Lecture 9 The Binomial Distribution Math 1107 Introduction to Statistics

Math 1107 Binomial Distribution If we are interested in determining the probability of events with binary outcomes (only 2 possibilities), we use the binomial distribution. Examples of outcomes: –Yes/no –Success/failure –Respond/No Respond

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. 4. The probabilities must remain constant for each trial. Math 1107 Binomial Distribution

P(x) = p x q n-x ( n – x ) ! x ! n !n ! 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 (Note that x and p must indicate the same event) q = probability of failure in any one trial (q = 1 – p) Math 1107 Binomial Distribution

Lets go through a few examples… Example 1: You are presented with a multiple choice test in nuclear physics. The test has 4 questions, with 5 answers for each question. Given that you know nothing about nuclear physics, what is the probability that you would pass the test and answer 3 out of 4 answers correctly?

Math 1107 Binomial Distribution First step is to verify that the binomial distribution makes sense to apply to this question. Does it? Why or why not? 1. The number of “trials” is fixed (4); 2. The 4 trials are independent of each other; 3. Each of the 4 trials have only 2 possible outcomes; 4. The probability of a correct answer is the same for each trial.

Math 1107 Binomial Distribution The second step is to identify the values of n,x,p and q. In this example: n= 4 X= 3 p=.2 q= 1-p or.8

P(x) = p x q n-x ( n – x ) ! x ! n !n ! for x = 0, 1, 2,..., n Math 1107 Binomial Distribution Now, we plug the numbers into the formula: And we get: P(x) = ( 4 – 3 ) ! 3 ! 4 !4 !

Math 1107 Binomial Distribution Our result is: P(3) = (4!/1!3!)*.008*.8 =.0256 Therefore, you have a 2.56% chance of passing the test if you are guessing at random. You had better study!

Math 1107 Binomial Distribution Example 2: You are presented with a multiple choice test in nuclear physics. The test has 4 questions, with 5 answers for each question. Given that you know nothing about nuclear physics, what is the probability that you would fail the test and score 0, 1, 2 out of a possible 4?

Math 1107 Binomial Distribution In this example: n= 4 x= 0, 1 and 2 p=.2 q=.8 So… [4!/((4-0)!*0!)]*.2 0 *.8 4 =.4096 [4!/((4-1)!*1!)]*.2 1 *.8 3 =.4096 [4!/((4-2)!*2!)]*.2 2 *.8 2 = =.9728 You had better study!

Math 1107 Binomial Distribution Fun EXCEL Exercise

Math 1107 Binomial Distribution Example 3: Across all airlines, the on time arrival percentage is 74%. If you commute to NY for work, and fly 10 times a month, what is the chance that 8 of those 10 flights will be on time?

P(x) = p x q n-x ( n – x ) ! x ! n !n ! for x = 0, 1, 2,..., n Math 1107 Binomial Distribution In this example: n= 10 x= 8 p=.74 q=.26 So… [10!/((10-8)!*8!)]*.74 8 *.26 2 =.2735