Participation Question 📊
Suppose you roll two fair six-sided dice. How many possible outcomes are there for rolling the two dice?
Answer the question at PollEv.com/erinhowardstats
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Example
Suppose you roll two fair six-sided dice. How many possible outcomes are there for rolling the two dice?
Sample space: the set of all possible outcomes
Probability
The proportion of times an outcome would occur if we observed the random process an infinite number of times.
\[P(A) = \frac{\text{the number of ways A can occur}}{\text{the number of total possible outcomes}}\]
Participation Question 📊
Suppose you roll two fair six-sided dice. What is the probability that the sum of the two dice is 7?
Answer the question at PollEv.com/erinhowardstats
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Probability Distribution
All possible outcomes and their probabilities.
Rules for Probability Distributions
Outcomes must be disjoint.
Probabilities must be between 0 and 1, inclusive.
The sum of all probabilities in the distribution must be 1.
Law of Large Numbers
As more observations are collected for a random process, the proportion of an outcome converges to the theoretical/true probability of that outcome.
Law of Large Numbers Simulation in Week 2 Slides & Additional Materials page
Complement
The complement of event A is the set of all outcomes in the sample space that don’t contain A.
Participation Question 📊
Suppose you roll two fair six-sided dice. Let A represent the event that neither die rolled shows an odd number. What is the complement of A?
Answer the question at PollEv.com/erinhowardstats
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Intersection - “A and B”
\(P(A \text{ and } B) = P(A \cap B)\)
Disjoint Events
Also referred to as mutually exclusive events.
Events that can’t happen simultaneously.
\(P(A \text { and } B) = 0\)
Participation Question 📊
Consider rolling one fair six-sided die. Each option below describes two events that may occur when the die is rolled. Which pairs of events are disjoint?
a) Rolling an odd number and rolling a three.
b) Rolling an odd number and rolling an even number.
c) Rolling a prime number and rolling an even number.
d) Rolling a 4 and rolling a 5.
Answer the question at PollEv.com/erinhowardstats
Union - “A or B”
\(P(A \text{ or } B) = P(A \cup B)\)
General Addition Rule
\(P(A \text{ or }B) =\)\(P(A) + P(B)\)\(-P(A \text{ and } B)\)
Participation Question 📊
Consider rolling one fair six-sided die. Let A = rolling an odd number and B = rolling a prime number. What is the probability of A or B occurring?
Answer the question at PollEv.com/erinhowardstats
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Independent Events
Events are considered independent if the outcome of one event does not impact the outcome of the other.
Formal definition of independent events:
\(A\) and \(B\) are independent if and only if \(P(A \text{ and } B) = P(A)\times P(B)\).
OPTIONAL - Conditional Probabilities
Example
An aircraft emergency locator transmitter (ELT) is a device designed to transmit a signal in the case of a crash. Suppose there are only four manufacturers that produce all ELTs. We’ll call these manufacturers Companies W, X, Y, and Z.
| Company W manufactures 60% of all ELTs. |
P(W)=0.6 |
| Company X manufactures 20% of all ELTs. |
P(X)=0.2 |
| Company Y manufactures 15% of all ELTs. |
P(Y)=0.15 |
| Company Z manufactures 5% of all ELTs. |
P(Z)=0.05 |
Conditional Probability
“\(A\) given \(B\)”
\(P(A|B)=\)\(\frac{P(A \text{ and } B)}{P(B)}\)
Conditional probabilities redefine the sample space to only include outcomes that are in the conditional subset.
Example
An aircraft emergency locator transmitter (ELT) is a device designed to transmit a signal in the case of a crash. Suppose there are only four manufacturers that produce all ELTs. We’ll call these manufacturers Companies W, X, Y, and Z.
| Of ELTs produced by Company W, 2% are defective. |
\(P(Def|W) = 0.02\) |
| Of ELTs produced by Company X, 2% are defective. |
\(P(Def|X) = 0.02\) |
| Of ELTs produced by Company Y, 4% are defective. |
\(P(Def|Y) = 0.04\) |
| Of ELTs produced by Company Z, 6% are defective. |
\(P(Def|Z) = 0.06\) |
General Multiplication Rule
\(P(A \text{ and }B)=\)?
Recall the definition of a conditional probability:
\(P(A|B) = \frac{P(A \text{ and }B)}{P(B)}\)
Additionally,
\(P(B|A) = \frac{P(A \text{ and }B)}{P(A)}\)
Therefore,
\(P(A \text{ and }B)= P(A|B)P(B) = P(B|A)P(A)\)
Independent Events
Events are considered independent if the outcome of one event does not impact the outcome of the other.
\(P(A|B) = P(A)\) if and only if \(A\) and \(B\) are independent. \(P(B|A) = P(B)\) if and only if \(A\) and \(B\) are independent.
Apply the general multiplication rule to independent events: \(P(A \text{ and }B)= P(A|B)P(B) = P(A)P(B)\)
Formal definition of independent events:
\(A\) and \(B\) are independent if and only if \(P(A \text{ and } B) = P(A)\times P(B)\).
Example
Suppose an ELT has been randomly selected from the general population of all ELTs produced.
Each quadrant of the room has been assigned a company. Based on your assigned company, calculate the probability that the randomly selected ELT is from your company and is defective.
| P(W) = 0.6 |
P(Def|W) = 0.02 |
| P(X) = 0.2 |
P(Def|X) = 0.02 |
| P(Y) = 0.15 |
P(Def|Y) = 0.04 |
| P(Z) = 0.05 |
P(Def|Z) = 0.06 |
| P(W and Def) = 0.6(0.02) = 0.012 |
| P(X and Def) = 0.2(0.02) = 0.004 |
| P(Y and Def) = 0.15(0.04) = 0.006 |
| P(Z and Def) = 0.05(0.06) = 0.003 |
Example
| P(W and Def) = 0.6(0.02) = 0.012 |
| P(X and Def) = 0.2(0.02) = 0.004 |
| P(Y and Def) = 0.15(0.04) = 0.006 |
| P(Z and Def) = 0.05(0.06) = 0.003 |
The sum of the four values above yields the probability of what event?
The probability a randomly selected ELT from any of the four companies is defective, P(Def)
Calculate this probability. P(Def)=0.012+0.004+0.006+0.003=0.025
Law of Total Probability
The marginal probability of event \(A\) is equal to the sum of the probabilities of all disjoint events that contain \(A\).
\[P(A) = P(A \text{ and } B_1) + P(A \text{ and } B_2) + ... + P(A \text{ and } B_k)\]
where \(B_1, B_2, ..., B_k\) are disjoint events.
For example,
P(Def)=P(W and Def)+P(X and Def)+P(Y and Def)+P(Z and Def)
Example
Suppose we have randomly selected an ELT and found it to be defective. What is the probability the ELT was produced by company X?
\(P(X|Def)=\)\(\frac{P(X \text{ and } Def)}{P(W \text{ and } Def) + P(X \text{ and } Def) + P(Y \text{ and } Def) + P(Z \text{ and } Def)}\)
\(P(X|Def) = \frac{0.004}{0.025}\)\(=0.16\)
Bayes’ Theorem
Let \(B_1,B_2,...,B_k\) be disjoint events.
\(P(B_1|A) =\)\(\frac{P(B_1 \text{ and } A)}{P(A)}\)\(=\frac{P(A|B_1)P(B_1)}{P(A \text{ and } B_1) + P(A \text{ and } B_2) + ... + P(A \text{ and } B_k)}\)\(=\frac{P(A|B_1)P(B_1)}{P(A|B_1)P(B_1) + P(A|B_2)P(B_2) +...+ P(A|B_k)P(B_k)}\)
Bayes Theorem:
\(P(B_1|A) = \frac{P(A|B_1)P(B_1)}{P(A|B_1)P(B_1) + P(A|B_2)P(B_2) +...+ P(A|B_k)P(B_k)}\)
