The classical definition and its rules

What you will be able to do

• find a probability by counting equally likely outcomes;
• use the three rules of probability and the complement;
• find the chance of at least one success with $1 - P(\text{none})$, and convert odds to a probability.

A number for a chance

A ticket is taken from a hat holding $40$ tickets, numbered $1$ to $40$. What is the chance that the number is a multiple of $7$? There are $5$ such numbers: $7$, $14$, $21$, $28$ and $35$. Out of $40$ equally likely tickets, $5$ are good. We say the probability is $\dfrac{5}{40} = \dfrac{1}{8}$.

The classical definition

Let a random experiment have a sample space $S$ of $n(S)$ outcomes, all equally likely. The probability of an event $E$ is

Here $n(E)$ is the number of favourable outcomes and $n(S)$ is the number of all outcomes.

Outcomes in $E$ are called favourable to $E$. The definition needs the outcomes to be equally likely, so first make sure that the sample space is the right one. For two dice that means the $36$ ordered pairs, as in the first course of this chapter.

For a card drawn from a pack, $P(\text{heart}) = \dfrac{13}{52} = \dfrac{1}{4}$ and $P(\text{face card}) = \dfrac{12}{52} = \dfrac{3}{13}$.

The three rules

Modern probability starts from rules and not from counting. A probability is a number $P(E)$ given to each event so that:

1. $P(E) \ge 0$ for every event $E$.
2. $P(S) = 1$.
3. If $A$ and $B$ are mutually exclusive, $P(A \cup B) = P(A) + P(B)$.

These are the axioms of probability. Counting satisfies them, and so does any other fair assignment. Everything else follows:

• $P(\varnothing) = 0$, because $\varnothing$ and $S$ are mutually exclusive and their union is $S$.
• $0 \le P(E) \le 1$ for every event.
• If $A \subseteq B$ then $P(A) \le P(B)$.
• The probability of an event is the sum of the probabilities of the simple events in it.

Outcomes that are not equally likely. A loaded die shows $1$ with probability $0.1$, $2$ with $0.1$, and each of $3$, $4$, $5$ and $6$ with $0.2$. The six numbers are at least $0$ and add up to $1$. So they are a fair assignment, and $P(\text{even}) = 0.1 + 0.2 + 0.2 = 0.5$. The counting definition does not apply here, but the axioms still do.

The complement

An event and its complement are mutually exclusive, and together they make $S$. By the third rule, $P(E) + P(E') = 1$, so

At least one

"At least one" is the hardest event to count directly, because it holds many cases. Its complement, "none", is one case. So

A die is thrown three times. The chance of no six is $\dfrac{5}{6} \times \dfrac{5}{6} \times \dfrac{5}{6} = \dfrac{125}{216}$, which can also be read as $5^3$ favourable triples out of $6^3$. So the chance of at least one six is $1 - \dfrac{125}{216} = \dfrac{91}{216}$.

Odds

Odds compare the cases for an event with the cases against it. If $P(E) = \dfrac{a}{a + b}$, the odds in favour of $E$ are $a : b$ and the odds against are $b : a$. Odds of $3 : 5$ in favour mean $3$ parts for and $5$ parts against, so $P(E) = \dfrac{3}{8}$. Conversely, $P(E) = 0.7$ gives odds of $7 : 3$ in favour. Odds are a ratio of two parts. A probability is the part for out of the whole. Some books write odds of $3 : 5$ as "$3$ to $5$". The meaning is the same.

What a probability means

A probability of $\dfrac{1}{6}$ for a six does not mean that every sixth throw is a six. It means that in a very long run of throws, the fraction of sixes settles near $\dfrac{1}{6}$. This is the idea of relative frequency. The last lesson of this chapter checks it with a simulation.

A worked example

A card is drawn from a pack of $52$ cards. Find the probability that it is not a spade, and the probability that it is an ace or a king.

1. Not a spade. There are $13$ spades, so $P(\text{spade}) = \dfrac{1}{4}$. By the complement, $P(\text{not a spade}) = 1 - \dfrac{1}{4} = \dfrac{3}{4}$.
2. An ace or a king. There are $4$ aces and $4$ kings, and no card is both. So the events are mutually exclusive, and $P = \dfrac{4}{52} + \dfrac{4}{52} = \dfrac{8}{52} = \dfrac{2}{13}$.

Practise with new numbers

Try it in Python

The program defines the classical probability of an event as a fraction, and tests the complement on three throws of a die.

Common mistakes

• Using outcomes that are not equally likely. The count of favourable outcomes over all outcomes works only when every outcome has the same chance.
• Adding chances for "at least one". Three throws do not give $3 \times \dfrac{1}{6}$. Use the complement.
• Mixing odds and probability. Odds of $3 : 5$ is a probability of $\dfrac{3}{8}$, not $\dfrac{3}{5}$.
• A probability above $1$ or below $0$. If you get one, there is a mistake in the count.

Remember these

• $P(E) = \dfrac{n(E)}{n(S)}$ for equally likely outcomes, and always $0 \le P(E) \le 1$.
• The axioms: $P(E) \ge 0$, $P(S) = 1$, and $P(A \cup B) = P(A) + P(B)$ for mutually exclusive events.
• $P(E') = 1 - P(E)$ and $P(\varnothing) = 0$.
• $P(\text{at least one}) = 1 - P(\text{none})$. Odds $a : b$ in favour give $P = \dfrac{a}{a + b}$.

Check yourself

Try each one on your own first. If you are stuck, open a hint. Then check with Show the answer.

1. A number is chosen from $1$ to $25$ at random. Find the probability that it is a perfect square.

2. The probability that a bus is late is $0.35$. What is the probability that it is on time?

3. A coin is tossed $5$ times. Find the probability of at least one head.

4. The odds against an event are $4 : 9$. Find its probability.

5. Can $P(A) = 0.6$ and $P(A') = 0.5$ both be true?

References

• NCERT, Mathematics for Class XI (earlier edition), Probability, sections on the axiomatic approach and the probability of an event.
• OpenStax, Introductory Statistics, Chapter 3, section on terminology.