Slides developed by Mine Çetinkaya-Rundel of OpenIntro
The slides may be copied, edited, and/or shared via the CC BY-SA license
To make a copy of these slides, go to File > Download as > [option],�as shown below. Or if you are logged into a Google account, you can choose Make a copy... to create your own version in Google Drive.
Introduction
to Probability
Slides developed by Mine Çetinkaya-Rundel of OpenIntro
The slides may be copied, edited, and/or shared via the CC BY-SA license
Some images may be included under fair use guidelines (educational purposes)
Random processes
Probability
There are several possible interpretations of probability but they (almost) completely agree on the mathematical rules probability must follow.
Frequentist interpretation:
Bayesian interpretation:
Practice
Which of the following events would you be most surprised by?
(a) exactly 3 heads in 10 coin flips
(b) exactly 3 heads in 100 coin flips
(c) exactly 3 heads in 1000 coin flips
Practice
Which of the following events would you be most surprised by?
(a) exactly 3 heads in 10 coin flips
(b) exactly 3 heads in 100 coin flips
(c) exactly 3 heads in 1000 coin flips
Law of large numbers
Law of large numbers states that as more observations are collected, the proportion of occurrences with a particular outcome, p̂n, converges to the probability of that outcome, p.
Disjoint and non-disjoint outcomes
Non-disjoint outcomes: Can happen at the same time.
Disjoint (mutually exclusive) outcomes: Cannot happen at the same time.
Union of non-disjoint events
What is the probability of drawing a jack or a red card from a well shuffled full deck?
Practice
What is the probability that a randomly sampled student thinks marijuana should be legalized or they agree with their parents' political views?
(a) (40 + 36 - 78) / 165
(b) (114 + 118 - 78) / 165
(c) 78 / 165
(d) 78 / 188
(e) 11 / 47
Practice
What is the probability that a randomly sampled student thinks marijuana should be legalized or they agree with their parents' political views?
(a) (40 + 36 - 78) / 165
(b) (114 + 118 - 78) / 165
(c) 78 / 165
(d) 78 / 188
(e) 11 / 47
Recap
General addition rule
P(A or B) = P(A) + P(B) - P(A and B)
Note: For disjoint events P(A and B) = 0, so the above formula simplifies to P(A or B) = P(A) + P(B)
Probability distributions
1. The events listed must be disjoint
2. Each probability must be between 0 and 1
3. The probabilities must total 1
A probability distribution lists all possible events and the probabilities with which they occur.
Practice
In a survey, 52% of respondents said they are Democrats. What is the probability that a randomly selected respondent from this sample is a Republican?
(a) 0.48
(b) more than 0.48
(c) less than 0.48
(d) cannot calculate using only the information given
Practice
In a survey, 52% of respondents said they are Democrats. What is the probability that a randomly selected respondent from this sample is a Republican?
(a) 0.48
(b) more than 0.48
(c) less than 0.48
(d) cannot calculate using only the information given
If the only two political parties are Republican and Democrat, then (a) is possible. However it is also possible that some people do not affiliate with a political party or affiliate with a party other than these two. Then (c) is also possible. However (b) is definitely not possible since it would result in the total probability for the sample space being above 1.
Practice
Complementary events are two mutually exclusive events whose probabilities that add up to 1.
Sample space is the collection of all possible outcomes of a trial.
S = {MM, FF, FM, MF}
S = { MM, FF, FM, MF }
Independence
Two processes are independent if knowing the outcome of one provides no useful information about the outcome of the other.
Practice
Between January 9-12, 2013, SurveyUSA interviewed a random sample of 500 NC residents asking them whether they think widespread gun ownership protects law abiding citizens from crime, or makes society more dangerous. 58% of all respondents said it protects citizens. 67% of White respondents, 28% of Black respondents, and 64% of Hispanic respondents shared this view. Which of the below is true?
Opinion on gun ownership and race ethnicity are most likely
(a) complementary
(b) mutually exclusive
(c) independent
(d) dependent
(e) disjoint
http://www.surveyusa.com/client/PollReport.aspx?g=a5f460ef-bba9-484b-8579-1101ea26421b
Practice
Between January 9-12, 2013, SurveyUSA interviewed a random sample of 500 NC residents asking them whether they think widespread gun ownership protects law abiding citizens from crime, or makes society more dangerous. 58% of all respondents said it protects citizens. 67% of White respondents, 28% of Black respondents, and 64% of Hispanic respondents shared this view. Which of the below is true?
Opinion on gun ownership and race ethnicity are most likely
(a) complementary
(b) mutually exclusive
(c) independent
(d) dependent
(e) disjoint
http://www.surveyusa.com/client/PollReport.aspx?g=a5f460ef-bba9-484b-8579-1101ea26421b
Checking for independence
If P(A occurs, given that B is true) = P(A | B) = P(A),�then A and B are independent.
P(randomly selected NC resident says gun ownership protects citizens, given that the resident is white)
= P(protects citizens | White) = 0.67
P(protects citizens | Black) = 0.28
P(protects citizens | Hispanic) = 0.64
P(protects citizens) = 0.58
P(protects citizens) varies by race/ethnicity, therefore opinion on gun ownership and race ethnicity are most likely dependent.
Determining dependence
based on sample data
We saw that P(protects citizens | White) = 0.67 and P(protects citizens | Hispanic) = 0.64. Under which condition would you be more convinced of a real difference between the proportions of Whites and Hispanics who think gun widespread gun ownership protects citizens?�n = 500 or n = 50,000
Determining dependence
based on sample data
We saw that P(protects citizens | White) = 0.67 and P(protects citizens | Hispanic) = 0.64. Under which condition would you be more convinced of a real difference between the proportions of Whites and Hispanics who think gun widespread gun ownership protects citizens?�n = 500 or n = 50,000
Product rule for independent events
P(T on the first toss) x P(T on the second toss)
= (1 / 2) x (1 / 2) = 1 / 4
You toss a coin twice, what is the probability of getting two tails�in a row?
P(A and B) = P(A) x P(B)
Or more generally, P(A1 and … and Ak) = P(A1) x … x P(Ak)
Practice
A recent Gallup poll suggests that 25.5% of Texans do not have health insurance as of June 2012. Assuming that the uninsured rate stayed constant, what is the probability that two randomly selected Texans are both uninsured?
(a) 25.52
(b) 0.2552
(c) 0.255 x 2
(d) (1 - 0.255)2
http://www.gallup.com/poll/156851/uninsured-rate-stable-across-states-far-2012.aspx
Practice
A recent Gallup poll suggests that 25.5% of Texans do not have health insurance as of June 2012. Assuming that the uninsured rate stayed constant, what is the probability that two randomly selected Texans are both uninsured?
(a) 25.52
(b) 0.2552
(c) 0.255 x 2
(d) (1 - 0.255)2
http://www.gallup.com/poll/156851/uninsured-rate-stable-across-states-far-2012.aspx
Disjoint vs. complementary
Do the sum of probabilities of two disjoint events always add up to 1?
Do the sum of probabilities of two complementary events always�add up to 1?
Yes, that's the definition of complementary, e.g. heads and tails.
Not necessarily, there may be more than 2 events in the sample space, e.g. party affiliation.
Putting everything together...
If we were to randomly select 5 Texans, what is the probability that at least one is uninsured?}
Putting everything together...
= 1 - (1 - 0.255)5
= 1 - 0.7455
= 1 - 0.23
= 0.77
At least 1:
P(at least one) = 1 - P(none)
Since the probability of the sample space must add up to 1:
P(at least 1 uninsured)
= 1 - P(none insured)
Practice
Roughly 20% of undergraduates at a university are vegetarian or vegan. What is the probability that, among a random sample of 3 undergraduates, at least one is vegetarian or vegan?
(a) 1 - 0.2 x 3
(b) 1 - 0.23
(c) 0.83
(d) 1 - 0.8 x 3
(e) 1 - 0.83
Practice
P(at least 1 from veg)
= 1 - P(none veg)
= 1 - 0.83
= 1 - 0.512 = 0.488
Roughly 20% of undergraduates at a university are vegetarian or vegan. What is the probability that, among a random sample of 3 undergraduates, at least one is vegetarian or vegan?
(a) 1 - 0.2 x 3
(b) 1 - 0.23
(c) 0.83
(d) 1 - 0.8 x 3
(e) 1 - 0.83