Chapter 1 - Discrete Probability Distributions

1.2 Discrete Probability Distributions

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Results

Suppose the outcome of an experiment depends on chance. The outcome is represented by an uppercase letter, such as $X$, called a random variable. The sample space, $\Omega$, of the experiment is the set of all possible outcomes. If it is finite or countable infinite, the random variable is said to be discrete.




Let $X$ be a random variable which denotes the outcome of an experiment with finite possible outcomes. Let $\Omega$ be the sample space of the experiment; the set of all possible values for $X$. A distribution function for $X$ is a real-valued function $m$ whose domain is $\Omega$ and which satisfies:

For any subset $E$ of $\Omega$, we define the probability of $E$ to be the number $P(E)$ given by, $$ P(E) = \sum_{\omega\in E}m(\omega). $$

Relationship between the probability and distribution function: $$ P(\{\omega\}) = m(\omega). $$




The probabilities assigned to events by a distribution function on a sample space $\Omega$ satisfy the following properties:




If $A_1,\ldots, A_n$ are pairwise disjoint subsets of $\Omega$, then $$ P(A_1\cup\ldots\cup A_n) = \sum_{i=1}^nP(A_i). $$




Let $A_1,\ldots, A_n$ be pairwise disjoint events with $\Omega = A_1\cup\ldots\cup A_n$ and let $E$ be any event. Then $$ P(E) = \sum_{i=1}^nP(E\cap A_i). $$




For any two events $A$ and $B$, $$ P(A) = P(A\cap B) + P(A\cap B^c). $$




If $A$ and $B$ are subsets of $\Omega$, then $$ P(A\cup B) = P(A) + P(B) - P(A\cap B). $$




The uniform distribution on a sample space $\Omega$ containing $n$ elements is the function $m$ defined by $$ m(\omega) = \frac{1}{n}, $$ for every $\omega\in\Omega$.

Exercise 1

Let $\Omega = \{a, b, c\}$ be a sample space. Let $m(a) = 1/2$, $m(b)= 1/3$, and $m(c) = 1/6$. Find the probabilities for all eight subsets of $\Omega$.

Answer
Note that $m(a) + m(b) + m(c) = 1$, so they fill out the entire sample space. We can also rewrite $1/2 = 3/6$ and $1/3 = 2/6$. Finding the probabilities. \begin{align} P(\Omega) &= P(\{a, b, c\}) = \frac{3}{6} + \frac{2}{6} + \frac{1}{6} = 1 \\ P(\emptyset) &= 0 \\ P(\{a\}) &= 1/2\\ P(\{b\}) &= 1/3\\ P(\{c\}) &= 1/6\\ P(\{a, b\}) &= 3/6 + 2/6 = 5/6 \\ P(\{a, c\}) &= 3/6 + 1/6 = 4/6 = 2/3 \\ P(\{b, c\}) &= 2/6 + 1/6 = 3/6 = 1/2 \end{align}

Exercise 2

Give a possible sample space for the following experiments. (Note to self: this is the space itself, not all possible outcomes, which is the underlying $\sigma$-algebra).

(a) An election decides between two candidates $A$ and $B$.

Answer
$$ \Omega = \{A, B\} $$

(b) A two-sided coin is tossed.

Answer
$$ \Omega = \{H, T\} $$

(c) A student is asked for the month of the year and the day of the week on which her birthday falls.

Answer
$$ \Omega = \{(d, m) \mid d\in\{\text{"Monday"},\ldots, \text{"Sunday"}\}, 1\leq m\leq 12\}, $$

(d) A student is chosen at random from a class of ten students.

Answer
$$ \Omega = \{s\mid s\in\{s_1, \ldots, s_{10}\}\}, $$ where each $s_i$ denotes one of the ten students in the class.

(e) You receive a grade in this course.

Answer
$$ \Omega = \{A, B, C, D, E, F\} $$

Exercise 3

For which of the cases in Exercise 2 would it be reasonable to assign the uniform distribution function?

Answer
The uniform distribution is suitable on (b), (c), and (d). In an election, or the possible grade for a student are not usually uniform by nature and are influenced by many external factors.

Exercise 4

Describe in words the events specified by the following subsets of $$ \Omega = \{HHH, HHT, HTH, HTT, THH, THT, TTH, TTT\}. $$ (a) $$ E = \{HHH,HHT,HTH,HTT\} $$ Answer
This events is all possible outcomes where the first coin toss is H.



(b) $$ E = \{HHH,TTT\} $$ Answer
All three coin tosses are the same kind.



(c) $$ E = \{HHT,HTH,THH\} $$ Answer
All outcomes where we get exactly two H.



(d) $$ E = \{HHT,HTH,HTT,THH,THT,TTH,TTT\} $$ Answer
All outcomes where we get at least one T.

Exercise 5

What are the probabilities of the events described in Exercise 4?

Answer
For a fair coin all the outcomes will have a uniform distribution, so all outcomes are equally likely with probability 1/8. We can simply count the number of occurrences in each event.

(a) $P(E) = 1/2$

(b) $P(E) = 1/4$

(c) $P(E) = 3/8$

(d) $P(E) = 7/8$

Exercise 6

A die is loaded in such a way that the probability of each face turning up is proportional to the number of dots on that face. (For example, a six is three times as probable as a two.) What is the probability of getting an even number in one throw?

Answer
Recalling the odds discussion. The sum of 1 to 6 is 21. The probability of rolling an $i$ therefore becomes: $P(i) = i/21$. The event of getting an even number is: $$ P(\{2, 4, 6\}) = \frac{2}{21} + \frac{4}{21} + \frac{6}{21} = \frac{12}{21}. $$

Exercise 7

Let $A$ and $B$ be events such that $P(A\cap B) = 1/4$, $P(A^c) = 1/3$, and $P(B) = 1/2$. What is $P(A\cup B)$?

Answer
Using some basic properties of probability. $$ P(A) = 1 - P(A^c) = 1 - 1/3 = 2/3. $$ Next, the following identity: $$ P(A\cup B) = P(A) + P(B) - P(A\cap B) = 2/3 + 1/2 - 1/4 = 2/3 + 1/4 = 8/12 + 3/12 = 11/12 $$

Exercise 8

A student must choose one of the subjects, art, geology, or psychology, as an elective. She is equally likely to choose art or psychology and twice as likely to choose geology. What are the respective probabilities that she chooses art, geology, and psychology?

Answer
Defining the sample space. $$ \Omega = \{A, P, G\}. $$ The total probability must be 1, so using that $A$ and $P$ are equally likely, and that $G$ is twice as likely as $A$: $$ P(A) + P(A) + 2P(A) = 1 \;\Longrightarrow\; 4P(A) = 1 \;\Longrightarrow\; P(A) = 1(4). $$ So, $P(A) = 1/4$, $P(P) = 1/4$ and $P(G) = 1/2$.

Exercise 9

A student must choose exactly two out of three electives: art, French, and mathematics. He chooses art with probability 5/8, French with probability 5/8, and art and French together with probability 1/4. What is the probability that he chooses mathematics? What is the probability that he chooses either art or French?

Answer
The student needs to select two classes, so we can define the sample space as: $$ \Omega = \{AF, AM, FM\}, $$ where $F$ is French, $A$ is arts, and $M$ is mathematics. As given: $$ P(F) = P(AF) + P(FM) = 5/8 $$ $$ P(A) = P(AF) + P(AM) = 5/8 $$ We also know that $P(AF) = 1/4 = 2/8$, and by using the above equations, we can calculate: $P(FM) = 3/8$ and $P(AM) = 3/8$. Note that the sum of all outcomes become 1.

The probability of selecting $M$ becomes: $$ P(M) = P(AM) + P(FM) = 3/8 + 3/8 = 6/8 = 3/4. $$ The probability of selecting $F$ or $A$ becomes 1, since one of them is included in all outcomes.

Exercise 10

For a bill to come before the president of the United States, it must be passed by both the House of Representatives and the Senate. Assume that, of the bills presented to these two bodies, 60 percent pass the House, 80 percent pass the Senate, and 90 percent pass at least one of the two. Calculate the probability that the next bill presented to the two groups will come before the president.

Answer
Define $P(H) = 0.6$, $P(S) = 0.8$ and $P(H\cup S) = 0.9$. We need to find the probability of $H\cap S$. This can be done with the following formula (a rewritten version of Theorem 1.4): $$ P(H\cap S) = P(H) + P(S) - P(H\cup S) = 0.6 + 0.8 - 0.9 = 0.5. $$ So the probability that a bill reaches the president is $1/2$.

Exercise 11

Skipped.

Exercise 12

Skipped.

Exercise 13

Skipped.

Exercise 14

Let $X$ be a random variable with distribution function $m_X(x)$ defined by $$ m_X(-1) = 1/5,\quad m_X(0) = 1/5,\quad m_X(1) = 2/5,\quad m_X(2) = 1/5 $$

(a) Let $Y$ be the random variable defined by the equation $Y = X+3$. Find the distribution function $m_Y(y)$ of $Y$.

Answer
Just adjust the values accordingly. $$ m_Y(2) = 1/5,\quad m_Y(3) = 1/5,\quad m_Y(4) = 2/5,\quad m_Y(5) = 1/5 $$

(b) Let $Z$ be the random variable defined by the equation $Z = X^2$. Find the distribution function $m_Z(z)$ of $Z$.

Answer
Notice that both $X = -1$ and $X = 1$ will be mapped into $Z = 1$, so it will get the sum of their probabilities. $$ m_Z(0) = 1/5,\quad m_Z(1) = 3/5,\quad m_Z(4) = 1/5 $$

Exercise 15

John and Mary are taking a mathematics course. The course has only three grades: A, B, and C. The probability that John gets a B is 0.3. The probability that Mary gets a B is 0.4. The probability that neither gets an A but at least one gets a B is 0.1. What is the probability that at least one gets a B but neither gets a C?

Answer
Let $XY$ be the grades they get, where $X$ is John's grade and $Y$ is Mary's grade. We define the sample space. $$ \Omega = \{AA, AB, AC, BA, BB, BC, CA, CB, CC\} $$ Define the provided events. $J_B$ are the outcomes where John gets a B. $$ P(J_B) = P(\{BA, BB, BC\}) = 0.3 $$ Similarly for Mary, we define the event $M_B$. $$ P(M_B) = P(\{AB, BB, CB\}) = 0.4 $$ And the final even, which we call $N$. $$ P(N) = P(\{BB, BC, CB\}) = 0.1 $$ Finally, the event we want to calculate the probability for, which is: $$ D = \{AB, BA, BB\} $$ The key observation here is to note that $J_B\cup M_B = D\cup N = \{AB, BA, BB, BC, CB\}$ and that $J_B\cap M_B = D\cap N = \{BB\}$. So: $$ P(J_B\cup M_B) = P(J_B) + P(M_B) - P(\{BB\}) $$ $$ P(D\cup N) = P(D) + P(N) - P(\{BB\}) $$ Since the events are equal, we can set these equal to each other, and eliminate $P(\{BB\})$. So: \begin{align} P(D\cup N) &= P(J_B\cup M_B) \\ P(D) + P(N) - P(\{BB\}) &= P(J_B) + P(M_B) - P(\{BB\})\\ P(D) + P(N) &= P(J_B) + P(M_B) \\ P(D) + 0.1 &= 0.3 + 0.4 \\ P(D) &= 0.6 \end{align} The observation was discovered when looking at a table with all possibilities and coloring the areas we want to find.

Exercise 16

In a fierce battle, not less than 70 percent of the soldiers lost one eye, not less than 75 percent lost one ear, not less than 80 percent lost one hand, and not less than 85 percent lost one leg. What is the minimal possible percentage of those who simultaneously lost one ear, one eye, one hand, and one leg?

Answer
So if there were 100 soldiers in the battle, at least 70 lost one eye etc. To calculate those who fall in all groups, we call it $A$: $$ P(A) = 0.70\cdot 0.75\cdot 0.80\cdot 0.85 = 0.357, $$ so around 35.7% of the soldiers lost everything.

Exercise 17

Assume that the probability of a "success" on a single experiment with n outcomes is 1/n. Let m be the number of experiments necessary to make it a favorable bet that at least one success will occur (see Exercise 1.1.5).

(a) Show that the probability that, in m trials, there are no successes is $(1 − 1/n)^m$.

Answer
The probability of success is $P(S) = 1/n$, so the probability of failure is $P(S^c) = 1 - 1/n$. For two consecutive fails, the probability becomes $P(S^c)P(S^c) = P(S^c)^2 = (1 - 1/n)^2$ and so on. The probability for $m$ failures is $(1 - 1/n)^n$.



(b) (de Moivre) Show that if $m = n\log 2$ then $$ \lim_{n\rightarrow\infty}\left(1 - \frac{1}{n}\right)^m = \frac{1}{2} $$ Hint: $$ \lim_{n\rightarrow\infty}\left(1 - \frac{1}{n}\right)^n = e^{-1} $$ Hence for large n we should choose m to be about $n\log 2$.

Answer
Selecting $m = n\log 2$. $$ \lim_{n\rightarrow\infty}\left(1 - \frac{1}{n}\right)^m = \lim_{n\rightarrow\infty}\left(1 - \frac{1}{n}\right)^{n\log 2} = \lim_{n\rightarrow\infty}\Bigg[\left(1 - \frac{1}{n}\right)^n\Bigg]^{\log 2} = e^{-\log 2} = \frac{1}{e^{\log 2}} = \frac{1}{2} $$



(c) Would DeMoivre have been led to the correct answer for de Méré’s two bets if he had used this approximation?

Answer
In this case, the probability of success will be: $$ \frac{1}{6^2} = \frac{1}{36} $$ so, $36\log 2 \approx 24.95$ which DeMoivre would round up to 25, which is exactly what de Méré found out he needed.

Exercise 18

(a) For events $A_1,\ldots, A_n$, prove that $$ P(A_1\cup\ldots \cup A_n) \leq P(A_1) + \ldots + P(A_n). $$

Answer
From Theorem 1.3, we have: $$ P(A_1\cup A_2) = P(A_1) + P(A_2) - P(A_1\cap A_2) \leq P(A_1) + P(A_2) $$ since all probabilities are greater or equal to 0. Define $B_2 = A_1\cup A_2$. Then, we can repeat the procedure again: $$ P(A_1\cup A_2\cup A_3) = P(B_2\cup A_3) = P(B_2) + P(A_3) - P(B_2\cap A_3) \leq P(B_2) + P(A_3) \geq P(A_1) + P(A_2) + P(A_3) $$ Continuing in this way, we can show the result.



(b) For events $A$ and $B$, prove that $$ P(A\cap B) \geq P(A) + P(B) - 1 $$

Answer
Rewriting the result from Theorem 1.4. Since $A\cup B\subset\Omega$, then $P(A\cup B)\leq P(\Omega)$, which also means $-P(A\cup B)\geq -1$. So: $$ P(A\cap B) = P(A) + P(B) - P(A\cup B) \geq P(A) + P(B) - 1. $$

Exercise 19

If A, B, C are any three events, show that: $$ P(A\cup B\cup C) = P(A) + P(B) + P(C) - P(A\cap B) - P(B\cap C) - P(C\cap A) + P(A\cap B\cap C) $$

Answer
This can be shown by repeated use of Theorem 1.4. Define $D = A\cup B$. \begin{align} P(A\cup B\cup C) &= P(D\cup C) \\ &= P(D) + P(C) - P(D\cap C) \\ &= P(A\cup B) + P(C) - P([A\cup B]\cap C) \\ &= P(A) + P(B) - P(A\cap B) + P(C)- P([A\cap C]\cup[B\cap C]) \\ &= P(A) + P(B) + P(C) - P(A\cap B) - P([A\cap C]\cup[B\cap C]) \\ &= P(A) + P(B) + P(C) - P(A\cap B) - \Big[P([A\cap C]\cup[B\cap C])\Big] \\ &= P(A) + P(B) + P(C) - P(A\cap B) - \Big[P(A\cap C) + P(B\cap C) - P(A\cap C\cap B\cap C\Big] \\ &= P(A) + P(B) + P(C) - P(A\cap B) - P(A\cap C) - P(B\cap C) + P(A\cap B\cap C\cap C) \\ &= P(A) + P(B) + P(C) - P(A\cap B) - P(A\cap C) - P(B\cap C) + P(A\cap B\cap C) \end{align}

Exercise 20

Explain why it is not possible to define a uniform distribution function (see Definition 1.3) on a countably infinite sample space. Hint: Assume $m(\omega) = a$ for all $\omega$, where $0 \leq a \leq 1$. Does $m(\omega)$ have all the properties of a distribution function?

Answer
In a uniform distribution every outcome has the same distribution, so every $P(\omega) = 1/n$ for some $1/n > 0$ for every $\omega\in\Omega$. If this space is countable infinite, then the probability of all possible outcomes becomes infinite instead of 1. $$ \sum_{i=1}^\infty P(\omega_i) = \sum_{i=1}^\infty \frac{1}{n} = \frac{1}{n} \sum_{i=1}^\infty 1 = \infty. $$ Therefore, it is impossible to make a uniform distribution on a countably infinite sample space.

Exercise 21

Example 1.13:
A coin is tossed until the first time that a head turns up. Let the outcome of the experiment, $\omega$, be the first time that a head turns up. Then the possible outcomes of our experiment are $$ \Omega = \{1, 2, 3, \ldots\}. $$

Find the probability that the coin turns up heads for the first time on the tenth, eleventh, or twelfth toss.

Answer
This is simply calculating the probability of sequences such as: TTTTTTTTTH. Let $X$ be the number of tosses until we get the first H. \begin{align} P(X = 10) &= \frac{1}{2^{10}} \\ P(X = 11) &= \frac{1}{2^{11}} \\ P(X = 12) &= \frac{1}{2^{12}} \end{align}

Exercise 22

A die is rolled until the first time that a six turns up. We shall see that the probability that this occurs on the $n$th roll is $(5/6)^{n−1}\cdot(1/6)$. Using this fact, describe the appropriate infinite sample space and distribution function for the experiment of rolling a die until a six turns up for the first time. Verify that for your distribution function $\sum_{\omega} m(\omega) = 1$.

Answer
The sample space becomes: $$ \Omega = \{1, 2, 3, \ldots\}. $$ As noted, the probability of the terms become: \begin{align} P(X = 1) &= \frac{1}{6} \\ P(X = 2) &= \left(\frac{5}{6}\right)\frac{1}{6} \\ P(X = 3) &= \left(\frac{5}{6}\right)^2\frac{1}{6} \\ &\vdots \\ P(X = n) &= \left(\frac{5}{6}\right)^{n-1}\frac{1}{6} \end{align} We are going to show that the following sum becomes 1. $$ \sum_{n=1}^\infty P(X = n) = \sum_{n=1}^\infty \left(\frac{5}{6}\right)^{n-1}\frac{1}{6} = \sum_{n=0}^\infty \left(\frac{1}{6}\right)\left(\frac{5}{6}\right)^n = \sum_{n=0}^\infty ar^n $$ where $a = 1/6$ and $r = 5/6 < 1$, which shows that this is a geometric series. Thus: $$ \sum_{n=0}^\infty ar^n = \frac{a}{1 - r} = \frac{1/6}{1 - \frac{5}{6}} = \frac{1/6}{1/6} = 1. $$

Exercise 23

Let $\Omega$ be the sample space $$ \Omega = \{0, 1, 2, \ldots\} $$ and define a distribution function by $$ m(j) = (1 - r)^jr, $$ for some fixed $r$, $0 < r < 1$, and for $j=0,1,2, \ldots$. Show that this is a distribution function for $\Omega$.

Answer
Note that $m(j)\geq 0$ for all $j$, which is the first requirement. Next, we must show that all infinite terms sum up to 1. We can show that this is a geometric series, with $a = r$ and $q = 1-r$ such that $|q| < 1$. \begin{align} \sum_{k=0}^\infty m(k) &= \sum_{k=0}^\infty (1-r)^k r \\ &= \sum_{k=0}^\infty aq^k \\ &= \frac{a}{1 - q} \\ &= \frac{r}{1 - (1-r)} \\ &= \frac{r}{r} \\ &= 1 \end{align} which verifies that this is a valid distribution function.

Exercise 24

Our calendar has a 400-year cycle. B. H. Brown noticed that the number of times the thirteenth of the month falls on each of the days of the week in the 4800 months of a cycle is as follows:

Monday	685
Tuesday	685
Wednesday	687
Thursday	684
Friday	688
Saturday	684
Sunday	687

From this he deduced that the thirteenth was more likely to fall on Friday than on any other day. Explain what he meant by this.

Answer
He simply made the observation that Fridays occurs most frequently on the 13th day of the month. In other words, for any random 13th day of the month, there is a slightly higher probability that it will be a Friday. (Bad luck for the superstitious!)

Exercise 25

Tversky and Kahneman asked a group of subjects to carry out the following task. They are told that:


The subjects are then asked to rank the likelihood of various alternatives, such as:


Tversky and Kahneman found that between 85 and 90 percent of the subjects rated alternative (1) most likely, but alternative (3) more likely than alternative (2). Is it? They call this phenomenon the conjunction fallacy, and note that it appears to be unaffected by prior training in probability or statistics. Is this phenomenon a fallacy? If so, why? Can you give a possible explanation for the subjects’ choices?

Answer
We will define the events $F$: Linda is in the feminist movement, and $B$: Linda is a bank teller. It is not possible for (3) to be more likely than (2) since $F\cap B\subset B$, and so $P(F\cap B)\leq P(B)$.

This phenomenon is a phallacy, and the reasoning is that people only identify the very probably event $F$ in the statement and fail to recognize that we are dealing with an intersection.

Exercise 26

Two cards are drawn successively from a deck of 52 cards. Find the probability that the second card is higher in rank than the first card. Hint: Show that 1 = P(higher) + P(lower) + P(same) and use the fact that P(higher) = P(lower).

Answer
Define the events $H$ as higher, $L$ as lower, and $S$ as same. There are only three possibilities, so: $$ P(H) + P(L) + P(S) = 1 $$ Because of the symmetry, we get $$ P(H) = P(L). $$ Finally, once we have drawn a card, we can find the explicit probability for the next card being the same - there are only three possible cards in the remaining 51-card deck: $$ P(S) = \frac{3}{51} $$ Putting all of this together: $$ P(H) + P(H) + \frac{3}{51} = 1 $$ $$ 2P(H) = \frac{48}{51} $$ $$ P(H) = \frac{24}{51} $$ which is approximately 0.4705882.

Exercise 27

A life table is a table that lists for a given number of births the estimated number of people who will live to a given age. In Appendix C we give a life table based upon 100,000 births for ages from 0 to 85, both for women and for men. Show how from this table you can estimate the probability m(x) that a person born in 1981 would live to age x. Write a program to plot m(x) both for men and for women, and comment on the differences that you see in the two cases.

Answer

Exercise 28

Here is an attempt to get around the fact that we cannot choose a "random integer."

(a) What, intuitively, is the probability that a "randomly chosen" positive integer is a multiple of 3?

(b) Let $P_3(N)$ be the probability that an integer, chosen at random between 1 and N, is a multiple of 3 (since the sample space is finite, this is a legitimate probability). Show that the limit $$ P_3 = \lim_{N\rightarrow\infty} P_3(N) $$ exists and equals 1/3. This formalizes the intuition in (a), and gives us a way to assign "probabilities" to certain events that are infinite subsets of the positive integers.

(c) If A is any set of positive integers, let A(N) mean the number of elements of A which are less than or equal to N. Then define the "probability" of A as $$ P(A) = \lim_{N\rightarrow\infty} A(N)/N $$ provided this limit exists. Show that this definition would assign probability 0 to any finite set and probability 1 to the set of all positive integers. Thus, the probability of the set of all integers is not the sum of the probabilities of the individual integers in this set. This means that the definition of probability given here is not a completely satisfactory definition.

(d) Let A be the set of all positive integers with an odd number of digits. Show that P(A) does not exist. This shows that under the above definition of probability, not all sets have probabilities.



Answer

(a)
By definition, every third integer is divisible by 3:
1, 2, 3, 4, 5, 6, 7, 8, 9, ...
If we select some random integer, then intuitively, it will have 1/3 probability of being divisible by 3.

(b)
The limit exists if the sequence converges. View the first few values of $P_3(N)$. \begin{align} P_3(1) &= 0 \\ P_3(2) &= 0 \\ P_3(3) &= 1/3 \\ P_3(4) &= 1/4 \\ P_3(5) &= 1/5 \\ P_3(6) &= 2/6 = 1/3 \\ P_3(7) &= 2/7 \\ P_3(8) &= 2/8 = 1/4 \\ P_3(9) &= 3/9 = 1/3 \\ \end{align} From this it is clear that we get the following three cases, where $N=3K$, $N = 3K + 1$ and $N = 3K+2$ for $N\geq 3$: $$ P_3(N) = \frac{K}{3K},\quad P_3(N) = \frac{K}{3K+1},\quad P_3(N) = \frac{K}{3K+2} $$ When $N\rightarrow\infty$, then $K\rightarrow\infty$, and in either case, it is clear that: $$ \lim_{N\rightarrow\infty} P_3(N) = 1/3. $$

(c)
If we set $A = \{5, 23, 85\}$, then for $N = 1000$ we would get $$ P(A) = \frac{A(N)}{N} = \frac{3}{1000}. $$ When we take the limit, and $|A| = K$ for some finite $K$, then $$ P(A) = \lim_{N\rightarrow\infty}\frac{A(N)}{N} = \lim_{N\rightarrow\infty}\frac{K}{N} = 0. $$ When $A$ is countable infinite, then $$ P(A) = \lim_{N\rightarrow\infty}\frac{A(N)}{N} = \lim_{N\rightarrow\infty}\frac{N}{N} = 1. $$

(d)
Let $A$ be the set of all numbers with an odd number of digits. We can partition the set as follows: $$ A = A_1\cup A_3\cup A_5\cup\ldots \cup A_{2k+1}\cup\ldots $$ where $A_p$ is the set of all numbers with $p$ digits, where $p$ is odd. Since these sets are disjoint: $$ P(A) = P(A_1) + P(A_3) + P(A_5) + \ldots = 0 + 0 + 0 + \ldots $$ which become 0 since they are finite sets. Any sets we specify become 0, so we cannot calculate this limit.

Exercise 29

Skipped. What is a cloverleaf intersection?

Exercise 30

Skipped. What is a cloverleaf intersection??!

Exercise 31

A reader of Marilyn vos Savant’s column wrote in with the following question:


(a) Is the answer 1/16?

(b) The following question was asked of a class of students. “I was driving to school today, and one of my tires went flat. Which tire do you think it was?” The responses were as follows: right front, 58%, left front, 11%, right rear, 18%, left rear, 13%. Suppose that this distribution holds in the general population, and assume that the two test-takers are randomly chosen from the general population. What is the probability that they will give the same answer to the second question?

Answer
(a)
No - the probability is 1/4 that they will answer the same. There are a total of 16 possible answers, but 4 of them will be cases where both students answer the same 'correct' answer. (Provided that they are actually guessing and that they didn't actually have a flat tire.)

(b)
Define the tires as 1, 2, 3, 4. The sample space becomes: $$ \Omega = \{11, 12, 13, 14, 21, 22, 23, 24, 31, 32, 33, 34, 41, 42, 43, 44\} $$ The event where the students give the same answer is $$ E = \{11, 22, 33, 44\} \;\Rightarrow\; P(E) = P(\{11\}) + P(\{22\}) + P(\{33\}) + P(\{44\}) $$ These probabilities can be calculated by the product of each individual probability. (All combinations are verified to sum up to 1, so it is a proper distribution). This gives: $$ P(E) = P(\{11\}) + P(\{22\}) + P(\{33\}) + P(\{44\}) = 0.58^2 + 0.11^2 + 0.18^2 + 0.13^2 = 0.3978, $$ so around 39.78% of answering corectly.