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Political Science Math Camp:
Problem Set 1
Solution Set
1. Assume q, r, s and t are mutually exclusive and collectively exhaustive with P (q) = 0.23,
P (r) = .15, and P (s) = 0.46. What is the joint probability of q ∪ t?
Because q, r, s and t are mutually exclusive and collectively exhaustive,
P (q) + P (r) + P (s) + P (t) = 1.
Therefore, 0.23 + 0.15 + 0.46 + P (t) = 1, or P (t) = 0.16.
Because q and t are mutually exclusive (q ∩ t), the joint probability of
q ∪ t = P (q) + P (t) = 0.23 + 0.16 = 0.39.
2. Assume e and w are independent events. Which of the following are true?
• P (e ∩ w) = P (e)P (w): True because events are independent.
• P (e|w) = P (e) + P (e)P (w): False. The events are independent. Therefore, P (e|w) is simply
P (e).
• P (w|e) = P (w): True, see above.
3. Let P (e) = 0.03 and P (e ∪ w) = 0.5. Find P (w) assuming both events are independent.
Explain whether or not you need independence to show this.
P r(e ∪ w) = P r(e) + P r(w) − P r(e ∩ w)
P r(e ∪ w) = P r(e) + P r(w) − P r(e) · P r(w)
We can do this because P r(e ∩ w) = P r(e) · P r(w|e) and, by independence, P r(w|e) = P r(w).
0.5 = 0.03 + P r(w) − (0.03) · P r(w)
0.5 − 0.03 = (1 − 0.03) · P r(w)
0.47 = 0.97 · P r(w)
P r(w) = 0.485
Yes, you need independence to solve this question. In step 2, we are able to use independence to say
P r(w|e) = P r(w). Without independence, we could not solve this question because we do not know
P r(w|e).
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4. Prove that P (AC ) = 1 − P (A).
P r(A ∪ AC ) = 1
by P r(Ω) = 1
P r(A) + P r(AC ) = 1
(1)
because A and AC are disjoint; thus, A ∪ AC = P (A) + P (AC ) (2)
P r(AC ) = 1 − P r(A)
(3)
5. Assume that 4 cards are independently drawn from a 52 card deck with replacement.
What is the probability that the ace of spades is drawn exactly once if all four selected
cards were spades?
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p = 13
because there are 13 spades in a deck, and we know that all four cards that are drawn are
spades. Given that there is one ace of spades, the probability of drawing the ace of spades given that
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any one card is a spade is 13
.
N = 4 because there are four cards drawn.
k = 1 because we only want 1 successful ace of spades draw.
Success is defined as drawing the ace of spades once, and the draws are independent of one another.
We can thus find the probability of getting exactly one ace using Bernoulli’s formula, which helps us
determine the probability of k successful events in n trials. Remember that the Bernoulli formula is
defined as:
N
· pk (1 − p)N −k
k
Using the values above:
4 1 1 12 3
Pr(1 ace in 4 spades) =
( ) ( )
1 13 13
12
1
= 4 · ( )1 ( )3
13 13
= 0.242
6. In the city of Not-Yet-Polarized, 30% of the citizens are conservatives, 30% are liberals,
and 40% are independents. In a recent election, 50% of conservatives voted, 40% of
liberals voted, and 30% of independents voted.
(a) What is the probability that a person voted?
P r(Voted) = 0.3 · 0.5 + 0.4 · 0.3 + 0.3 · 0.4
= 0.39
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(b) If the person voted, what is the probability that the voter was a conservative?
P r(Voter—Conservative)P r(Conservative)
P r(Voter)
(0.50)(0.30)
=
0.39
= 0.38
P r(Conservative—Voter) =
(c) Liberal?
P r(Voter—Liberal)P r(Liberal)
P r(Voter)
(0.40)(0.30)
=
0.39
= 0.31
P r(Liberal—Voter) =
(d) Independent?
P r(Voter—Independent)P r(Independent)
P r(Voter)
(0.30)(0.40)
=
0.39
= 0.31
P r(Independent—Voter) =
7. Suppose there were a test for cancer with a property that 90% of those with cancer reacted positively wehreas 5% of those without cancer reacted positively. Assume that 1%
of the patients in a hospital have cancer. What is the probability that a patient selected
at random who reacts positively to this test actually has cancer?
P r(Positive—Cancer)P r(Cancer)
P r(Positive)
P r(Positive—Cancer)P r(Cancer)
=
P r(Positive—Cancer)P r(Cancer) + P r(Positive—No Cancer)P r(No Cancer)
0.9 · 0.01
=
(0.9 · 0.01) + (0.05 · 0.99)
0.009
=
0.0585
= 0.15
P r(Cancer—Positive) =
8. (a) Four draws are going to be made at random with replacement from a box with five
tickets in it. The tickets are labelled 1, 2, 2, 3, 3, respectively. Find the chance that 2 is
drawn at least once.
On each draw, 2 is drawn with probability 2/5; with probability 3/5, a different number is drawn.
.
Thus, the probability that a different number appears on each of four draws is (3/5)4 = 81/625 = 0.13.
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(Remember, the draws are independent as we are drawing with replacement, so you can multiply the
probabilities). Thus, the chance that 2 appears on at least one of the four draws is 1 − 0.13 = 0.87.
Note: to figure out the chance of an event, it is sometimes useful to figure out the chance of its opposite;
then subtract from 100%.
Repeat (a), if the draws are made at random without replacement.
2 must be drawn after four draws: the probability is 1.
9. There are currently 18 students registered for our class. (a) Suppose I take a survey of
the 18 students to find out who is left-handed. Is the number of left-handers a random
variable?
A random variable is a chance procedure for generating a number—such as drawing tickets at random
from a box. When we draw tickets, we can describe the chances that different numbers will be drawn,
based on what is in the box. (In practice, we may not know what’s in the box—but the unobserved
values of the tickets in the box still determine the chances).
Here, there’s no such chance procedure. We are interviewing all 18 students in the class: it is as if we
are observing every ticket in the box. There is no chance procedure that assigns students to be lefthanded. Nor are we drawing the 18 class members at random from a larger well-defined population.
Thus, we can’t readily describe the chances of different outcomes occurring in terms of the features of
some underlying box.
10. You are thinking about playing a lottery. The rules: you buy a ticket, choose 3 different
numbers from 1 to 100, and write them on the ticket. The lottery has a box with 100 balls
numbered from 1 through 100. Three balls are drawn at random without replacement.
If the numbers on the balls are the same as the numbers on your ticket, you win. (Order
doesn’t matter). If you decide to play, what is your chance of winning? Briefly explain
your answer.
The probability that the first number drawn is any of three numbers x, y, or z (e.g., 1, 2, or 3) is
3/100. If one of these numbers is drawn, the probability that the second draw turns up one of the
remaining two numbers is 2/99. Finally, if two of the numbers are drawn on the first two draws, the
probability that the last draw is the remaining number is 1/98. Thus, the probability that all three
numbers are drawn is (3/100) ∗ (2/99) ∗ (1/98) = 6/970, 200.
Notes: in the background, you are using the rule of summing the probabilities of mutually exclusive
events: for example, the probability that the first ball is x, y, OR z (each mutually exclusive events)
is the sum of their individual probabilities: 1/100 + 1/100 + 1/100 = 3/100. After you have the
probabilities for each draw, you multiply the conditional probabilities. If this is unfamiliar, read FPP,
Chapters 13-14.
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