Probability Individual Solutions
FAMAT State Convention 2017
1. Answer: b
Solution: The expected value of a die roll is 3.5. If Donna rolls less than 3.5 on her first
roll, she should roll again. Rounded up to the faces of a die, that number is 4.
2. Answer: b
Solution: If the die is rolled six times and no value is rolled more than once, then all
values must be rolled exactly once. In other words, we have a six-die straight. There are
6! different permutations of such a straight, and there are 66 total possible sequences of
six rolls.
3. Answer: e
Solution: 100% - 0.5 * (30% + 60%) = 55%
4. Answer: b
Solution: In order for 𝑓 to be a valid PDF, it must integrate to 1. Its shape over the
interval (1, 2) is that of a trapezoid with area 0.75. Since 𝑓 takes the form of a rectangle
with height α over (0, 1), we need for α to be 1/4 in order for X to be well-defined.
5. Answer: e
Solution: The first parenthesized event means that A occurred but B did not. The second
means that B occurred but A did not. In either case, one event occurs and the other
does not. The answer choice d is incorrect because it is too inclusive. Consider for
instance the event AC ∩ BC . Do we wish to count this event? (No.)
6. Answer: c
Solution: The probability of there being exactly one box is 1/2. The probability of there
being exactly two boxes is 1/2 * 2/3. Similarly, for three boxes: 1/2 * 1/3 * 3/4.
Following this pattern and using the simple definition of expectation, we arrive at c.
7. Answer: b
Solution: In brief, two of the four green sides belong to the all-green card.
More rigorously, let GG represent the all-green card and allow the asterisk to serve as a
wildcard so that G* represents a card with a visible green side and an unknown other
side. Then we have via expansion of conditional probability:
P(GG|G*) = P(GG ^ G*) / P(G*) = P(GG) / P(G*) = 1/4 / (1/4 + 2/4 * 1/2) = 1/2.
More intuitively, let’s first agree that the all-red card is a bit of a red herring (since we
couldn’t possibly have drawn it, given G*). That leaves three cards that are relevant for
Probability Individual Solutions
FAMAT State Convention 2017
our purposes. Since only one of the three is the all-green card, it’s tempting to conclude
that P(GG|G*) is 1/3. But that’s a bit like saying there’s a 50% chance of me winning the
presidential election: there are two cases to consider, I either win or I lose, and so…well,
50-50. The problem with this line of reasoning (the reason it’s usually called the naïve
definition of probability) is that it treats all events in the sample space as equally likely.
In this problem, though our sample space might be written as {GG, GR, RG}, the event
GG is in fact twice as likely (given our observed evidence) as either of the other two
events. It’s as if GG’s vote counts twice.
And if you still aren’t convinced, let’s consider an analogy. Suppose we sample a random
tall 25 year-old from the US, ask him to play basketball, and—before we know it—Mr.
Tall Random Sample has made 80 three-pointers in a row. Prior to observing him play,
we might have thought it quite unlikely that we had happened to sample a professional
basketball player, but after learning of his remarkable talents, the odds of him being an
NBA player increase dramatically—who else could possibly make 80 three-pointers in a
row? Just as in the card problem, the evidence we’ve observed actually tells us a great
deal—and our quantified belief changes significantly from the a priori base rate.
8. Answer: d
Solution: 32 - 16 + 8 = 24. No four-way Venn diagrams needed. The 32 figure counts the
dual nationals twice and the triple nationals three times. We subtract the dual nationals
once in order to account for them once in our final figure. But since the 16 figure counts
the triple nationals three times, we now haven’t accounted for them at all. We need to
add them back once in order to arrive at the final figure. This is known formally as the
inclusion-exclusion principle. The idea is to avoid overcounting by cleverly adding and
subtracting overlapping regions.
9. Answer: d
Solution: To make this problem more tractable, let’s do some case analysis and focus on
the four possible outcomes of the next two points: WW, LW, WL, LL. The key insight is
that if Harvey wins one point but loses the other, he is back at deuce and the problem is
the same as the one with which we started.
Letting p denote the probability of Harvey winning the game from a deuce position, we
can encode the four possible outcomes of the next two points as an equation:
p = 0.6 * 0.6 * 1 + 0.4 * 0.6 * p + 0.6 * 0.4 * p + 0.4 * 0.4 * 0
 0.52p = 0.36
 p = 9 / 13
Notice how we make use of the “back to square one” effect of WL and LW—that’s what
causes p to show up on the right-hand side. This is why it was so important for us to
condition on the outcome of the next two points. Partitioning the problem into four
Probability Individual Solutions
FAMAT State Convention 2017
distinct pieces in this way is made possible by the Law of Total Probability. This is a
version of the classic Gambler’s Ruin Problem.
10. Answer: b
Solution: Consider the urn right before the first draw. There are two cases: either there
is one black marble and one white marble in the urn, or there are two white ones. A
priori, these two cases are equally likely, but the moment we observe some evidence
(the fact that the first marble drawn is white), we can use Bayes’ Rule to update our
beliefs about the probabilities involved.
One way to think about this problem is to picture four marbles: one black one and three
white ones. We drew a white marble, and only one of the three white marbles belongs
to the “world” in which the urn originally contained a black marble, which means there
is a one third probability the remaining marble is black.
A version of this problem was first posed by Lewis Carroll in 1873.
11. Answer: b
Solution: This type of permutation where no element is allowed to appear in its original
position is known as a derangement. A quick way to solve this problem is to list the 3
derangements which involve patron one taking seat two, and then to realize from
symmetry that there must likewise be 3 derangements with patron one in seat three
and 3 more with patron one in seat four—for a total of 9. An even cleverer approach
would make use of the inclusion-exclusion principle.
In any case, it is important to understand why the answer is not simply 3!. It might seem
that there are 3 open seats for the first patron (since she can’t choose seat one) and 2
open seats for the second patron (excluding seat two and the seat taken by the first
patron). But this turns out to be wrong: if the first patron takes seat two, the second
patron has not 2 but 3 open seats to choose between! Thus the answer is in fact larger
than 3!.
12. Answer: c
Solution: Imagine that Elinor reaches her right hand into the dark. Now, if we wish for all
three of the friends to join their own two hands, we certainly need for Elinor to join her
own two hands. Once Elinor has managed to grab her own left hand (with probability
1/5), Marianne must manage to do the same (with probability 1/3). Colonel Brandon
faces no open hands but his own.
1/5 * 1/3 = 1/15
Probability Individual Solutions
FAMAT State Convention 2017
13. Answer: e
Solution: Imagine again that Elinor reaches her right hand into the dark. There are 5
open hands (including her left one) she may grab. If she grabs one of the Colonel’s
hands (probability 2/5), she’s succeeded. If she grabs one of Marianne’s hands, Elinor’s
only chance is to grab one of the Colonel’s hands using her left hand (probability 2/3).
2/5 + 2/5 * 2/3 = 10/15 = 2/3
Alternatively, combinatorics tells us there are 15 equally likely pairings and that 10 of
these involve Elinor holding hand(s) with the Colonel.
14. Answer: c
Solution: Let’s first find the probability that we have the “HHT” coin:
(1/2 * 2/3) / (1/2 * 2/3 + 1/2 * 1/3) = 2/3.
Next, we simply treat the two cases separately (either we have the “HHT” coin or the
“TTH” coin): 2/3 * 1/3 + 1/3 * 2/3 = 4/9
15. Answer: a
Solution:
I. The fact that we rolled a two or a three does not tell us whether we rolled an even
number.
II. Knowing that we rolled doubles tells us there is a 1/6 chance that we rolled at least
one five. Without this information, there is an 11/36 chance.
III. If we’ve rolled a one, the other die has to be a two in order for us to have rolled
consecutive numbers. If instead we rolled, for example, a four, we would have an
“open-ended straight draw”—a three or a five would seal the deal. The one limits our
options for the straight and thus makes it less likely for us to roll consecutive numbers.
16. Answer: c
Solution: If A is a subset of B, and B is a subset of A, then it follows that A and B are one
and the same—B contains all of the members of A and no more. The set of all subsets of
a given set A is called the power set of A. In this case, since A contains 5 members, its
power set contains 2^5 = 32 members.
For each of the five elements of A, there are two possibilities: that element is either in B
or not. The multiplication rule tells us to multiply these twos together.
Another way to see this is to imagine lining up the 5 members of A and walking down
the line, stopping at each member and considering the two cases: I can add this element
to my subset B or leave it out. However you choose to look at it, the only one of the 32
possible subsets of A which is also a superset of A is the set A itself.
Probability Individual Solutions
FAMAT State Convention 2017
17. Answer: c
Solution: The expected value of David’s annual claims is 0.8 * 1000 + 0.2 * 200 = 840.
18. Answer: d
Solution: Notice that E (
X
) = E(
X+Y
X+Y−Y
X+Y
Y
since we know from symmetry that E (
E(
X
X+Y
1
Y
Y
) = E (1 − X+Y) = 1 − E (X+Y). And
X
) = E (X+Y), we have immediately that
X+Y
) = 2.
And since var(X) = E(X 2 ) − [E(X)]2 , it follows from var(X) = 1 and E(X) = 0 that
E(X 2 ) = 1.
19. Answer: b
Solution: This is an inverse take on a classic conditional probability problem. Instead of
giving the evidence and asking about the probability, this question gives the probability
and asks for the evidence which would give rise to that probability.
After observing one tails, we have:
(1/5) / (4/5 * 1/2 + 1/5) = (2/10) / (6/10) = 1/3.
After observing two tails, we have:
(1/5) / (4/5 * 1/2 * 1/2 + 1/5) = (1/5) / (2/5) = 1/2.
And so Famat would have to flip two tails in a row in order to be 50% certain that he
originally chose the unfair coin.
20. Answer: d
Solution: This question can be solved using Markov’s Inequality.
Alternatively, consider: The total income of the neighborhood is $1,000,000. The mean
income at Barton Cottage can be no greater than $200,000 since that would amount to
a total neighborhood income greater than $1,000,000.
Simply put, since Barton Cottage is responsible for 1/4 of the neighborhood population,
its mean income can be no larger than 4 times the neighborhood mean income.
21. Answer: d
Solution: No matter how he chooses to play, Jeeves will have to take exactly 95 strokes
in order to win the game, 45 of which are forehands and 50 are backhands. The only
choice he really faces is how to arrange these strokes. Choosing which of the 95 strokes
Probability Individual Solutions
FAMAT State Convention 2017
are forehands completely determines which of them are backhands, and so Jeeves has
95 choose 45, or equivalently 95 choose 50, possible game-winning plans.
22. Answer: a
Solution: This question asks about the mean of a geometric random variable with p =
0.5. The expectation is simply the reciprocal of p.
23. Answer: c
Solution: Imagine a row of ten seats numbered 1-10. Assign the people in the first four
seats to the group of four and the people in the next three seats to a group of three.
That leaves the three people in seats 8-10 for the other group of three. There are 10!
possible ways to arrange 10 people into 10 seats. But since it doesn’t matter how the
people are seated within each group, we need to adjust this figure. We divide by 4! once
and by 3! twice to correct for permutations within groups, similarly to how we might go
about finding the permutations of the word MISSISSIPPI. However, since it also makes
no difference which of the two groups of three are chosen first, we need to divide the
final figure by 2! to correct for permutations between groups.
Alternatively:
6
(10
4 )(3)
2
24. Answer: d
Solution:
I. This is true if X and Y are identically distributed. We don’t know that they are.
II. This holds true because we are told Y and Z are identically distributed.
III. Two RVs can have the same mean and variance yet be distinct.
IV. Two RVs can have the same distribution yet be distinct. Example: two fair coins.
V. This follows from the definition of variance, the fact that X and Y have identical
means and variances, and the fact that Y and Z are identically distributed.
25. Answer: d
Solution: The first three rolls do not bring Ana any closer to her objective (rolling a six).
The relevant expectation is therefore 3 + E(N) = 3 + 6 = 9. This property is referred to as
memorylessness in probability and statistics circles. The geometric distribution is in fact
the only discrete probability distribution with this property. See the exponential
distribution for the only continuous example.
26. Answer: b
Solution: 1/3 * 1/2 + 1/3 * 1/4 + 1/3 * 1/4 = 1/3
Probability Individual Solutions
FAMAT State Convention 2017
27. Answer: a
Solution: This is really another Gambler’s Ruin problem in disguise. Letting p denote the
probability of Pushkin winning, we have:
p = 1/6 + p/6 + 1/12 + p/12 + 1/12 + p/12
 2p/3 = 1/3
 p = 1/2,
which also follows from symmetry.
28. Answer: d
Solution:
a. The fair mixed strategy wins half of the time (as found in question 24).
b. The “reciprocal” strategy can be shown to win 6 out of 11 times.
c. Pushkin’s strategy would beat Pushkin half of the time.
d. The pure strategy (0, 1, 0) wins two thirds of the time.
29. Answer: c
Solution: The Bernoulli and hypergeometric distributions describe discrete phenomena.
30. Answer: a
Solution: The abbé must “dodge” the bracelet on the first, second, and third scoop. The
probability of dodging the bracelet in a scoop is 499/500.