Independent events
Two events are independent if knowing that one event is true or
has happened does not change the probability of the other event.



“male” and “get head when flipping a coin”  indep.
“male” and “taller than 6 ft”  dep.
“male” and “pregnant”  dep.
Sampling without replacement
Pick one frog at random from your target population
and don’t put it back. Then pick another frog, etc.
Artificial pond with 10 male and 10 female frogs.
P(1st frog is male) = 10/20 = 1/2.
If 1st frog is male, P(2nd frog is male) = 9/19.
 Here, successive picks are dependent.
Survey of a whole county with thousands of frogs (half males, half females).
P(1st frog is male) = 1/2
If 1st frog is male, P(2nd frog is male) = 499/999 ≈ 1/2.
 Here, successive picks are “nearly” indep.
Conditional probability
Conditional probabilities reflect how the probability of an event can be
different if we know that some other event has occurred or is true.
The conditional probability
of event B, given event A is:
(provided that P(A) ≠ 0)
P( A and B)
P( B | A) 
P( A)
When two events A and B are independent, P(B | A) = P(B).
No information is gained from the knowledge of event A.
Probabilities of hearing impairment and blue eyes among Dalmatian dogs.
HI = Dalmatian is hearing impaired
B = Dalmatian is blue eyed
Neither HI nor B
0.66
HI and not B
0.23
B and not HI
0.06
HI and B
0.05
P(HI and B) = .05 P(HI) =.23+.05=.28 P(B) = .05+.06 =.11
P(HI |B) = P(HI and B) / P(B) = .05/.11=5/11≈.46
P(HI | B) ≠ P(HI ), therefore HI and B are dependent.
Large-scale surveys indicate that 11% of the population smokes. Medical
researchers know that the probability that a smoker will get lung cancer is 0.34.
The probability that a person will get lung cancer if the person doesn’t smoke is
0.03.
The probability P(Lung cancer | Smoker) is .34
A) 0.03
B) 0.0394
C) 0.34
D) 0.583
E) 0.97
Are “Smoker” and “Lung cancer” independent? No because cond. prob.
above are diff.
A) Yes
B) No
Another approach: P(L)=.06 but P(L|S)=.34. therefore L and S are dependent.
Large-scale surveys indicate that 11% of the population smokes. Medical
researchers know that the probability that a smoker will get lung cancer is 0.34.
The probability that a person will get lung cancer if the person doesn’t smoke is
0.03.
Lung cancer
Non-smoker and no Lung cancer
Smoker
0.11
Multiplication rule
General multiplication rule:
The probability that any two events, A and B, both occur is:
P(A and B) = P(A)P(B|A)
Multiplication rule for independent events:
If A and B are independent, then:
P(A and B) = P(A)P(B)
Artificial pond with 10 male and 10 female frogs
Successive captures are not independent.
Probability of randomly capturing 2 male frogs in a row:
P(male and male) =(10/20)*(9/19) = .237
Blood donation center
Unrelated visitors are independent.
Probability that the next two unrelated visitors
are both type O:
P(O and O) = .45*.45 = .2025
Tree diagrams
Tree diagrams are used to represent probabilities graphically and
facilitate computations.
Probabilities of skin cancer among men and women by body locations.
0.44
Man
Head
0.15
0.41
In a random individual with skin cancer:
P(head) = .15
0.56
Woman
0.63
Man
0.37
Woman
0.20
Man
P(trunk) = .41
P(limbs) = .44
Trunk
0.44
% in each group who are women:
P(woman | head) = .56
P(woman | trunk) = .37
Limbs
0.80
Woman
P(woman | limbs) = .80
Bayes’s Theorem:
𝑃 𝐴𝑗 𝐵 =
𝑃 𝐵 ∩ 𝐴𝑗
𝑃(𝐴𝑗 )𝑃(𝐵|𝐴𝑗 )
= 𝑘
𝑃(𝐵)
σ𝑖=1 𝑃(𝐴𝑖 )𝑃(𝐵|𝐴𝑖 )
where A1, A2, … , Ak nonzero mutually exclusive & exhaustive events, B any other
event w/ P(B)≠0, P(B)≠1 but P(B) not given directly. P(Aj) prior probability; 𝑃(𝐴𝑗 |𝐵)
posterior probability.
Skin cancer example) P(H) = .15; P(H|M) = ??
Q) Mutually exclusive and exhaustive events that include H?
H, T, L
Q) Role of M is Event B. How to compute P(M) = P(M and H) + P(M and T) + P(M
and L) = P(H)*P(M|H) + P(T)*P(M|T) + P(L)*P(M|L)
P(M) = .15*.44+.41*.63+.44*.2 = .4123
P(H|M) = P(H and M)/P(M) = [P(H)*P(M|H)]/P(M) = [.15*.44]/.4123 = .16
≠ P(M|H) = .44
Example) The proportion of people in a given community who have a certain
disease is .005. A test is available to diagnose the disease. When a person has the
disease, the probability that the test will produce a positive signal is .99. When a
person does not have the disease, the probability that the test will produce a positive
signal is .01. When a person tests positive, what is the probability that the person
actually has the disease?
Answer) Let D represent the event that the person actually has the disease, and
let + represent the event that the test gives a positive signal. We wish to find
P(D|+). We are given the following probabilities:
𝑃 𝐷 = .005, 𝑃 + 𝐷 = .99, 𝑃 + 𝐷𝑐 = .01
Using the Bayes’ rule,
𝑃 + 𝐷 𝑃(𝐷)
.99 (.005)
P 𝐷+ =
=
𝑃 + 𝐷 𝑃 𝐷 + 𝑃 + 𝐷𝑐 𝑃(𝐷𝑐 )
.99 (.005) + .01 (.995)
= .332
Example) An automobile dealer has kept records on the customers who visited his
showroom. Forty percent of the people who visited his dealership were women.
Furthermore, his records show that 37% of the women who visited his dealership
purchased an automobile, while 21% of the men who visited his dealership purchased
an automobile. Suppose a customer visited the showroom and purchased a car. What
is the probability that the customer was a woman?
Answer) Define the events: A = customer is a woman, Ac = customer is a man,
B = customer purchased.
“40% of the people who visited were women.” => 𝑃 𝐴 = .4, 𝑃 𝐴𝑐 = .6
“37% of the women who visited purchased.” =>𝑃 𝐵 𝐴 = .37
“21% of the men who visited purchased.” =>𝑃 𝐵 𝐴𝑐 = .21
We need to compute 𝑃 𝐴 𝐵 .
𝑃 𝐴𝐵 =
𝑃
𝑃
𝐵𝐴
𝐵𝐴
𝑃 𝐴 +𝑃
𝑃 𝐴
𝐵 𝐴𝑐
𝑃 𝐴𝑐
=
.37 (.4)
.37 (.4)+ .21 (.6)
.148
= .148+.126 ≈ .54
Diagnosis tests
# individuals with disease
total # individuals
Disease
prevalence
P( pos | dis)
Diagnosis
sensitivity
Positive
True positive
Negative
False negative
Positive
False positive
Negative
True negative
Disease
Individual
No disease
If a person gets a positive test
result, what is the probability that
he/she actually has the disease?
This is the positive predictive value:
PPV = P(disease | positive test)
Diagnosis
specificity
P(neg | no dis)
HIV-AIDS: Suppose that about 1% of a large population has HIV-AIDS antibodies.
The enzyme immuno-assay test has sensitivity .9985 and specificity .9940.
Diagnosis
sensitivity
.9985
Disease
prevalence
Positive
True positive
Negative
False negative
Positive
False positive
Negative
True negative
HIV antibodies
.01
.0015
.99
.006
Random adult
taking the test
No HIV
antibodies
Incidence of HIV-AIDS in
the general population
.994
Diagnosis
specificity
HIV-test performance
If a person who takes this test gets a positive test result, what is the probability
that he or she actually has HIV-AIDS, P(HIV-AIDS | positive test)?
What’s the positive
Disease
incidence
predictive value of the
enzyme immuno-assay
.01
.99
Incidence of HIVAIDS in general
population
Positive
HIV
antibodies
Negative
False negative
.0060
Positive
False positive
.9940
Negative
.0015
Random adult
taking the test
test for HIV-AIDS?
PPV = P(disease | positive)
Diagnosis
sensitivity .9985
No HIV
antibodies
Diagnosis
specificity
HIV-test
performance
P(true positive) = P(disease and positive) = P(disease)P(positive | disease)
= .01*.9985 = .009985
P(false positive) = P(no disease and positive) = P(no disease)P(positive | no disease)
= .99*.0060 = .00594
PPV = P(disease | positive) = P(disease and positive) / P(positive)
= P(true positive) / P(true or false positive)
=.009985/(.009985 + .00594) = .627
What is the rate of prostate
cancer among mature men?
P(cancer)= (3+1)/100
=P(high PSA and cancer)
+P(normal PSA and cancer) = (10/100)*(3/10) +
(90/100)*(1/90) = 4/100.
How “sensitive” is the PSA test? P(+ | cancer)=
=P(+ and Cancer)/P(Cancer) = (.1*.3)/.4 = 3/4
How “specific” is the PSA test? P(– | no cancer)=
P(- and no cancer)/P(no cancer)= .90*(89/90)/.96
= 89/96
If a man gets an elevated PSA test (+), what is the
probability that he has prostate cancer?
PPV = P(cancer | +)= .3