Stat330 - Solution to Homework 1
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Alpha to Omega
Using the greek alphabet sheet to transcribe the letters gives the following sayings:
Data is a lot like humans: It is born. Matures. Gets married to other data, divorced. Gets old. One
thing that it does not do is die. It has to be killed.
Arthur Miller
Round numbers are always false.
Samuel Johnson
Smoking is one of the leading causes of statistics.
Fletcher Knebel
The following letters are the one we are going to use in class. Write down their names and draw them
yourself:
ΣασλθβχΦ
The letters’ names are:
Sigma alpha sigma lambda theta beta epsilon chi Phi
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Math Stuff
Math turns out to be one of the major stumbling blocks later on in the class - to prepare you and prevent
the math from getting into your way, here are a couple of typical examples of what you should be able to
to do. If you find you have difficulties with this type of problems - have a look at your analysis text books
again!
(a) Differentiate the following functions with respect to x:
(a) f (x) = x2 (1 − x2 )
f 0 (x) = 2x − 4x3
(b) g(x) = eαx+β
g 0 (x) = αeαx+β
(c) h(x, t) = ln(x + 2t)
h0 (x) =
1
x + 2t
(b) Decide for which of the following functions, you can find a value for C ∈ R, such that the integral has
a value of 1. If possible, give the value of C.
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(a)
R1
0
Cx2 (1 − x2 )dx
Z
1
Cx2 (1 − x2 )dx = C
Z
1
x2 − x4 dx =
0
0
= C(1/3x3 − 1/5x5 ) |10 = C(1/3 − 1/5) =
= 2/15C
(b)
To get a result of 1, we have to pick C = 15/2 = 7.5.
R∞
Cdx
0
Z
∞
Z
∞
1dx = Cx |10 = ∞C
Cdx = C
0
0
We do not find a value for C to make this integration result a 1.
R∞
(c) 1 Cx−2 dx
Z
∞
−2
Cx
Z
∞
x−2 dx = C − 1/x |1∞ = C(0 + 1) = C
dx = C
1
1
For C = 1 the integral is 1.
R∞
(d) 0 Ce−3t dt
Z
∞
Ce−3t dt
Z
= C
0
∞
e−3t dt = C − 1/3e−3t |0∞ = C(0 + 1/3) = 1/3C
0
For C = 3 the integral is 1.
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Sample Spaces
Determine the sample spaces for each of the following activities:
(a) Rolling a die.
The sample space with the highest resolution is (for a regular six sided die):
Ω = {1, 2, 3, 4, 5, 6}
If the die is fair, all elementary outcomes have the same probability.
(b) Tossing a coin three times.
tossing a coin three times: the sample space with the highest resolution is
Ω = {HHH, HHT, HT H, T HH, HT T, T HT, T T H, T T T }
If the coin is fair, each elementary outcome is equally likely.
(c) Counting the number of heads in five coin tosses.
the sample space for this experiment is
Ω = {0, 1, 2, 3, 4, 5} number of heads
We get a 0, if the coin showed tails five times in a row; we get a 5, if the coin showed heads five times
in a row. If we have a fair coin, both of these outcomes are fairly unlikely compared to 2 or 3 counts
of heads in a five toss sequence - i.e. in the above sample space not all elementary outcomes have the
same probability.
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(d) Rolling a pair of dice and taking their sum.
For the sum of two dice we get a sample space of
Ω = {2, 3, 4, ..., 12}
Again, not all elementary outcomes have the same probability (under the assumption that we are dealing
with fair dice) - e.g. we get a 2 as a sum only, if both dice show a 1, whereas a 7 could be the result of
a (1,6), (2,5), (3,4), (4,3), (5,2), or (6,1).
(e) Drawing a card from a standard deck.
A standard deck of cards has 52 different cards, Ω is a listing of all these possibilities. If we draw a
card at random, we can assume, that each card appears with the same probability.
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Playing Sports
Out of a group of 40 students, all play at least one of badminton, volleyball or table tennis.
8 students play all three games, 10 students play badminton and table tennis
20 students play table tennis and volleyball, 12 students play badminton and volleyball
30 students play table tennis, 25 students play volleyball.
Draw a Venn Diagram of the situation.
(a) How many of the students play only badminton?
(b) How many of the students play badminton?
Assume, one student is picked at random out of this group. What is the probability that he/she
(c) plays badminton?
(d) plays at least two sports (out of badminton, volleyball and table tennis)?
Assume the student you’ve picked is a volleyball player.
(e) What is now the probability that he/she is a badminton player?
(f) Why is the probability for picking a badminton player in (e) different from the probability in (b)?
For this example, find two events that are
(g) disjoint
(h) independent
With events V, B, and T defined as:
V := a student plays volleyball,
B := a student plays badminton,
T := a student plays table tennis,
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a Venn Diagram of the above situation looks like:
5
B
1
4
V
8
12
2
T
0
8
Ω
With that, we can answer questions (a) and (b):
(a) the number of students playing volleyball only is 1.
(b) 19 students play badminton.
Since a student is picked at random, each student has the same probability to be picked. We can assume a
sample space Ω = {s1 , s2 , s3 , ..., s40 } and the events V, B, and T are subsets of students playing this particular
sport.
(c) the probability that the student plays badminton therefore is
P (B) =
19
|B|
=
= 0.475.
|Ω|
40
(d) the probability that he plays at least two sports is the probability to play volleyball and badminton, or
badminton and table tennis, or table tennis and volleyball or all three sports. by looking at the venn
diagram, it’s easy to see, that we are dealing with the gray shaded area:
B
5
1
4
V
8
12
2
0
T
8
Yielding a probability of P ( at least two sports ) =
Ω
4+8+2+12
40
= 0.65.
Using a set notation, it’s slightly more difficult to come up with the same result:
P ( at least two sports )
=
=
=
=
P ((V ∩ B) ∪ (V ∩ T ) ∪ (B ∪ T ) ∪ (B ∩ T ∩ V )) =
... =
P (V ∩ B) + P (V ∩ T ) + P (B ∪ T ) − 2P (B ∩ T ∩ V ) =
0.3 + 0.5 + 0.25 − 2 · 0.2 = 0.65.
The ... part is fairly lengthy.
Now we found out that the student picked is a volleyball player. That changes the probability, that he is a
badminton player, too:
(e) we know the student is a badminton player, so we are reducing our sample space to V only (he’s one
out of 25 volleyball players), 12 out of the 25 volleyball players also play badminton. Therefore the
probability, that he plays badminton is 12/25 = 0.48.
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(f ) the different probabilities for playing badminton are due to our gain in knowledge. For the second
part, we already knew more, we therefore were able to reduce the sample space. Question (e) asks for
a conditional probability: the conditional probability of playing badminton given that a student plays
volleyball.
P (B | V ) =
P (B ∩ V )
0.3
=
= 0.48.
P (V )
0.625
For this example, find two events that are
(g) disjoint Disjoint events do not have a common element, so we could, for example take the set of students
that only plays table tennis and the set of students that only plays volleyball. These two sets correspond
to events that are disjoint.
(h) independent two events are independent, if
P (A ∩ B) = P (A) · P (B)
If we want to be sneaky, we can pick one of those two events to be the zero - e.g. B = ∅. Then it does
not matter how we choose A - the two events A and B are always independent.
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