This document discusses random variables, expected values, and geometric probability distributions. It defines discrete and continuous random variables and how to calculate expected value. It then explains the geometric setting where the variable of interest is the number of trials to obtain the first success. Finally, it provides two examples, one calculating the probability that a child's third birth has blood type O, and another calculating the probability that a hard drive company's fifth unit tested is the first defective.

1. Module 6 Lesson 6
Random Variables and Expected Values

2. Random Variables: a variable whose value is
determined by the outcome of an experiment
• Discrete Random Variables
These can be counted.
Examples:
• the number of cars going through
a drive thru
• The number of shoes you own
• Continuous Random Variables
These can be measured.
Examples:
• the amount of water in a cup
• your height

3. Expected Value: the average outcome of a
probability experiment
How to calculate the expected
value:
1. Multiply each outcome by its
corresponding probability
2. Add up all the products
Example: Find the expected value
of X from the given probability
distribution table.
1 0.22 + 2 0.2 + 3 0.15 + 4(0.11)
= 1.51
x P(x)
1 0.22
2 0.2
3 0.15
4 0.11

4. Geometric Probability Distributions
The Geometric Setting
4. The variable of interest is the number of trials
required to obtain the 1st success.
1. Each observation falls into one of two categories,
SUCCESS or FAILURE.
2. The probability of success is the same for each
observation.
3. The observations are all independent.

5. Developing the Geometric Formula
Number of
Rolls to get a 4 Probability
1
2
3
4
What is the probability of rolling a “4” on a dice?
p = the probability of success l
trials
x = number of trials
P(x) = (1 – p)x -1 · (p)1
6
= 0.167
5
6
∙
1
6
= 0.139
5
6
∙
5
6
∙
1
6
= 0.116
5
6
∙
5
6
∙
5
6
∙
1
6
= 0.096

6. Example 1:
• Blood type is inherited. If both parents carry genes for the O and A
blood types, each child has probability 0.25 of getting two O genes
and so of having blood type O. Different children inherit
independently of each other. We wish to find the probability that the
first child these parents have with type O blood is their third child.
P(O blood) = 0.25
P(other blood type) = 1-P(O blood) = 1- 0.25 = 0.75
P(3rd child has O blood): failure, failure, success
P(3rd child) = 0.75 ∙ 0.75 ∙ 0.25 = 0.141
The parents have a 14.1% chance of their 3rd child have
type O blood.
image: “blood” by qimono at pixabay.com, license CCO Public Domain

7. Example 2:
• Suppose we have data that suggest that 3% of a company’s hard disc
drives are defective. You have been asked to determine the
probability that the first defective hard drive is the fifth unit tested.
Probability (defective) = 0.03
Probability (works) = 1-P(defective) = 1-.03 = 0.97
P(5th defective) = works ∙ work ∙ work ∙ work ∙ defective
P(5th defective) = 0.97 ∙ 0.97 ∙ 0.97 ∙ 0.97 ∙ 0.03 = 0.027
There is a 2.7% chance that the fifth hard drive will be the
first defective.
image: “girl” by Concord90 at pixabay.com, license CCO Public Domain