Answer to all of these mathematic exercises.

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Mathematics

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- Correlations - Probability, probability distribution - Random variables


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Explanation & Answer

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Math Exercises
March 13, 2018

1

Exercise 1

By definition, the probability of any event must be between 0 and 1. Therefore,
all numbers not in this range cannot be probabilities. These are
1, 2;

2


77 1
; − ; 5; 2
75 2

Exercise 2

Each person has the same probability of being picked. Moreover, we know that
someone must be picked, so the sum of probabilities must be 1. Then if p is the
probability per person and there are n people, np = 1 =⇒ p = n1 . Since we
have n = 30, then the probability of any one person getting picked (namely, the
1
tallest) is 30
.

3

Exercise 3

We follow the same reasoning as for question 2; there are 5 possible choices and
one of the choices must be correct. Therefore, picking randomly among them
should generate the right answer with probability 15 .

4

Exercise 4

We assume that the randomly selected 400 drivers represent the total population. Therefore, the probability of choosing someone in the general population
who has been in a car accident is equal to the proportion of people who have
beeen in a car accident in our smaller sample. This is 136
400 ≈ 34%.

5

Exercise 5

The probability of choosing an unlisted phone number is equal to the proportion
of unlisted phone numbers in the randomly generated sample. Therefore, the
1

probability of picking an unlisted number is

6

56
144+56

=

7
25 .

Exercise 6

a)
There are 2 possible outcomes for each question, ”T” or ”F”, representing
”True” or ”False”, respectively. Therefore, for 3 questions there are 23 = 8
possibilities:
TTT; TTF; TFT; TFF; FTT; FTF; FFT; FFF

b)
There is only one correct solution out of a possible 8 total solutions, so the
probability of guessing all three correctly is 81 .

c)
To answer all three questions incorrectly, we need to pick the wrong choice every
time. There is one way to do this for each of the three questions, so there is
13 = 1 outcome that produces 3 incorrect answers out of a possible 8 outcomes.
Therefore, the probability of all three guesses being wrong is 81 .

d)
We notice that for every way we can guess two questions correctly, there is a
way we can guess two questions incorrectly, by flipping our answer choices. For
example, if T T F has two correct answers, then F F T has two incorrect answers,
or one correct answer. Similarly, for every way we can answer three questions
correctly, there is a way to answer all three incorrectly (or answer 0 correctly).
This means in exactly half the outcomes, we guess at least 2 answers correctly,
so the probability is 2! .

7

Exercise 7

a)
Assuming binary labeling, a child can either be a B or a G. Then the possible
outcomes are
BB; BG; GB; GG
Where the letter on the left represents the first child.

2

b)
Since each child is equally likely to be a boy or girl, the outcomes are all equally
likely. Then since we only have one outcome GG with two girls out of 4, the
probability of two girls is 14 .

c)
There are two outcomes out of four that produce one of each sex, BG and GB.
Therefore, the probability is 24 = 12 .

8

Exercise 8

There are 6 outcomes for each die roll. Since the die rolls are independent, we
multiply the number of outcomes for each die roll to get the total number of
outcomes for two dice rolls, 62 = 36.

a)
Let (a, b) represent rolling two dice with a being the number rolled on the first
die and b being the number rolled on the second. Then there are six ways to
roll a sum of 7 between the two dice:
(1, 6); (2, 5); (3, 4); (4, 3); , (5, 2); , (6, 1)
Note that we count both (3, 4) and (4, 3) because they correspond to different
rolls: the former means we rolled a 3 then a 4 while the latter means we rolled
a 4 then a 3.
Then if we have six desired outcomes out of 36 equally likely outcomes, our
6
probability of getting a sum of 7 is 36
= 16 .

b)
We can get an 11 in 2 ways:
(5, 6), (6, 5)
Since there are two de...


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