1. Chapter 7 Lesson 7.5 Random Variables and Probability Distributions
7.5: Binomial and Geometric Distributions

2. Special Distributions
Two Discrete Distributions:
Binomial and Geometric
One Continuous Distribution:
Normal Distributions

3. Properties of a Binomial Experiment
There are a fixed number of trials
Each trial results in one of two mutually exclusive outcomes. (success/failure)
Outcomes of different trials are independent
The probability that a trial results in success is constant.
The binomial random variable x is defined as
x = the number of successes when a binomial experiment is performed
We use n to denote the fixed number of trials.

4. Are these binomial distributions?
Toss a coin 10 times and count the number of heads
Yes
Deal 10 cards from a shuffled deck and count the number of red cards
No, probability does not remain constant
The number of tickets sold to children under 12 at a movie theater in a one hour period
No, no fixed number

5. Binomial Probability Formula:
Let
n = number of independent trials in a binomial experiment
p = constant probability that any trial results in a success
Where:
Technology, such as calculators and statistical software, will also perform this calculation.

6. What is the probability of “success”?
Let’s record the gender of the next 5 newborns at Huntington Memorial Hospital and see how many girls we get.
Is this a binomial experiment?
Yes, if the births were not multiple births (twins, etc).
Define the random variable of interest.
x = the number of females born out of the next 5 births
What are the possible values of x? x
What is the probability of “success”?

7. Newborns Continued . . .
What is the probability that exactly 2 girls will be born out of the next 5 births?
What is the probability that less than 2 girls will be born out of the next 5 births?

8. What is the mean number of girls born in the next five births?
Newborns Continued . . .
Let’s construct the discrete probability distribution table for this binomial random variable:
What is the mean number of girls born in the next five births? x 1 2 3 4 5
p(x)
.03125
.15625
.3125
Notice that this is the same as multiplying
n × p
Since this is a discrete distribution, we could use:
mx = 0(.03125) + 1(.15625) + 2(.3125) + 3(.3125) + 4(.15625) + 5(.03125)
=2.5

9. Formulas for mean and standard deviation of a binomial distribution

10. Newborns Continued . . .
How many girls would you expect in the next five births at a particular hospital?
What is the standard deviation of the number of girls born in the next five births?

11. The Binomial Model (cont.)
Binomial Probability Model
n = number of trials
p = probability of success
q = 1 – p = probability of failure
X = # of successes in n trials
P(X = x) = nCx px qn-x

12. Independence
One of the important requirements is that the trials be independent.
When we don’t have an infinite population and we are sampling without replacement, the trials are not independent.
But, there is a rule that allows us to pretend we have independent trials:
The 10% condition: If the trials are not independent, it is still okay to proceed as long as the sample is smaller than 10% of the population.

14. How is this question different from a binomial distribution?
Newborns Revisited . . .
Suppose we were not interested in the number of females born out of the next five births, but
which birth would result in the first female being born?
How is this question different from a binomial distribution?

15. Properties of Geometric Distributions:
There are two possible outcomes: a success or failure
Each trial is independent of the others
The probability of success is constant for all trials.
A geometric random variable x is defined as
x = the number of trials UNTIL the FIRST success is observed ( including the success). x
So what are the possible values of x
To infinity
How far will this go?
. . .

16. Probability Formula for the Geometric Distribution
Let
p = constant probability that any trial results in a success
Where x = 1, 2, 3, …

17. Suppose that 40% of students who drive to campus at your school or university carry jumper cables. Your car has a dead battery and you don’t have jumper cables, so you decide to stop students as they are headed to the parking lot and ask them whether they have a pair of jumper cables.
Let x = the number of students stopped before finding one with a pair of jumper cables
Is this a geometric distribution?
Yes

18. Jumper Cables Continued . . .
Let x = the number of students stopped before finding one with a pair of jumper cables
p = .4
What is the probability that third student stopped will be the first student to have jumper cables?
What is the probability that at most three student are stopped before finding one with jumper cables?
P(x = 3) =
(.6)2(.4) = .144
P(x < 3) = P(1) + P(2) + P(3) =
(.6)0(.4) + (.6)1(.4) + (.6)2(.4) = .784

19. The Geometric Model (cont.)
Geometric probability model for Bernoulli trials:
p = probability of success
q = 1 – p = probability of failure
X = # of trials until the first success occurs
P(X = x) = qx-1p

20. Homework
Pg.439: #7.44, 47, 52-54, 56, 58, 60, 61