Bowdoin Math 1200, Spring 2017 – Midterm Exam #1 – Solutions
Instructor: Mario Micheli
Date/time: Friday, March 3, 2017, 10:00 p.m. – 11:25 p.m.
Place: Searles 213, Bowdoin College
• You have 85 minutes to complete this midterm. This exam has 7 pages, including this cover sheet.
• There are 6 problems. I suggest that you start with the problems that you find the easiest.
If you are stuck, move on to the next problem!
• No books, no printed material, and no homework are allowed. You may use a calculator, though you may
leave your answers in terms of sums and multiplications of fractions, factorials, and binomial coefficients.
Besides pencil, paper, and a calculator, only a one-sided, hand-written, letter-sized sheet of paper with your
own choice of definitions, results and formulas is allowed.
• Show your work. Ambiguous or otherwise unreadable answers will be marked incorrect.
So: write clearly and provide only one answer to each question.
Your Name (LAST, First):
I certify that the work appearing on this exam is completely my own.
(Your signature)
Problem #
Points
1
18
2
20
3
12
4
12
5
14
6
24
Total
100
Your score
c Copyright 2017 by Mario Micheli
1
Problem 1. (6 + 12 = 18 points total) Parts (a) and (b) may be solved INDEPENDENTLY.
(a) (6 points) A professor scaled (curved) the scores on an exam by multiplying the students’ raw score by
1.2, and then adding 15 points. The mean and standard deviation of the raw scores before the curve
were 51 and 5, respectively. What are the mean and standard deviation of the scaled scores?
Answer: By the results of Problem 2D in HW#2, given two data sets X = {X1 , X2 , . . . , Xn } and
2 , and σ = |a| σ .
Y = {Y1 , Y2 , . . . , Yn } such that yi = axi + c, then we have µY = aµX + c, σY2 = a2 σX
Y
X
In the case of this problem we have that: a = 1.2, c = 15, µX = 51 and σX = 5. Therefore:
new mean = µY = (1.2)(51) + 15 = 76.2 ,
new standard deviation = σY = (1.2)(5) = 6 .
(b) (4 × 3 = 12 points total) The scores of female (F) and male (M) students on a statistics exam are
displayed in the following boxplots. The pluses (+) indicate the location of the means.
Answer the following questions:
• Approximately, what are the IQRs of the two score distributions for male and female students?
IQRF ' 90 − 55 = 35 ,
IQRM ' 90 − 45 = 45 .
• Are the histograms for the two score distributions skew left, skew right, or symmetric? Why?
Answer: They are both skew left : it can be seen by the fact that, in both cases, the mean, indicated by a plus (+), is less than the median, indicaded by the central vertical bar in the box plot.
• The top-scoring 75% of the female students scored more than a, and the bottom-scoring 50% of
the of the male students scored less than b, where:
a = 55 ,
b = 70 .
• Do you have enough information to compute the mean score for the entire class? (Justify!)
Answer: No , because we do not know how many female students and how many male students
there are: we need these numbers to compute proportions—see Problem 1B, part (a), in HW#1. 2
Problem 2. (20 points total) Suppose P (A) = 0.1 and P (B) = 0.5.
(a) (2 pts.) If P (A|B) = 0.1, what is P (A ∩ B)?
(b) (2 pts.) If P (A|B) = 0.1, are A and B independent?
(c) (2 pts.) If P (A ∩ B) = 0, are A and B independent?
(d) (2 pts.) If P (A ∪ B) = 0.55, are A and B mutually exclusive?
(e) (4 pts.) If A and B are mutually exclusive, what is P (Ac ∪ B)?
(f ) (4 pts.) If A and B are independent, what is P (Ac ∪ B)?
(g) (4 pts.) If A and B are independent, what is the probability that exactly one of the two events A
and B occurs, but not both? Hint: Draw a suitable Venn Diagram.
Justify your answers precisely.
Answers:
(a) By the Multiplication Rule: P (A ∩ B) = P (A|B)P (B) = (0.1)(0.5) = 0.05 .
(b) Yes . By part (a) we have: P (A ∩ B) = 0.05, and P (A)P (B) = (0.1)(0.5) = 0.05 = P (A ∩ B).
(c) No . If they were independent we would have P (A ∩ B) = P (A)P (B) = (0.1)(0.5) = 0.05, which
contradicts the assumption that P (A ∩ B) = 0.
(d) No . We have P (A∪B) = P (A)+P (B)−P (A∩B), which in this case gives 0.55 = 0.1+0.5−P (A∩B),
and solving for P (A ∩ B) yields P (A ∩ B) = 0.05. If A and B were mutually exclusive (A ∩ B = ∅)
we would have P (A ∩ B) = P (∅) = 0.
(e) If A∩B = ∅ then B ⊆ Ac and Ac ∪B = Ac (see Figure 1), so P (Ac ∪B) = P (Ac ) = 1−P (A) = 0.9 .
(f ) In this case:
(∗)
(∗∗)
P (Ac ∪ B) = 1 − P (Ac ∪ B)c = 1 − P (A ∩ B c ) = 1 − P (A)P (B c )
= 1 − P (A) 1 − P (B) = 1 − (0.1)(0.5) = 1 − 0.05 = 0.95 ;
in (∗) we have used DeMorgan’s law, and in (∗∗) the fact that if A and B are independent, so are the
events A and B c (see Problem 4D in HW#4). Alternatively, by direct inspection of Figure 2, we have
P (Ac ∪B) = 1−P (A)+P (A∩B) = 1−P (A)+P (A)P (B) = 1−0.1+(0.1)(0.5) = 0.9+0.05 = 0.95 .
(g) We have that “exactly one of the two events A and B occurs, but not both” = (A ∩ B c ) ∪ (Ac ∩ B)
(see Figure 3 below). The two events A ∩ B c and Ac ∩ B are disjoint (mutually exclusive), therefore:
(∗)
P (A ∩ B c ) ∪ (Ac ∩ B) = P (A ∩ B c ) + P (Ac ∩ B) = P (A)P (B c ) + P (Ac )P (B)
= P (A) 1 − P (B) + 1 − P (A) P (B)+ = (0.1)(0.5) + (0.9)(0.5) = 0.05 + 0.45 = 0.5 ,
where in the step (∗) we have sed the fact that if A and B are independent, so are the pair of
events Ac and B, and A and B c (again, see Problem 4D in HW#4).
\
Figure 1.
Figure 2.
3
Figure 3.
Problem 3. (12 points) Suppose that, in a particular city, airport A handles 50% of all airline traffic, and
airports B and C handle 30% and 20%, respectively. The detection rates for weapons at the three airports
are 0.9, 0.8, and 0.85, respectively. If a passenger at one of the airports is found to be carrying a weapon
through the boarding gate, what is the probability that the passenger is using airport A?
Answer: The tree diagram that describes the problem is illustrated below:
where the events are the following:
A = “the passenger goes through airport A”,
B = “the passenger goes through airport B”,
C = “the passenger goes through airport C”,
D = “the passenger’s weapon is detected”.
Now, by Bayes’ rule it is the case that:
P (D|A)P (A)
P (D|A)P (A)
=
P (D)
P (D|A)P (A) + P (D|B)P (B) + P (D|C)P (C)
0.45
(0.9)(0.5)
=
=
' 0.5232 .
(0.9)(0.5) + (0.8)(0.3) + P (0.85)(0.2)
0.86
P (A|D) =
where to compute the denominator we have used the Total Probability Theorem.
4
Problem 4. (12 points total) Consider a family that has two children, and the following events:
G1 =“the first (oldest) child is a girl”
G2 =“the second (youngest) child is a girl”
B1 =“the first child is a boy”
B2 =“the second child is a boy”
Assume that having a girl or a boy are equally likely, and that the gender of the first and second children
are independent.
(a) (4 points) What is the probability that both children are girls, given that the first child is a girl?
(b) (4 points) What is the probability that at least one of the two children is a girl?
(c) (4 points) What is the probability that both children are girls, given that at least one is a girl?
Answer: We have that P (B1 ) = 1/2, P (B2 ) = 1/2, P (G1 ) = 1/2, P (G2 ) = 1/2, and in fact G1 = B1c and
G2 = B2c . Moreover B1 and B2 are independent and (by Problem 4D in HW#4), so are the following pairs
of events: B1 and G2 , G1 and B2 , G1 and G2 .
(a) The probability that both children are girls, given that the first child is a girl is
P (G2 |G1 ) =
P (G1 ∩ G2 ) (∗) P (G1 )P (G2 )
1
=
= P (G2 ) =
,
P (G1 )
P (G1 )
2
where in (∗) we have used the indepednence of the events G1 and G2 .
(b) The probability that at least one of the two children is a girl:
P (G1 ∪ G2 ) = P (G1 ) + P (G2 ) − P (G1 ∩ G2 )
(∗∗)
= P (G1 ) + P (G2 ) − P (G1 )P (G2 ) =
1
3
1 1 1 1
+ − · =1− =
,
2 2 2 2
4
4
where, once again, in (∗∗) we have used the indepednence of the events G1 and G2 .
Another way to find the same answer is by using the binomial distribution, namely:
2 1 1 1 1
2 1 2 1 0
1 1 1
1 1
3
+
=2· · + = + =
.
1 2
2
2 2
2
2 2 4
2 4
4
(c) Finally, the probability that both children are girls, given that at least one is a girl is
P (G1 ∩ G2 | G1 ∪ G2 ) =
P ((G1 ∩ G2 ) ∩ (G1 ∪ G2 )) (∗∗∗) P (G1 ∩ G2 )
P (G1 )P (G2 )
1/4
1
=
=
=
=
.
P (G1 ∪ G2 )
P (G1 ∪ G2 )
P (G1 ∪ G2 )
3/4
3
Note that G1 ∩ G2 ⊆ G1 ∪ G2 , therefore (G1 ∩ G2 ) ∩ (G1 ∪ G2 ) = G1 ∩ G2 , which we used in (∗∗∗).
Remark: Some of you solved the problem recurring to a tree diagram, which leads to the same answers. 5
Problem 5. (14 points) This is the trickiest problem. An urn contains 10 marbles: 4 red and 6 blue. A
second urn contains 16 red marbles and an unknown number (call it n) of blue marbles. A single marble
is drawn from each urn, in an independent manner. The probability that both marbles are the same color
is 0.44. Calculate the number of blue marbles in the second urn.
Hint: consider the events
R1 = “the marble drawn from the first urn is red”
R2 = “the marble drawn from the second urn is red”
B1 = “the marble drawn from the first urn is blue”
B2 = “the marble drawn from the second urn is blue”
We have that: “both marbles are the same color” = (R1 ∩ R2 ) ∪ (B1 ∩ B2 ). (Why?) Also, note that the
events R1 and R2 are independent; similarly, B1 and B2 are independent.
Setting up the equation correctly will give you most of the credit, even if you do not solve for n.
Answer: The situation is effectibely described by the following graph (note that it is not a tree, because we
pick a marble from both Urn 1 and Urn 2—we do not choose the urn at random):
Note that the event (R1 ∩ R2 ) ∪ (B1 ∩ B2 ) may be read as “either the marbles are both red (i.e. R1 ∩ R2 ) or
the marbles are both blue (i.e. B1 ∩ B2 )”: in other words, “both marbles are the same color”. So we have:
0.44 = P “both marbles are the same color” = P (R1 ∩ R2 ) ∪ (B1 ∩ B2 )
4
16
6
n
= P (R1 ∩ R2 ) + P (B1 ∩ B2 ) = P (R1 )P (R2 ) + P (B1 )P (B2 ) =
+
,
10 16 + n 10 16 + n
where in (∗) we have used the fact that the events (R1 ∩ R2 ) and (B1 ∩ B2 ) are mutually exclusive, whereas
in (∗∗) we have used the fact that the events R1 and R2 are independent, and the events B1 and B2 are also
independent. In conclusion, we have the following equation, where n is an unknown:
4
16
6
n
+
= 0.44 .
10 16 + n 10 16 + n
Solveing for n we get:
64 + 6n
10(16 + n)
=⇒
64 + 6n = (0.44)(10)(16 + n)
=⇒
640 + 60n = 44 (16 + n)
=⇒
16n = 64
=⇒
=⇒
n=4 .
6
=⇒
64 + 6n = 4.4 (16 + n)
640 + 60n = 704 + 44n
Problem 6. (3 points each, 24 points total.) Short answer questions.
6.1. Let µ and σ be the mean and standard deviation of a data set. Typically, how much of the data lies
in the interval between µ − σ and µ + σ?
Answer: Typically, around 68% (the answer “70%” is fine too).
6.2. You come across the following data set: X = {7, 4, 10, 5, 9, 2, 9, 5}. Compute median and IQR.
Answer: Ordered set: 2, 4, 5, 5, 7, 9, 9, 10.
4+5
5+7
= 6 ,
Q1 =
= 4.5,
median =
5
2
Q2 =
9+9
= 9,
2
IQR = 9 − 4.5 = 4.5 .
6.3. CIRCLE which ones of the following statistics of
a data set are robust. (No justification is required.)
(a) Minimum
(b) Q3
(c) Mean
Note: both (a) and (c) are sensitive to extreme values in the data set.
6.4. CIRCLE one statement which is TRUE. (No justification is required.)
(a) 0 ∈ {2, 3}
Note: 0 is not an element of the set {2, 3}.
(b) ∅ ⊆ {2, 3}
Note: The empty set ∅ is a subset of any set.
(c) {2} ∈ {2, 3} Note: {2} is not an element, but a set. We should write {2} ⊆ {2, 3}.
(d) 2 ⊆ {2, 3}
Note: 2 is not set, not an element. We should write 2 ∈ {2, 3}.
6.5. Three cards are drawn from an ordinary 52-card deck without replacement. Compute the following:
P (1st and 2nd card are Hearts, and 3rd one is not Hearts)
Answer: P (H1 ∩ H2 ∩ H3c ) = P (H1 )P (H2 |H1 )P (H3c |H1 ∩ H2 ) =
13 12 39
·
·
' 0.04588 .
52 51 50
6.6. CIRCLE one statement which is FALSE. (No justification is required.)
(a) (A ∪ B)c = Ac ∩ B c
(b) P (A) = P (A|B)P (B)
Note: the correct statement is P (A ∩ B) = P (A|B)P (B) .
(c) If P (A|B) = P (A), then A and B are independent.
(d) P (A ∪ B ∪ C) = P (A) + P (B) + P (C) − P (A ∩ B) − P (A ∩ C) − P (B ∩ C) + P (A ∩ B ∩ C)
6.7. Assume that the chances of being born girl is p = 0.49. If a family has five children, what is the
probability that at most two of these are girls?
5
5
5
0
5
1
4
Answer:
(.49) (.51) +
(.49) (.51) +
(.49)2 (.51)3 ' 0.5147 .
0
1
2
6.8. How many distinct words (that is, how many different letter arrangements) can we form with the letters
in the word MISSISSIPPI? (Feel free to leave the answer in terms of factorials.)
Answer: There are 11 letters in the word MISSISSIPPI, with some repetitions (four S’s, four I’s, and
two P’s). There are 11! permutations of 11 letters: however, we have to ‘factor out’ the permutations
of the repeated letters, otherwise we would count the same word multiple times. The answer is:
11!
= 31, 650 .
4! 4! 2!
Note that there
is another line of reasoning that leads to the same answer: we have
11 ‘slots’ to fill.
7
There are 11
ways
of
placing
the
four
S’s;
now
there
are
7
slots
left.
There
are
4
4 ways of placing
3
the four I’s; now there are 3 slots left. There are 2 ways of placing the four P’s; now there is only
one slot left, and there is only 11 = 1 way of placing there the letter M. In total, the number of ways
11 7 3 1
11!
7!
3!
1!
11!
.
we have to arrange the letters is given by:
=
·
·
·
=
4
4 2 1
4! 7! 4! 3! 2! 1! 1! 0!
4! 4! 2!
7