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MATH 3C: FINAL REVIEW
JOE HUGHES
1. Counting
1. Two three-person committees need to be formed from ten people.
a. How many ways can this be done if no one can serve on more than one committee?
Solution: There are 10
3 ways to choose the people to serve on the first committee. Once
the first committee has been formed, there are seven people left from which to form the
second committee, hence there are 73 ways to form the second committee. So in total
there are
10 7
3
3
ways to form the two committees.
b. How many ways can the committees be formed if people can serve on both committees?
Solution: There are 10
3 ways to choose each of the two committees, so
2
10
3
ways to choose the two committees.
*2. How many ways are there to put n balls into 3 bins?
Solution: There is a trick to solving this problem, which is often called “stars and bars”.
Write out n stars to represent the balls (try it with n = 10). Then draw two large bars
which separate the stars into 3 groups. The number of ways to put n balls into 3 bins is
then the number of stars and bars choose the number of bars, i.e.
n+2
2
3. How many words can be formed from the word defenseless?
Solution: There are 11 letters in this word, so if we regard all of the letters as being distinct
(say by numbering the e’s and s’s), then there are 11! permutations of the word. We then
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must divide by the number of ways to rearrange the e’s and s’s, so there are
11!
4!3!
words which can be formed from defenseless.
2. Probability
1. (Monty Hall Problem) A famous game show in the 1960s featured the following scenario:
There are three doors. Behind one of the doors is a car, while behind the other two is a
goat. The contestant picks a door. The host (Monty Hall) then opens one of the doors
that the contestant did not pick, revealing a goat, and gives the contestant the option of
switching the door they picked. Is the contestant better off switching their guess or staying
with the original guess?
Solution: Draw a tree diagram. Switching wins with probability 23 .
2. Suppose that you have three dice. The first die is fair, the second die lands on k with
1
if 4 ≤ k ≤ 6, and the third
probability 41 if 1 ≤ k ≤ 3 and lands on k with probability 12
1
1
die lands on k with probability 4 if k is even and with probability 12
if k is odd.
You choose a die at random, roll it, and get a 3. What is the probability that you chose
the fair die?
Solution: Let A be the event that you chose the fair die, and B the event that you roll a
3. We want to find P (A|B).
By definition, P (A|B) = P P(A∩B)
(B) . Using a tree diagram shows that P (A ∩ B) =
1 1
1 1
1
6
while P (B) = 3 · 6 + 3 · 4 + 13 · 12
= 36
= 61 .
Thus P (A|B) =
6
18
1
3
· 16 =
1
18 ,
= 13 .
3. Discrete Random Variables
1. Imagine that your friend asks you to play the following game: you pick a number
between 1 and 6, and she rolls a fair die. If the number that comes up is greater than or
equal to your guess, then you win the amount of money equal to your guess (e.g. if you
guessed 2 and win, you win 2 dollars). What is the best number to guess?
Solution: Guessing 1 guarantees a win, but you only win one dollar, so the expected return
5
is 1. Guessing 2 wins with probability 56 , so the expected return is 10
6 = 3.
Guessing 3 wins with probability 46 = 23 , so the expected return is 2. Guessing 4 wins with
probability 36 = 12 , so the expected return is again 2.
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Guessing 5 wins with probability 62 = 13 , so the expected return is 53 . Finally, guessing 6
wins with probability 61 , so the expected return is 1.
Thus either 3 or 4 is the best bet.
4. Binomial, Multinomial, and Geometric distributions
1. Imagine that you begin with the function f (x) = e2x . You then flip a fair coin. If
the coin comes up heads, you differentiate the function from the previous step, while if it
comes up tails then you take the antiderivative (ignore the constant of integration). If you
flip the coin 10 times, what is the probability that you end up with the original function
f (x)?
Solution: The only way to end up with the original function is if the number of times you
differentiate is equal to the number of times you differentiate. Thus the number of heads is
equal to the number of tails. Let X be the random variable counting the number of heads;
we want P (X = 5). Using the binomial distribution,
10 1 10
P (X = 5) =
2
5
2. A college class contains 40 men and 60 women. Of the 60 women, 25 are from California
and 35 are not. Of the 40 men, 20 are from California and 20 are not.
a. During the lecture, the instructor periodically chooses a name at random and calls on
the person. If the instructor does this 8 times, what is the probability that she chooses 4
women from California, 2 women not from California, one man from California, and one
man not from California?
Solution: This is a multinomial distribution problem. Let X1 be the random variable
counting the number of women from California chosen, X2 the random variable for women
not from California, X3 for men from California, and X4 for men not from California.
The corresponding probabilities are p1 =
Since there are 8 trials, it follows that
25
100
= 14 , p2 =
P (X1 = 4, X2 = 2, X3 = 1, X4 = 1) =
35
100
=
7
20 ,
and p3 = p4 =
20
100
= 51 .
8! 1 4 7 2 1 1 1 1
4!2!1!1! 4
20
5
5
b. In the same scenario as part (a), what is the probability that the instructor calls on at
least one man?
Solution: Now we are ignoring whether or not the students are from California, so there
are just two outcomes (man or woman). Therefore we should use the binomial distribution.
Let X be the random variable counting the number of men called on, then
3 8
10
40 0 60 8
P (X ≥ 1) = 1 − P (X = 0) = 1 −
=1−
0
100
100
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3. Suppose that I have two packs of gum, one in my left pants pocket and one in my right.
Each pack starts out with 10 pieces of gum in it. Every now and then, I choose a pocket
at random and take a piece of gum out of the corresponding pack. After a while, I reach
into a pocket and discover that the pack of gum is empty.
Let X be the random variable which counts the number of pieces of gum remaining in the
other pack. Find P (X = k).
Solution: Suppose that I have found the left pack to be empty and that the right pack still
has k pieces remaining. View taking a piece of gum from the left pack to be a success, and
taking a piece from the right pack to be a failure. Then this event occurs if exactly 10 − k
failures occur before my 11th success, so has probability
20 − k 1 21−k
10
(Why is the binomial coefficient
is a success).
20−k
10
and not
2
21−k
11
? Because I know that the last choice
Since there is nothing special about the left pack versus the right pack, multiply by 2 to
get
20 − k 1 20−k
20 − k 1 21−k
=
P (X = k) = 2
2
10
2
10
4. You are given a chance to play the following game: you begin by choosing a positive
integer n. You then roll n dice simultaneously until they all (simultaneously) land on 6,
at which point you are given 6n dollars. If you have to pay a dollar for each roll, which
integer n should you choose to maximize your expected return?
Solution: There are 6n possible outcomes for rolling n dice, only one of which is the desired
outcome (all 6’s). The probability of success is 61n , so by the geometric distribution, you
should expect to roll the dice 6n times, so your expected return is 6n − 6n .
The function f (x) = 6x − 6x has derivative f 0 (x) = 6 − ln(6)6x < 0 for x ≥ 1, so is
maximized on the interval [1, ∞) by taking x = 1. Therefore you should choose n = 1, in
which case your expected return is 6 − 6 = 0.
5. The Poisson distribution
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1. A certain genetic disorder affects 1000
of the population. Find the probability that 10
people out of a population of 3000 have the disorder.
Solution: Use the Poisson approximation to the binomial distribution, with parameter
3000
λ = np = 1000
= 3. Then
310
P (X = 10) = e−3
10!
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6. Continuous Random Variables
1. Let f (x) =
sin x π2 ≤ x ≤ 3π
Is f (x) the density function of a continuous random
0
else
variable?
Solution: The first condition a density function must satisfy is that its total integral must
be one. And
Z 3π
Z ∞
h
i3π
π
f (x) dx =
sin x dx = − cos x π = − cos(3π) + cos( ) = 1
π
2
−∞
2
2
This may lead you to believe that f is a density function. But remember that a density
function must also be non-negative. Since sin(x) < 0 on the interval (π, 2π), f (x) cannot
be a density function.
1
x≥a
2x3
Find the constant a > 0 making f a density function,
2. Define f (x) =
0
x<a
and then determine the expected value of the corresponding random variable.
Solution: f (x) is non-negative since 2x13 > 0 for all x > 0, so we only need to ensure that
the total integral of f is 1. And
Z R
Z ∞
Z ∞
h
dx
1 iR
dx
1
−
=
lim
=
lim
= 2
f (x) dx =
3
3
2
R→∞ a 2x
R→∞
2x
4x a
4a
a
−∞
Therefore
1=
1
4a2
or a = 12 .
The expected value is
Z ∞
Z
xf (x) dx =
−∞
∞
1
2
dx
= lim
2x2 R→∞
Z
R
1
2
h
dx
1 iR
= lim −
=1
2x2 R→∞
2x 12
7. Normal, uniform, and exponential distributions
1. Suppose that the scores on an exam are normally distributed with mean µ = 70 and
standard deviation σ = 10. Find the probability that a student got between an 80 and a
90 on the exam.
Solution: The density function is f (x) =
1
√
e−
10 2π
P (80 ≤ X ≤ 90) =
1
√
10 2π
(x−70)2
2·102
Z
90
80
, hence
e−
(x−70)2
2·102
dx
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Make the change of variables t = x−70
10 , so that the integral becomes
Z 2
Z 2
Z 1
t2
t2
t2
1
1
1
√
e− 2 dt = √
e− 2 dt − √
e− 2 dt
2π 1
2π −∞
2π −∞
So on an exam, you would look up these two integrals in the table, and subtract them to
get the desired probability.
2. Suppose that X is uniformly distributed on the interval (1, 2), and that x1 , x2 are real
numbers such that P (X < x1 ) = 14 and P (X > x2 ) = 14 . Find P (x1 ≤ X ≤ x2 ).
1 −1
= x1 −1
Solution: There are two ways to solve this problem. First, since P (X < x1 ) = x2−1
x2 −1
5
3
for 1 < x1 < 2, it follows that x1 = 4 . Similarly, 4 = P (X ≤ x2 ) = 2−1 = x2 − 1, so
x2 = 74 . Therefore
7
−5
1
P (x1 ≤ X ≤ x2 ) = 4 4 =
2−1
2
Alternately, note that the sets {X < x1 }, {x1 ≤ X ≤ x2 } and {X > x2 } partition the
probability space, so
1 1
1
P (x1 ≤ X ≤ x2 ) = 1 − P (X < x1 ) − P (X > x2 ) = 1 − − =
4 4
2
3. A restaurant is running a promotion where you can get a free burger (worth 5 dollars)
if you wait in a special line. The waiting time is exponentially distributed with parameter
λ = 2 in units of hours−1 . If you value your time at 20 dollars an hour, is this a good
deal?
Solution: Since λ = 2, the expected waiting time is µ = 21 , which at a rate of 20 dollars per
hour would cost you 10 dollars. The burger is only worth 5 dollars, so this is not a good
deal.