Question 1 of 12
A doctor (Michael) has gathered the following statistics on on a medical test.
He knows that, given out of a 100 individuals who have the medical condition (let's call it
condition A), he will observe 35 individuals with the symptoms (let's call it symptom B). He also
knows that out of 100 individuals who do not have the medical condition, he will observe
symptom B 20 times. The doctor also knows, that in the overall population, 1% of individuals
have the condition.
Michael performs the test on Suad and the test is positive (symptom B is observed). What is the
probability of the condition given the symptom?
The question tells us that
P(B|A) = 35%
P(B|no A) = 20%
P(A) = 1%.
We are interested in P(A|B) = P(A) * P(B|A) / P(B)
We know P(A) * P(B|A) but do not know (yet) P(B), the probability of the symptoms.
P(B) = P(A) * P(B|A) + P(no A) * P(B|no A) = 1% * 35% + (1-1%) * 20% =20.15%
P(A|B) = P(A) * P(B|A) / P(B) = 1% * 35% / 20.15% =1.74%
Notice here that the probability of the condition given the symptoms is much lower
than the probability of the symptoms given the condition. A doctor should be
interested in that former probability (1.74%). In plain English, the symptom is not
very informative, as 98.26% of the individuals with the symptoms do not have the
condition.
Question 2 of 12
At the roulette in a Macau Casino, John observes that the probability that the number 35 is drawn
is 2.7% that a red number is drawn with probability 50%. The probability that the number 35 is
drawn or that a red number is drawn is 52.7%. Numbers are either red or black. Therefore:
A)
The number 35 is a black number.
B)
The number 35 is a red number.
C)
The number 35 may be either red or black.
Points: 1 out of 1
If 35 were a red number then the probability of “drawing a red number of drawing 35” would be
equal to the probability of “drawing a red number” and thus the probability would be 50%.
If 35 is a black number then the events “drawing a red number” and “drawing 35” do not overlap
(the probability of “drawing a red number and drawing 35” is zero). Thus the formula applies and
P(“drawing a red number” or “drawing 35”) is the sum of the probabilities of these events: 52.7%
Question 3 of 12
The probability that an NYUAD graduate graduates is 98%. Conditional on graduation, the
probability that an NYUAD student at the end of the 4th year earns more than $45,000 is 45%.
Therefore the probability that an NYUAD student graduates and earns more than $45,000 at the
end of the 4th year is:
A)
34.5%
B)
20.7%
C)
10.9%
D)
44.1%
We use the rule P(A and B) = P(A) * P(B|A). We know that the probability of “earning
more than $45,000” conditional on “graduating” is 45%. The probability of “graduating”
is 98%. Thus the probability of “graduating and earning more than $45,000” is
98%*45%=44.1%.
Question 4 of 12
The probability that the value of an Apple share is higher than $400 tomorrow and that the value
of the same Apple share is higher than $500 in a year from now is 34%. The probability that the
value of an Apple share is higher than $400 tomorrow is 45%.
Tomorrow, Abdulqadir observes that the Apple share has a value of $450. Therefore, he infers
that the probability that the Apple share is higher than $500 in a year from now is:
A)
92.1%
B)
43.1%
C)
75.6%
D)
89.8%
We are interested in the P(A|B) where A is “Apple share has a value above $500 in a year from
now” conditional on “Apple share has a value above $400 tomorrow”. We know P(A and
B)=34% and we know P(B)=45% thus P(A|B)=P(A and B)/P(B) = 34%/45%=75.6%
Question 5 of 12
A judge in Switzerland realizes that out of 100 individuals convicted, 20 are innocent, and out of
100 individuals not convicted, 50 are innocent.
The ministry of justice realizes that a judge can thus make two kinds of errors: he can either
convict an innocent, or not convict an individual who is not innocent.
The ministry is thus interested in the probability of conviction given that an individual is not
innocent. Out of 100 individuals facing trial, 40 are convicted.
What is the probability of conviction given that the individual is not innocent?
A)
10%
B)
52%
C)
98%
This is presented in recitation #7’s handout on the website.
Question 6 of 12
The probability of an event A can take all possible values between -1 inclusive and 1 inclusive.
A)
True
B)
False
That’s false, probabilities take values between 0 and 1 inclusive.
Question 7 of 12
Anton finds that the probability of event A "plus" the probability of event B is 120% (1.20). Thus,
A)
The events overlap, P(A and B) is nonzero.
B)
The events are independent, P(B given A)= P(A).
C)
The events are unrelated.
D)
The probability of A or B is 1 (100%).
As the probability of an event is always lower or equal to 1, the probability of “A or B” cannot be
120%. Hence the formula P(A or B)=P(A)+P(B) cannot apply. The only case where such formula
does not apply is when P(A and B) is non zero, i.e. the events overlap.
Question 8 of 12
A random variable Y takes values y1, y2, y3, y4, ... , yK, where K is the number of values that the
random variables can take. The expected value of Y is:
A)
E(Y) = the sum of yk times P(Y=yk) for k = 1,2,...,K.
B)
E(Y) = the sum of P(Y=yk) for k = 1,2,...,K.
C)
E(Y) = the sum of the squared deviation of yk times P(Y=yk) for k = 1,2,...,K.
Straight from the course slides.
Question 9 of 12
What is the correct formula?
A)
P(A|B) = P(B|A) P(A) / P(B)
B)
P(A|B) = P(B|A) P(B) / P(A)
C)
P(A|B) = P(B|A) P(B)
D)
P(A|B) = P(B|A) P(A)
This formula was used in question 1. We get it simply from the rule P(A and B)= P(A) P(B|A):
indeed, P(A and B)=P(B and A)=P(B) P(A|B). By equating P(A)P(B|A)=P(B)P(A|B) we find the
formula.
Question 10 of 12
Joseph takes the Berlin U Bahn metro every day.
He knows that when he gets stopped by an U Bahn agent, he has bought a ticket about 60% of the
time. He gets stopped by an U Bahn agent about 10 rides out of 100. When he is stopped by an U
Bahn agent and does not have a ticket, he gets a fine with probability 1.
A ticket costs 5 euros, and the fine is 60 euros.
Purely based on the expected net monetary payment to the U Bahn company (Joseph has no moral
issue here), should Joseph buy an U Bahn ticket?
A)
Yes, his expected net monetary payment to the U Bahn is higher when buying a ticket than when
not buying a ticket.
B)
Yes, his expected net monetary payment to the U Bahn is lower when buying a ticket than when
not buying a ticket.
C)
No, his expected net monetary payment to the U Bahn is higher when buying a ticket than when
not buying a ticket.
D)
No, his expected net monetary payment to the U Bahn is lower when buying a ticket than when
not buying a ticket.
First, as Joseph only cares about how much he will pay (and is not concerned about civility,
respect of the laws, etc), Joseph will buy a ticket if and only if the expected net monetary payment
to the U Bahn system when buying a ticket is lower than when not buying a ticket.
The net payment is a random variable: when crossing the U Bahn entrance, Joseph does not know
whether he will be inspected or not. Thus the value of the net payment is not yet realized and is a
random variable.
We find that the net payment is the price of the ticket when buying a ticket (5 euros). The net
payment is the price of the fine times the probability of the fine when not buying a ticket (60 euros
times 10%=6 euros). Thus Joseph should buy a ticket.
Question 11 of 12
The random variable "height" has a normal distribution. Check which properties apply to the
normal distribution:
A)
Asymmetric
B)
Symmetric
C)
Mode=mean
D)
Mean=Median
E)
P(mean - 3 sd < height < mean + 3 sd) = 99.7%
F)
P(mean - 3 sd < height < mean + 3 sd) = 98.2%
Symmetric, mode=mean=median, and empirical rule where “almost” is 99.7%.
Question 12 of 12
In the Major League Baseball season, the probability that the Kansas City Royals enter the World
Series is 22%, while the probability that the San Francisco Giants enter the World Series is 32%.
The probability that the Kansas City Royals and the San Francisco Giants enter the World Series
is 12%. Therefore:
A)
The event that the Kansas City Royals and the event that the San Francisco Giants enter the World
Series are independent.
B)
The event that the Kansas City Royals and the event that the San Francisco Giants enter the World
Series are not independent.
C)
The event that the Kansas City Royals and the event that the San Francisco Giants enter the World
Series do not overlap.
If the events were independent the probability of both teams entering the World Series would be
the product of the probabilities (22%*32%). Since such probability is not 7%, the events are not
independent.