OpenIntro 3 Flashcards
(27 cards)
probability
There are several possible interpretations of probability but they (almost) completely agree on the mathematical rules probability must follow. I P(A) = Probability of event A I 0 <= P(A) >= 1
Law of large numbers
H H H H H H H H H H ? The probability is still 0.5, or there is still a 50% chance that another head will come up on the next toss. P(H on 11th toss) = P(T on 11th toss) = 0.5 The coin is not “due” for a tail.
Disjoint and non-disjoint outcomes
Disjoint (mutually exclusive) outcomes: Cannot happen at the same time. Non-disjoint outcomes: Can happen at the same time.
calculating probability
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
A probability distribution lists all possible events and the probabilities with which they occur. Rules for probability distributions: 1. The events listed must be disjoint 2. Each probability must be between 0 and 1 3. The probabilities must total 1
x/median = 0
1 x/median < 1
1 x/median > 1
Symmetric = equal mean and median
left skew, mean is bigger
right skew, meadian is bigger
sample space
list of all the possible outcomes {}
independent and dependant probability
independent does not influence the outcome (eg. one coin toss does not influence the next)
calculate the probability of inependent events
What is the Conditional propability?
P of A given B
If we were to randomly select 5 Texans, what is the probability that
at least one is uninsured?
If we were to randomly select 5 Texans, the sample space for
the number of Texans who are uninsured would be:
S = {0, 1, 2, 3, 4, 5}
We are interested in instances where at least one person is
uninsured:
S = {0, 1, 2, 3, 4, 5}
So we can divide up the sample space into two categories:
S = {0, at least one}
P(at least one) = 1 − P(none)
Since the probability of the sample space must add up to 1:
Prob(at least 1 uninsured) = 1 − Prob(none uninsured)
= 1 − [(1 − 0.255)5]
= 1 − 0.7455
= 1 − 0.23
= 0.77
mutually exclusive events or disjoint events
Two outcomes are called disjoint or mutually exclusive if they cannot both happen
What is the probability that this woman has cancer if this second
mammogram also yielded a positive result?
(a) 0.0936
(b) 0.088
(c) 0.48
(d) 0.52
Bayes’ Theorem (inverting probabilities)
Bayes’ theorem allows you to update predicted probabilities of an event by incorporating new information
Imagine there is a drug test that is 98% accurate, meaning 98% of the time it shows a true positive result for someone using the drug and 98% of the time it shows a true negative result for nonusers of the drug.
Next, assume 0.5% of people use the drug. If a person selected at random tests positive for the drug, the following calculation can be made to see whether the probability the person is actually a user of the drug.
(0.98 x 0.005) / [(0.98 x 0.005) + ((1 - 0.98) x (1 - 0.005))] = 0.0049 / (0.0049 + 0.0199) = 19.76%
Bayes’ theorem shows that even if a person tested positive in this scenario, it is actually much more likely the person is not a user of the drug.
MARGINAL AND JOINT PROBABILITIES
If a probability is based on a single variable, it is a marginal probability. The probability of
outcomes for two or more variables or processes is called a joint probability.
If a photo is actually about fashion, what is the chance the ML classier correctly identied the
photo as being about fashion?
We can estimate this probability using the data. Of the 309 fashion photos, the ML algorithm
correctly classied 197 of the photos:
P(mach learn is pred fashion given truth is fashion) = 197/309= 0,638
We call this a conditional probability
P(truth is fashion given mach learn is pred fashion) =
197/219 = 0:900
We call this a conditional probability because we computed the probability under a condition:
the ML classier prediction said the photo was about fashion.
two parts to a conditional probability
There are two parts to a conditional probability, the outcome of interest and the condition.
It is useful to think of the condition as information we know to be true, and this information usually
can be described as a known outcome or event.
P(AjB) =
P(A and B) / P(B)
Complement of an Event
All outcomes that are NOT the event. So the Complement of an event is all the other outcomes (not the ones we want). And together the Event and its Complement make all possible outcomes.
Write out, in formal notation, the probability a randomly selected person who was not inoculated
died from smallpox, and nd this probability.
P(result = died j inoculated = no) =
P(result = died and inoculated = no) / P(inoculated = no)
= 0,1356/0,9608 = 0,1411
We are given only two pieces of information: 96.08%
of residents were not inoculated, and 85.88% of the residents who were not inoculated ended up
surviving.
How could we compute the probability that a resident was not inoculated and lived?
(General Multiplication Rule for independent process)
We want to determine
P(result = lived and inoculated = no)
and we are given that:
P(result = lived j inoculated = no) = 0:8588
P(inoculated = no) = 0:9608
(Among the 96.08% of people who were not inoculated, 85.88% survived)
P(result = lived and inoculated = no) = 0,8588 * 0,9608 = 0.8251
This is equivalent to the General Multiplication Rule
Dics. tree
Guessing on an exam.
In a multiple choice exam, there are 5 questions and 4 choices for each
question (a, b, c, d). Nancy has not studied for the exam at all and decides to randomly guess the answers.
What is the probability that:
(a) the rst question she gets right is the 5th question?
(b) she gets all of the questions right?
(c) she gets at least one question right?
a ) (multiplication, disjoint) 81/1024 = 7,8 %
b ) (multiplication, disjoint) 1/1024 = 0,01 %
c ) (oppisite (read up on it)) 1 – 0.75^5 = 1 – 0.237 = 0.763 = 76.3%
