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Probability and Expected Value Problems
(Answers are on the third page)
1. You roll two dice. What is the probability of:
a. Two sixes?
b. Exactly one six?
c. No sixes?
2. You roll two dice. What is the expected number of sixes that will show?
3. You are taking a 30 question, multiple choice test (five choices per question). The
directions say that your grade will equal the number of correct answers minus
one-fourth of the number of wrong answers. You are sure you have 20 of the
answers correct. On each of the remaining 10 questions, you can definitely
eliminate two of the choices. If you choose from the remaining three responses at
random, what is your expected grade for the entire test?
4. You’re already late for work, and you will be docked $1 for additional minute
that you delay in getting to work. When you get to your local subway station,
there’s a local train (say, the C-train) in the station and you can take it. The local
will take 15 minutes to get to your destination. Alternatively, you can wait for
the express (say, the A-train), which comes regularly every 5 minutes, takes 12
minutes to get to the destination, but is not in the station (and you can’t tell from
how many people are waiting for it how long they’ve been waiting). Should you
take the local or wait for the express in order to minimize your penalty?
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5. Let’s go back to the Aids model, with the same probabilities that we had on the
board, namely:
a. (the prior): the prevalence of Aids in the population: P(A) = 0.01
b. (the false negative): the chance that someone with the disease will test
negative: P(-|A) = 0.1
c. (the false positive): the chance that someone without the disease will test
positive: P(+|not-A) = 0.1
On the board, we derived that, given the probabilities above, the probability
that a person testing positive will in fact have Aids is only: P(A|+) = 1/12 =
8.3%.
This is an unfortunately low percentage, and is due to the fact that the false
positive error rate is 0.1, which is not good enough. The false positive rate
should be much lower – the test should be making fewer mistakes and
therefore telling the truth more often. Suppose we were the manufacturers of
the test itself, and we wanted the probability of a person testing positive for
Aids to actually have Aids to be much higher: instead of 8.3%, we want it to
be 90% -- in other words, we want the accuracy of the test to rise from 8.3% to
90% in the case of a positive test result. What should the false-positive rate be
for our new test to accomplish that, given that the false-negative rate can
remain the same (we’re not going to work on improving that)?
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Answers:
1. There are 36 possible outcomes when throwing 2 dice. For instance, you can get a 1
on the first die and a 2 on the second die (chance = 1/36), or vice versa, etc.
1a. there’s a 1/36 chance of two sixes.
1b. there are 10 ways of getting exactly one 6: 1-6, 2-6, 3-6, 4-6, 5-6, 6-1, 6-2, 6-3, 6-4, 6-5,
so the answer is 10/36 = 5/18
1c. No sixes? Well, either there are 2 sixes, or 1 six or no sixes. Since we’ve calculated
the probability of 2 sixes (1/36) and 1 six (10/36), all the rest must be the probability of
no sixes (25/36).
2. The expected value is the sum of the products of the probabilities and their
corresponding values. And that’s
25
12
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E(sixes)  361 (2)  10
36 (1)  36 (0)  36  3
3. You’re going to get 20 points for the 20 answers you’re confident of. For each answer
you’re not confident of, there’s a probability of 1/3 that you’ll get it right (you’ll get one
point), and 2/3 that you’ll get it wrong (you’ll lose ¼ point). Your expected number of
points/answer for those questions is:
E (points)  13 (1)  32 ( 14 )  16
So, you’ll get an average of 1/6 point for each of the non-confident questions, giving
you 20 + 10(1/6) = 21 2/3 points total.
4. What is this penalty business? -- I think you should get another job. But assuming
you can’t do that today, we need to figure out what the expected travel time is on the
express. Since the express comes every 5 minutes and is not currently in the station, it
will come in 1 minute, 2 minutes, 3 minutes or 4 minutes with equal probability. The
)
expected waiting time is (
. So, by waiting for the express,
you’ll get to your destination in 12 minutes (the trip) + 2.5 minutes (expected waiting
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time) = 14.5 minutes, which is better than the 15 minutes that the local would take. So,
Take the A-Train.
5. Our original formula was:
P ( A | ) 
P( | A) P( A)
P( | A) P( A)  P( | A) P( A)
where P( A)  0.01 and P( | A)  0.9 and P(  | A)  0.1
What should the false-positive be, so that P(A|+)=0.9 instead of 0.083? Solving the
equation for the false-positive:
 1

P( | A) P( A) 
 1
 P( A | ) 
P( | A) 
P( A)
Given: P( | A)  0.9 and the desired P( A | )  0.9, we get:
 1

0.9 0.01 
 1
 0.9   .001
P( | A) 
0.99
In other words, the test has to be improved from a false-positive rate of 0.1 to a falsepositive rate of 0.001 (a factor of 100) so that someone getting a positive result on the
test will be 90% sure that they have Aids.