solutions

Chalmers TU course ESS011, 2014: mathematical statistics homework
Week 2
These are the solutions of the homeworks for course week 2 sessions (26.& 28.3.2014), related to topics discussed
on weeks 1 and 2.
1. Sample spaces
Write down the sample space and the event of interest in the following random experiments:
a) You bet that you can get at least one die to score above 4 when rolling two dice.
Solution: Sample space S = {(1, 1), (1, 2), (1, 3), (1, 4), (1, 5), (1, 6), ...(6, 2), (6, 3), (6, 4), (6, 5), (6, 6)} with a
total of 36 entries. The event of interest A = {(1, 5), (1, 6), (2, 5), (2, 6), ..., (5, 1), (5, 2), (5, 3), ..., (6, 5), (6, 6)}
with a total of 20 entires.
b) You have jar of candy with 3 red, 5 green and 4 blue jelly beans. You don’t like green ones. You are about to
randomly pick one: will you like it?
Solution: S = {R, G, B}, A =”Pick a bean you don’t like”= {G}, P (A) =
5
12
c) You arrive to the tram stop. You only know that the tram comes once per hour, so from your perspective its
arrival time is random. What is the chance the tram arrives in the next 10 minutes? (in minute resolution)
Solution: S = 1, 2, 3, 4, ..., 58, 59, 60, A = {”T ram arrives in the next ten minutes”} = 1, 2, 3, ..., 8, 9, 10.
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The probability becomes, P (A) = 10
60 = 6 .
2. Combinatorics The formula for the number of combinations n Ck is also called the binomial coefficient,
n!
denoted by nk :=
.
k!(n − k)!
a) Evaluate 6!, 6 P2 and 6 P4 .
Solution: 6! = 720, 6 P2 = 30 and 6 P4 = 360
b) Find n when n2 = 28
n(n − 1)
n(n − 1)
n!
=
. Solve
= 28 giving n = 8
Solution: Note that n2 =
2!(n − 2)!
2
2
c) Evaluate 6i for i = 0, ..., 6.
Solution: 60 = 1, 61 = 6, 62 = 15, 63 = 20, 64 = 15, 65 = 6, 66 = 1.
d) How many 4 letter word are there that begin with ’b’ or ’s’ with the rest made of 2 ’e’s and 1 ’d’ ? (hint: use
multiplication rule and permutations of indistinguishable objects)
Solution: First note that there are two possibilities for the first letter in the word. For the last three letters we
3!
need to use the formula for permutations of indistinguishable objects (see page 6 of the book) giving: 2!1!
= 3.
The multiplication principle (page 10) gives 2 ∗ 3 = 6 possible words.
Notice the symmetry in a). Do you think it has a meaning in terms of probabilities of the complements?
3. Dead batteries A flashlight operates on two batteries which need to be replaced. Nine batteries are available,
but three are dead. In a random selection of batteries, what is the probability that exactly one dead battery will
be selected? (Hint: does order matter?)
Solution: Order does not matter. The total number of ways to select two batteries from nine is 92 = 36. The
number of ways to select exactly one dead battery is 31 61 = 18. The probability to select exactly one dead batter
is P ({”Exactly one dead”}) = 18
36 = 0.5.
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Chalmers TU course ESS011, 2014: mathematical statistics homework
Week 2
4. The duke’s dice problem Galileo Galilei once wrote a book about gambling, called ”Thoughts about Dice
Games”. In it he describes a solution to a problem imposed on him by his patron, the grand duke of Tuscany. The
duke was a passionate gambler, and he was troubled by the following question: Why is it that when you throw 3
dice at once, number 10 appears more frequently than number 9 as the sum of the dice scores?
The dukes confusion was this: If you can have the sum of 9 with the 6 combinations
{1, 2, 6}, {1, 3, 5}, {1, 4, 4}, {2, 2, 5}, {2, 3, 4} and {3, 3, 3},
and the sum 10 with also 6 combinations
{1, 3, 6}, {2, 2, 6}, {1, 4, 5}, {2, 3, 5}, {2, 4, 4} and {3, 3, 4},
meaning we can have both sums in 6 ways, why are the probabilities not the same?
Solve the duke’s dice problem by computing the probabilities P ({sum of 3 dice is 9}) and P ({sum of 3 dice is 10})
using the knowledge of the multiplication, permutation and combination.
Solution: First using the formula for permutations of indistinguishable objects (page 6) we can calculate that
there is only one way to get {3, 3, 3}, three ways to get {2, 2, 5} and six ways to get {2, 3, 4}. That is the number
of ways to get a specific combination is smaller when there are identical numbers in that combination. Since the
combinations listed are all mutually exclusive we can simply add them together to get the total number of ways
possible. For {sum of 3 dice is 9} we have 6 + 6 + 3 + 3 + 6 + 1 = 25 possible ways. For {sum of 3 dice is 10} we
have 6 + 3 + 6 + 6 + 3 + 3 = 27 possible outcomes. Divide by the total number of outcomes (63 = 216) to get
25
27
P ({sum of 3 dice is 9}) = 216
and P ({sum of 3 dice is 10}) = 216
5. Axioms of probability Prove the following theorems using the three Axioms of Probability:
Note: The solutions below are suggestions, there are multiple ways to show these results.
a) P (∅) = 0
Solution: Note that P (S) = 1 and S ∪ ∅ = S and that S and ∅ are mutually exclusive. General addition rule
yields P (S ∪ ∅) = P (S) + P (∅) = 1 which implies that P (∅) = 0.
b) P (A0 ) = 1 − P (A) for any event A
Solution: Note S = A∪A0 and that A and A0 are mutually exclusive. General addition rule yields P (A∪A0 ) =
P (A) + P (A0 ) = 1 giving P (A0 ) = 1 − P (A).
c) P (A) ≤ 1 for all events A
Solution: By the reasoning in b) note that P (A) + P (A0 ) = 1. The second axiom of probability gives that
P (A0 ) ≥ 0 which implies that P (A) ≤ 1
d) Let events A and B be mutually exclusive. Axiom 3 says that P (A ∪ B) = P (A) + P (B). Show that general
addition rule yields the same result.
Solution: The general addition rule states P (A ∪ B) = P (A) + P (B) − P (A ∩ B). Since A and B are mutually
exclusive P (A ∩ B) = 0 yielding the desired result.
6. Basic computations with probabilities
a) Lecture 2, slide 17, blood types: What is the probability that the patient is not of type A or B?
Solution: P (Not A or B) = P ((A ∪ B)0 ) = 1 − P (A ∪ B) = 1 − P (A) − P (B) = 1 − .44 − .12 = 044. Note that
A and B are mutually exclusive.
b) Let P (A) = 0.3 and P (A ∩ B) = 0.2. Compute P (B|A).
Solution: P (B|A) =
(P (A∩B)
P (B)
=
0.2
0.3
≈ 0.667
c) Let P (A|B) = 0.9 and P (B 0 ) = 0.3. Compute P (A ∩ B).
Solution: Reorganization of the formula for conditional probability gives P (A ∩ B) = (1 − P (B))P (A|B) =
(1 − 0.3)0.9 = 0.63 .
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Chalmers TU course ESS011, 2014: mathematical statistics homework
Week 2
d) Lecture 2, slide 19, toxins: What are the probabilities P (A0l |A0m ) and P (A0m |A0l )?
Solution: P ((A0l ∩A0m ) = P ((Al ∪Am )0 ) = 1−0.38 = 0.62 , P (A0l ) = 1−0.32 = 0.68 , P (A0m ) = 1−0.16 = 0.84
0.62
0
0
P (A0l |A0m ) = 0.62
0.84 ≈ 0.738 , P (Am |Al ) = 0.68 ≈ 0.912
7. Horrible illness Horrible illness ravaged the medieval country side in west Sweden. Roughly 40% of children
of age 15 or less got the illness, and of those only 10% survived.
You are a historian and are now studying death records of that era and area. You notice that on average, 37% of
the population died at the age of 15 or less. What are the chances that a child in the records with a missing cause
of death died of the illness?
Solution: Define the events I : Had the illness, S : Survived. We are given P (I) = 0.4, P (S|I) = 0.1 and P (S 0 ) =
0.37. Through the complement we also know that P (S 0 |I) = 0.9. We seek the probability that a child who died did
0
(I)
= 0.9∗0.4
so because of the disease, i.e. P (I|S 0 ). Bayes theorem (page 36) gives us P (I|S 0 ) = P (SP |I)P
(S 0 )
0.37 ≈ 0.972.
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