ECO 227Y1
Sample Examination Questions
12 October 2010
1. Two six-sided dice are thrown sequentially, and the face values that come
up are recorded.
(a) List the sample space.
Ω ≡ {(n1 , n2 ) : n1 , n2 ∈ {1, . . . , 6}}.
(b) List the elements that make up the following events:
i. A ≡ the sum of the two values is at least 5;
A ≡ {(1, 4), (1, 5), (1, 6), (2, 3), (2, 4), (2, 5), (2, 6), (3, 2), (3, 3),
(3, 4), (3, 5), (3, 6), (4, 1), (4, 2), (4, 3), (4, 4), (4, 5), (4, 6),
(5, 1), (5, 2), (5, 3), (5, 4), (5, 5), (5, 6), (6, 1), (6, 2), (6, 3),
(6, 4), (6, 5), (6, 6)} .
ii. B ≡ the value of the first die is higher than the value of the second;
B ≡ {(2, 1), (3, 2), (3, 1), (4, 1), (4, 2), (4, 3), (5, 1), (5, 2), (5, 3),
(5, 4), (6, 1), (6, 2), (6, 3), (6, 4), (6, 5)} .
iii. C ≡ the first value is 4.
C ≡ {(4, 1), (4, 2), (4, 3), (4, 4), (4, 5), (4, 6)} .
1
(c) List the elements of the following events:
i. A ∩ C;
A ∩ C = C.
ii. B ∪ C;
B ∪ C ≡ {(2, 1), (3, 2), (3, 1), (4, 1), (4, 2), (4, 3), (4, 4), (4, 5), (4, 6),
(5, 1), (5, 2), (5, 3), (5, 4), (6, 1), (6, 2), (6, 3), (6, 4), (6, 5)} .
iii. A ∩ (B ∪ C).
A ∩ (B ∪ C) = {(3, 2), (4, 1), (4, 2), (4, 3), (4, 4), (4, 5), (4, 6), (5, 1),
(5, 2), (5, 3), (5, 4), (6, 1), (6, 2), (6, 3), (6, 4), (6, 5)} .
2. Two dice are rolled, and the sum of the face values is six. What is the
probability that at least one of the dice came up a three?
P (at least one 3| sum of face values is six)
P ({(3, 3)})
=
P ({(1, 5), (2, 4), (3, 3), (4, 2), (5, 1)})
=
1
36
5
36
=
1
.
5
3. How many distinct letter arrangements can be obtained from the letters of
the word “excruciating”, using all the letters?
Twelve letters in total, including two C’s and two I’s. So
rangements.
12!
2!2!
distinct ar-
4. How many distinct letter arrangements can be obtained from the letters of
the words “harrowing ordeal”, using all the letters?
Fifteen letters in total, including two A’s, three R’s and two O’s. So
distinct arrangements.
2
15!
3!2!2!
5. An elevator containing five people can stop at any of seven floors. What is
the probability that no two people get off at the same floor? Assume that the
occupants act independently and that all floors are equally likely for each
occupant.
If no two people get off at the same floor, then they all each get off at
different floors. There are 7 × 6 × 5 × 4 × 3 different sequences of floors
corresponding to the order in which the five people leave the elevator. The
probability that each of the people in the elevator get off at different floors
is accordingly
7×6×5×4×3
.
75
6. Two cards are selected at random from a deck of 52 playing cards. The
cards in the deck are divided into 13 possible denominations (i.e., “face
values”) in 4 different suits.
(a) What is the probability that the two cards selected constitute a “pair”,
i.e., they are of the same denomination?
Pr [ pair ] = Pr [ 2nd card has same face value as first card]
3
=
.
51
(b) What is the conditional probability that the two cards selected constitute a pair given that they come from different suits?
Pr [ pair & different suits ]
Pr [ different suits ]
Pr [ pair ]
=
Pr [ different suits ]
Pr [ pair | different suits] =
=
3
51
39
51
=
1
.
13
7. Bill and George go “stalking” together in the highlands of Scotland. Both
take shots at a stag at the same time. Suppose that Bill hits the stag with
7
probability 10
, while George, independently, has a probability of hitting the
animal with probability 25 .
3
(a) Given that exactly one shot hits the stag, what is the probability that it
came from George’s gun?
Let G denote the event of George hitting. Let B denote the event of
Bill hitting. Thus Gc and B c denote the events of George and Bill
missing, respectively. The question tells us that G and B are independent.
P [G ∩ B c ]
P [ only 1 hit]
P [G ∩ B c ]
=
P [G ∩ B c ] + P [Gc ∩ b]
2
· 3
= 2 35 107 3
· + 10 · 5
5 10
2
=
.
9
P [G| exactly one hit] =
(b) Given that the stag is hit, what is the probability that George hit it?
P [G ∩ { hit}]
P [ hit ]
P [G]
=
P [ hit ]
P [G| hit] =
=
1−
20
=
.
41
2
5
3
10
·
3
5
8. Consider predicting the outcome of the 2008 Stanley Cup Final, which will
involve (for the sake of argument) the Ottawa Senators facing the Detroit
Red Wings. The final consists of a series of games played according to a
best-of-seven format, which means that the winning team is the first team to
win four games. Suppose that the probability of Ottawa winning each game
is p = .55, and that sequential games are considered to be independent
events. What is the probability that the Senators win the series? Provide an
explicit formula that indicates clearly your ability to answer this question
using a calculator.
4
Consider a four-game series resulting in an Ottawa win. There is only one
4
such series, with a probability of p4 = (.55)
(5) ≈
( ).0915. A five-game series
resulting in an Ottawa win can occur in 4 − 44 = 5 − 1 = 4 ways (i.e.,
the number of ways to distribute 4 wins in a sequence of 5 wins or losses
minus the one sequence that would involve a four-game sweep), and each
such series has a probability of p4 (1 − p) = (.55)4 × (.45) ≈ .0412. Similar
reasoning allows one to deduce that there exist
( ) {( ) ( )} ( )
6
5
4
4
−
−
−
= 15 − 4 − 1 = 10
4
4
4
4
and
( ) [( ) {( ) ( )} ( )] {( ) ( )} ( )
7
6
5
4
4
5
4
4
−
−
−
−
−
−
−
4
4
4
4
4
4
4
4
= 35 − 10 − 4 − 1
= 20
six and seven-game series, respectively, that lead to a Stanley Cup for Ottawa. Each six-game series has a probability of p4 (1 − p)2 ≈ .0185, while
the corresponding probability for a seven-game series is p4 (1−p)3 ≈ .0083.
Putting these results together, we obtain a probability of an Ottawa win
equal to
p4 + 4p4 (1 − p) + 10p4 (1 − p)2 + 20p4 (1 − p)3
≈ .6073.
(Note that this is not a binomial probability.)
9. An admissions director of a prestigious law school receives 50 applications
for admission, 22 of which are from men and 28 of which are from women.
The admission director is only permitted by the school to grant admission
to 20 people.
(a) Suppose that students are admitted completely randomly to the law
school. What is the probability that equal numbers of men and women
will be admitted?
(22)(28)
(50)10 .
10
20
5
(b) Suppose again that the admission process is completely random. What
is the probability that no women will be admitted?
(22)
)
(20
50 .
20
10. Suppose that a particular personal trait—like eye colour or handedness—is
classified on the basis of one pair of genes. Let d denote a dominant gene
and r a recessive one. A person with dd genes is “purely dominant”, one
with rr genes “purely recessive”, and one with rd genes a “hybrid”. Purely
dominant and hybrid individuals are alike in appearance with respect to the
trait in question. Children receive one gene from each parent. If, with respect to a particular trait, two hybrid parents have a total of four children,
what is the probability that exactly three of the four children have the outward appearance of the dominant gene? Assume that each child is equally
likely to inherit either of two genes from each parent.
The probability that a child of two hybrid parents is dd is 14 from the assumption that the probability of inheriting a gene of either type from a parent is
1
. Similarly, the probability that a child of two hybrid parents is rr is 14 , and
2
the probability that the same child is rd is 12 .
Since an individual will have the outward appearance of the dominant gene
if he or she is either dd or rd, the probability of a child of two hybrid parents
expressing the dominant gene is 34 . We are told that the two hybrid parents
have four children. As such,
( 3 )the number of children expressing the dominant gene is a binomial 4, 4 random variable. The desired probability is
accordingly
( ) ( )3 ( )
4
3
1
27
= .
3
4
4
64
11. Suppose that an airplane engine will fail in flight with probability 1 − p.
Assume that engine failures are independent events across engines, and that
once a failure has occurred, it is of a sufficiently serious nature that the
crew will not be able to restart the engine while the airplane is still flying.
Suppose also that an airplane will make a succesful flight if at least 50
percent of its engines do not fail. For what values of p is a four-engine plane
preferable—from a safety perspective, of course—to a two-engine plane?
Each engine fails or stays running in flight independently of what happens
to the other engines. The number of engines that do not fail, therefore, is a
6
binomial random variable. If the airplane has four engines, its probability
of making a successful flight is
( )
( )
( )
4 2
4 3
4 4
2
p (1 − p) +
p (1 − p) +
p (1 − p)0
2
3
4
= 6p2 (1 − p)2 + 4p3 (1 − p) + p4 .
Similarly, the probability of a two-engine plane making a sucessful flight is
( )
( )
2
2 2
p(1 − p) +
p
1
2
= 2p(1 − p) + p2 .
The four engine plane is safer if
6p2 (1 − p)2 + 4p3 (1 − p) + p4 ≥ 2p(1 − p) + p2 ,
which is equivalent to
6p(1 − p)2 + 4p2 (1 − p) + p3 ≥ 2(1 − p) + p = 2 − p,
i.e.,
(p − 1)2 (3p − 2) ≥ 0.
It is clear that (p − 1)2 ≥ 0 is both obvious and not particularly helpful, so
we must have that
3p − 2 ≥ 0.
Therefore a four-engine plane is safer when p ≥ 32 . A two-engine plane is
safer when p < 23 .
12. A committee from the International Skating Union is forming a panel of
eight judges for an upcoming international figure-skating competition. They
will be choosing judges from a pool that is comprised of 3 Canadians, 3
Americans, 4 Frenchmen, 4 Russians and 4 Norwegians.
(a) How many different panels of judges are possible?
There is a pool of 18 available judges from which the panel can be
formed, so there are
( )
18
18!
= 43758.
=
8
8!10!
different possible panels consisting of 8 judges.
7
(b) How many panels will involve all three of the available Americans?
( )
15
15!
= 3003.
=
5
5!10!
(c) If the judges are selected randomly, then what is the probability that
the panel will consist of at least one person from each of the five countries represented in the pool?
It is important to recognize that
P ( all 5 countries in the pool are represented in the panel )
= 1 − P ( panel excludes at least one of the 5 countries ) .
Now note that
P ( panel excludes at least one of the 5 countries )
=
3
∑
P ( panel excludes i countries ) ,
i=1
since it is not possible to exclude 4 countries and still have a sufficient
number of candidates to serve on the panel of 8 judges.
Consider each of the possibilities for excluding countries from the
panel:
( )
• If only one country is excluded, then there are 15
ways to form
8
( )
the panel if the excluded country is Canada or the U.S.A., and 14
8
possible panels if France, Russia or Norway is excluded. Therefore there are a total of
( )
( )
15
14
2×
+3×
8
8
possible panels involving the representation of 4 of the 5 countries
represented in the pool of judges.
( )
• If two countries are excluded, then there 12
possible
8
( )panels if
the two excluded countries are Canada and the U.S., 11
possible
8
panels if either Canada or the (U.S.
) is excluded along with one of
10
France, Russia or Norway and 8 panels if two of France, Russia
or Norway is excluded. Summing the possibilities for excluding
two countries from the panel, we find that there are
( )
( )
( )
12
11
10
+2×3×
+3×
8
8
8
8
possible panels of eight judges involving the representation of
only 3 countries.
• (Finally,
if three countries are excluded, then there are only 3 ×
)
8
=
3
possible
panels that can be formed—here each panel will
8
consist of representatives of two countries in the set
{ France, Russia, Norway } .
Gathering previous results, we have that
P ( all 5 countries in the pool are represented in the panel )
{
( )
(14)} {(12)
( )
(10)}
2 × 15
+ 8 + 6 × 11
+3
8 +3× 8
8 +3× 8
(18)
= 1−
8
≈ .4629.
13. The number of times that a certain Russian ice-dance pair falls during their
routine in an upcoming international figure-skating competition is modelled
as a binomial random variable with n = 5 trials, each having a “success
probability” of p = .3. Note that here p denotes the probability of falling.
The number of times that a competing Canadian ice-dance pair falls during
their routine also follows a binomial distribution, but this time with n = 4
and p = .2. Assume that the number of times the separate pairs fall are
independent events.
(a) What is the probability that both pairs do not fall during their routines?
The assumption of independence allows us to multiply the binomial
probabilities corresponding to each pair not falling at all during their
routines, so
( )
( )
5
4
0
5
(.3) (.7) ×
(.2)0 (.8)4 ≈ .0688.
0
0
(b) What is the probability that the Russian pair experiences a greater
number of falls than the Canadian pair?
Let R and C denote the number of falls experienced by the Russians
and the Canadians, respectively. We want to calculate
P (R > C)
= P (R = 1, C = 0) + P (R = 2, C = 0) + · · · + P (R = 5, C = 4)
9
=
5
4
∑
∑
P (R = r, C = c)
r=c+1 c=0
=
5
4
∑
∑
P (R = r) P (C = c)
r=c+1 c=0
( )
5
4 ( )
∑
∑
5
4
r
5−r
=
(.3) (.7)
×
(.2)c (.8)4−c .
r
c
r=c+1 c=0
14. A set of 150 examination scores has a distrribution with a mean of 75 and a
standard deviation of 7. Approximately how many of the scores would you
expect to lie in the interval from 61 to 89? Which assumptions did you make
concerning the distribution of scores in order to answer this question?
Note that the interval (61, 89) includes approximately all scores within two
standard deviations of the mean. If one assumes that the distribution of
scores is approximately normal in the sense of corresponding to a symmetric
and mound-shaped relative frequency histogram, then the interval (61,89)
will contain approximately 95% (or about 142) of the scores.
15. A group of three male and five female Criminal Court judges are available
to try various different cases. If four judges are to be randomly selected from
this group to hear cases on a given day, find the probability that exactly two
male judges will be amongst the four chosen.
(8)
Note that there are
(3)(4 5)= 70 ways to choose four judges from the group of
eight. There are 2 2 = 30 ways to choose exactly 2 male judges from
the available three and 2 female judges from the five available. Random
selection implies that each sample point is equally likely, so the probability
30
is 70
= 37 = .4286.
16. A famous shipwreck is presumed to have an equal probability of being found
in any one of three stretches of ocean. Supposing that the wreck is actually
in region i (where i ∈ {1, 2, 3}), let 1−pi denote the conditional probability
of finding the wreck after a search of region i. What is the conditional
probability that the wreck is actually in:
(a) region 1, given that a search of region 1 is unsuccessful?
For i ∈ {1, 2, 3}, let Fi denote the event that the shipwreck is found
after a search of region i. Let Wi denote the event that the wreck is
10
actually in region i. Then P [ Fi | Wi ] = 1 − pi , while P [Wi ] =
i = 1, 2, 3. Then
1
3
for
P [ W1 | F1c ]
P [F1c | W1 ] P [W1 ]
P [F1c | W1 ] P [W1 ] + P [F1c | W2 ] P [W2 ] + P [F1c | W3 ] P [W3 ]
p1 · 31
=
p1 · 13 + 13 + 13
p1
=
.
p1 + 2
=
(b) region 2, given that a search of region 1 is unsuccessful?
Similar reasoning yields
P [ W2 | F1c ] =
1
.
p1 + 2
(c) region 3, given that a search of region 1 is unsuccessful?
P [ W3 | F1c ] =
1
.
p1 + 2
17. A rental agency for construction equipment leases a heavy backhoe by the
day. A check of company records shows that that piece of equipment is
leased on average only one day out of every five. Assume that the probability of the equipment being rented on any given day is independent of the
probability of rental on any other day. Let X be the number of days between
successive rentals. Find the density function of X.
Suppose the backhoe is rented on a given day. Then the backhoe is rented
on the next day with probability .2 and not rented with probability .8, which
implies that P [X = 0] = .2. Similar reasoning yields P [X = 1] = .2 ×
.8 = .16 and P [X = 2] = .2 × .82 = .128. In general, X has support equal
to the set of whole numbers and density function given by
{
.2 × .8x , x = 0, 1, 2, . . .
pX (x) =
0
,
otherwise.
18. A 1980 article in the Aeronautical Journal1 models the probability of detecting a crack in an airplane wing as the product of p1 , the probability of
1
Goranson, U.G. and J. Hall (1980). “Airworthiness of Long-Life Jet Transport Structures”.
Aeronautical Journal 84:279–280.
11
inspecting a plane with a wing crack; p2 , the probability of inspecting the
location in the wing in which the crack is located; and p3 , the probability of
detecting the damage.
(a) What assumptions justify the multiplication of p1 , p2 and p3 ?
The three events associated with p1 , p2 and p3 need to be mutually
independent.
(b) Suppose p1 = .9, p2 = .8 and p3 = .5 in connection with a given
airline’s fleet. Suppose three aircraft are inspected from this fleet. Find
the probability that a crack will be detected on the wing of at least one
of these airplanes.
Let X denote the number of airplanes in the group of three in which
wing cracks are detected. Then X is binomial with n = 3 and p =
p1 p2 p3 = .9 × .8( ×
that P [X ≥ 1] = 1 −
) .5 =0 .36. It follows
3
3
P [X = 0] = 1 − 0 (.36) (1 − .36) = 1 − (.64)3 = .7379.
12