Lecture Notes for Introductory Probability
These notes were started in January 2009 with help from Christopher Ng, a student in
Math 135A and 135B classes at UC Davis, who typeset the notes he took during my lectures.
This text is not a treatise in elementary probability and has no lofty goals; instead, its aim is
to help a student achieve the proficiency in the subject required for a typical exam and basic
real-life applications. Therefore, its emphasis is on examples, which are chosen without much
redundancy. A reader should strive to understand every example given and be able to design
and solve a similar one. Problems at the end of chapters and on sample exams (the solutions
to all of which are provided) have been selected from actual exams, hence should be used as a
test for preparedness.
I have only one tip for studying probability: you cannot do it half-heartedly. You have to
devote to this class several hours per week of concentrated attention to understand the subject
enough so that standard problems become routine. If you think that coming to class and reading
the examples while also doing something else is enough, you’re in for an unpleasant surprise on
the exams.
This text will always be available free of charge to UC Davis students. Please contact me if
you spot any mistake. I am thankful to Marisano James for numerous corrections and helpful
suggestions.
Copyright 2010, Janko Gravner
1 INTRODUCTION 1
1 Introduction
The theory of probability has always been associated with gambling and many most accessible
examples still come from that activity. You should be familiar with the basic tools of the
gambling trade: a coin, a (six-sided) die, and a full deck of 52 cards. A fair coin gives you Heads
(H) or Tails (T) with equal probability, a fair die will give you 1, 2, 3, 4, 5, or 6 with equal
probability, and a shuffled deck of cards means that any ordering of cards is equally likely.
Example 1.1. Here are typical questions that we will be asking and that you will learn how to
answer. This example serves as an illustration and you should not expect to understand how to
get the answer yet.
Start with a shuffled deck of cards and distribute all 52 cards to 4 players, 13 cards to each.
What is the probability that each player gets an Ace? Next, assume that you are a player and
you get a single Ace. What is the probability now that each player gets an Ace?
Answers. If any ordering of cards is equally likely, then any position of the four Aces in the
deck is also equally likely. There areparenleftbig524parenrightbigpossibilities for the positions (slots) for the 4 aces. Out
of these, the number of positions that give each player an Ace is 134: pick the first slot among
the cards that the first player gets, then the second slot among the second player’s cards, then
the third and the fourth slot. Therefore, the answer is 134(52
Here is how you can quickly estimate the second probability during a card game: give the
second ace to a player, the third to a different player (probability about 2/3) and then the last
to the third player (probability about 1/3) for the approximate answer 2/9 ≈ 0.22.
History of probability
Although gambling dates back thousands of years, the birth of modern probability is considered
to be a 1654 letter from the Flemish aristocrat and notorious gambler Chevalier de M´er´e to the
mathematician and philosopher Blaise Pascal. In essence the letter said:
I used to bet even money that I would get at least one 6 in four rolls of a fair die.
The probability of this is 4 times the probability of getting a 6 in a single die, i.e.,
4/6 = 2/3; clearly I had an advantage and indeed I was making money. Now I bet
even money that within 24 rolls of two dice I get at least one double 6. This has the
same advantage (24/62 = 2/3), but now I am losing money. Why?
As Pascal discussed in his correspondence with Pierre de Fermat, de M´er´e’s reasoning was faulty;
after all, if the number of rolls were 7 in the first game, the logic would give the nonsensical
probability 7/6. We’ll come back to this later.
1 INTRODUCTION 2
Example 1.2. In a family with 4 children, what is the probability of a 2:2 boy-girl split?
One common wrong answer: 15, as the 5 possibilities for the number of boys are not equally
likely.
Another common guess: close to 1, as this is the most “balanced” possibility. This repre-
sents the mistaken belief that symmetry in probabilities should very likely result in symmetry in
the outcome. A related confusion supposes that events that are probable (say, have probability
around 0.75) occur nearly certainly.
Equally likely outcomes
Suppose an experiment is performed, with n possible outcomes comprising a set S. Assume
also that all outcomes are equally likely. (Whether this assumption is realistic depends on the
context. The above Example 1.2 gives an instance where this is not a reasonable assumption.)
An event E is a set of outcomes, i.e., E ⊂ S. If an event E consists of m different outcomes
(often called “good” outcomes for E), then the probability of E is given by:
(1.1) P(E) = mn .
Example 1.3. A fair die has 6 outcomes; take E = {2,4,6}. Then P(E) = 12.
What does the answer in Example 1.3 mean? Every student of probability should spend
some time thinking about this. The fact is that it is very difficult to attach a meaning to P(E)
if we roll a die a single time or a few times. The most straightforward interpretation is that for
a very large number of rolls about half of the outcomes will be even. Note that this requires
at least the concept of a limit! This relative frequency interpretation of probability will be
explained in detail much later. For now, take formula (1.1) as the definition of probability.