063) = 0.035 b. There are 6 possible arrangements of the four questions: (WWCC), (WCWC), (WCCW), (CCWW), (CWCW), (CWWC) c. The probability of getting exactly 2 correct answers when four guesses are made is therefore: (6)(0.035)=0.211 Using the formula we have: P( x) =

n! 4! ⋅ p x ⋅ q n− x = ⋅ (0.25) 2 ⋅ (0.75) 2 = (6)(0.035) = 0.211 (n − x)! x! (2)!2!

Method 2: Using Microsoft Excel: We can use Microsoft excel’s “BINOMDIST” function to determine the binomial probability as well as a binomial distribution of a procedure. When using this function, you will need to enter the arguments (s, n, p, false) into the parentheses. Example: A tennis player makes a successful first serve 48% of the time. If she serves 9 times, what is the probability that she gets exactly 3 first serves in? Assume that each serve is independent of the others. Solution: We already calculated the probability to this problem as 0.184 using the binomial proability formula. Using excel we obtain: =BINOMDIST (3, 9, 0.48, false) Excel will calculate the answer. Verify that the probability is as expected. By creating a table of all the possible values of the random variable x, we can also create a probability distribution table for this problem. x 0 1 2 3 4 5 6 7 8 9

P(x) 0.003 0.023 0.085 0.184 0.254 0.235 0.144 0.057 0.013 0.001

Example: Using the binomial probability distribution table from the last example, determine the probability that she 3 or fewer serves in. Solution: Three or fewer would mean 3 serves or 2 serves or 1 serve or no serves. This would involve the use of the addition rule of probability and the probabilities from the table. P(3 or fewer)=P(3)+P(2)+P(1)+P(0)=0.295

Method 3: Using Table A-1 in Appendix A: Table A-1 in Appendix A has limited usefulness for determining probability distributions. Example: Use the table in Appendix A to determine the probability given n = 12, x = 3, p = 0.20, then verify using excel. The table has limited usefulness because probabilities are only multiples of 0.10 and frequent interpolation would then be required. Solution: From the table the probability is 0.236.

For each of the following examples, determine the probability using the binomial probability formula, then confirm using excel and Table A-1 when possible. Example: The brand name of a certain chain of coffee shops has a 54% recognition rate in the town of Coffleton. An executive from the company wants to verify the recognition rate as the company is interested in opening a coffee shop in the town. He selects a random sample of 10 Coffleton residents. Find the probability that exactly 4 of the 10 Coffleton residents recognize the brand name. Solution:

n! ⋅ p x ⋅ q n− x (n − x )! x! 10! = ⋅ (0.54) 4 ⋅ (0.46) 6 (6)!4!

P ( x) =

= 0.169

x 0 1 2 3 4 5 6 7 8 9 10

P(x) 0.000 0.005 0.026 0.082 0.169 0.238 0.233 0.156 0.069 0.018 0.002

Example: In a survey of 300 college graduates, 53% reported that they entered a profession closely related to their college major. If 9 of those survey subjects are randomly selected without replacement for a follow-up survey, what is the probability that 3 of them entered a profession closely related to their college major? Solution: P( x) =

n! ⋅ p x ⋅ q n− x ( n − x )! x!

9! ⋅ (0.53) 3 ⋅ (0.47) 6 (6)!3! = 0.135

=

x 0 1 2 3 4 5 6 7 8 9

P(x) 0.001 0.011 0.051 0.135 0.228 0.257 0.193 0.093 0.026 0.003

For each of the following examples, determine the probability using the most appropriate method learned in class.

Example: An airline estimates that 97% of people booked on their flights actually show up. If the airline books 70 people on a flight for which the maximum number is 68, what is the probability that the number of people who show up will exceed the capacity of the plane?

Solution: We need to find the probability that more than 68 people will show up; that is, P(69 or 70). To use the binomial probability formula would require doing a calculation for x = 69 and x = 70 and then using the addition property to add the probabilities. The excel BINOMDIST function will be much quicker. BINOMDIST (69,70,0.97,false) = 0.257 BINOMDIST (70,70,0.97,false) = 0.119 P(69 OR 70) = 0.257 + 0.119 = 0.375 Therefore, the probability that more than 68 people will show up is 0.375.

Example: A car insurance company has determined that 6% of all drivers were involved in a car accident last year. Among the 12 drivers living on one particular street, 3 were involved in a car accident last year. If 12 drivers are randomly selected, what is the probability of getting 3 or more who were involved in a car accident last year? Solution: Make a probability distribution table using excel. The cells highlighted in yellow must be added up to determine the required probability. x 0 1 2 3 4 5 6 7 8 9 10 11 12

P(x) 0.476 0.365 0.128 0.027 0.004 0.000 0.000 0.000 0.000 0.000 0.000 0.000 0.000

The probability of selecting 3 or more drivers who were involved in an accident is 0.32.