Grade 10

Events and probability

The chance that an outcome occurs, measured as a number between 0 and 1.

IntuitionIntuition: how likely is it?

Flip a coin, roll a die, or spin a wheel: each of these is a random experiment whose exact result cannot be predicted in advance, but whose set of possible results — the sample space $\Omega$ — is known ahead of time. A single possible result is called an outcome, and any collection of outcomes we care about, such as "the die shows an even number", is called an event $A$ . Probability assigns to it a number $P(A)$ between $0$ and $1$ that measures how likely the event is, with $P(A) = 0$ meaning impossible and $P(A) = 1$ meaning certain.

A spinner divided into equal arcs is spun, and the pointer stops at some angle

\theta

measured from the top. If event

A

is the arc spanning

90

degrees out of the full

360

degrees, then by symmetry

P(A) = \frac{90}{360} = \frac{1}{4}

, exactly the fraction of the circle that arc occupies.

SchoolSample space and classical probability

Definition: Sample space and event

The sample space $\Omega$ of a random experiment is the set of all possible outcomes. An event is any subset $A \subseteq \Omega$ of the sample space; it "occurs" exactly when the actual outcome of the experiment lies in $A$ .

P(A) = \frac{|A|}{|\Omega|}

Here $|A|$ denotes the number of outcomes in the event $A$ (assuming $\Omega$ is finite), and $|\Omega|$ denotes the total number of outcomes in the sample space. This formula — sometimes called the classical (or Laplace) definition of probability — only applies when every outcome in $\Omega$ is equally likely; for example, all $6$ faces of a fair die or all $52$ cards of a well-shuffled deck.

P(A \cup B) = P(A) + P(B) - P(A \cap B)

For two events $A$ and $B$ , this addition rule accounts for outcomes in $A \cap B$ being counted once in $P(A)$ and once in $P(B)$ , so the overlap $P(A \cap B)$ must be subtracted once to avoid double-counting. When $A$ and $B$ cannot happen at the same time, $A \cap B = \varnothing$ , so $P(A \cap B) = 0$ and the rule simplifies to $P(A \cup B) = P(A) + P(B)$ . A different relationship, independence, holds when knowing one event gives no information about the other; algebraically this is defined by $P(A \cap B) = P(A) \cdot P(B)$ .

Special relationships between two events
Relationship	Defining condition	Resulting formula
Mutually exclusive	$A \cap B = \varnothing$	$P(A \cup B) = P(A) + P(B)$
Independent	$P(A \cap B) = P(A) \cdot P(B)$	$P(A \cup B) = P(A) + P(B) - P(A) P(B)$
General (arbitrary) events	—	$P(A \cup B) = P(A) + P(B) - P(A \cap B)$

UndergraduateRigorous statements and proofs

Theorem: The addition rule (inclusion–exclusion for two events)

For any two events $A, B \subseteq \Omega$ in a finite sample space, $P(A \cup B) = P(A) + P(B) - P(A \cap B)$ .

Why is it true?

Simply adding $P(A)$ and $P(B)$ counts every outcome lying in both events twice — once inside $P(A)$ and once inside $P(B)$ — so the overlap $P(A \cap B)$ must be removed exactly once to correct the count.

Proof

Step 1 (partition the union into disjoint pieces): split $A \cup B$ into three pairwise disjoint pieces: $A \setminus B$ (outcomes only in $A$ ), $A \cap B$ (outcomes in both), and $A \cap B$ $B \setminus A$ (outcomes only in $B$ ). Every outcome of $A \cup B$ falls into exactly one of these three pieces.

Step 2 (count the union by the addition rule for disjoint sets): since the three pieces are pairwise disjoint, the number of outcomes satisfies $|A \cup B| = |A \setminus B| + |A \cap B| + |B \setminus A|$ .

Step 3 (express $|A|$ and $|B|$ using the same three pieces): similarly, $|A| = |A \setminus B| + |A \cap B|$ (splitting $A$ by whether it overlaps $B$ ) and $|B| = |B \setminus A| + |A \cap B|$ (splitting $B$ the same way). Adding these two equations gives $|A| + |B| = |A \setminus B| + |B \setminus A| + 2|A \cap B|$ .

Step 4 (combine and divide by $|\Omega|$ ): comparing Step 2 and Step 3, $|A| + |B| - |A \cap B| = |A \setminus B| + |B \setminus A| + |A \cap B| = |A \cup B|$ . Dividing both sides by $|\Omega|$ and applying the classical probability formula $P(A) = \frac{|A|}{|\Omega|}$ to each term gives exactly $P(A \cup B) = P(A) + P(B) - P(A \cap B)$ .

Theorem: The multiplication rule for independent events

Let $\Omega = \Omega_1 \times \Omega_2$ be a sample space built from two independent finite equally-likely sample spaces, with $m$ outcomes in the first and $n$ outcomes in the second. For events $A \subseteq \Omega_1$ and $B \subseteq \Omega_2$ , writing $A \cap B$ for the event that the first outcome lies in $A$ and the second lies in $B$ , $P(A \cap B) = P(A) \cdot P(B)$ .

Why is it true?

Choosing an outcome that satisfies both $A$ and $B$ independently is a two-step counting process (the multiplication rule for counting), so the number of favorable pairs is simply the product of the numbers of favorable outcomes in each component.

Proof

Step 1 (set up the counting): let $|A| = a$ and $|B| = b$ . An outcome of $\Omega = \Omega_1 \times \Omega_2$ satisfying both conditions is a pair $(\omega_1, \omega_2)$ with $\omega_1 \in A$ and $\omega_2 \in B$ .

Step 2 (apply the multiplication rule for counting): choosing $\omega_1$ can be done in $a$ ways and, independently of that choice, $\omega_2$ can be done in $b$ ways, so by the multiplication rule for counting there are $a \cdot b$ favorable pairs.

Step 3 (divide by the total number of outcomes): the total sample space has $m \cdot n$ equally likely outcomes, so $P(A \cap B) = \frac{ab}{mn}$ .

Step 4 (factor the fraction): rewriting $\frac{ab}{mn} = \frac{a}{m} \cdot \frac{b}{n} = P(A) \cdot P(B)$ , since $P(A) = a/m$ and $P(B) = b/n$ by the classical probability formula, which proves the claim.

UndergraduateReal-World Applications and Worked Examples

Classical probability and the addition and multiplication rules are the everyday arithmetic of quality control (chance that a batch has a defect), finance (chance that at least one of several independent investments loses money), genetics (chance of inheriting a trait from independent alleles), and computer science (chance that a hash collision or a random test failure occurs). The two examples below apply the addition rule and the multiplication rule to concrete numbers.

Example: A language-course survey

In a school of $200$ students, $120$ study French, $80$ study Spanish, and $30$ study both languages. Find the probability that a randomly chosen student studies French or Spanish.

Solution

Step 1: let $A$ be "studies French" and $B$ be "studies Spanish", so $|\Omega| = 200,\; |A| = 120,\; |B| = 80,\; |A \cap B| = 30$ .

Step 2: since "French or Spanish" is the event $A \cup B$ , the addition rule gives $P(A \cup B) = P(A) + P(B) - P(A \cap B)$ .

Step 3: substituting the classical probabilities, $P(A \cup B) = \frac{120}{200} + \frac{80}{200} - \frac{30}{200} = \frac{170}{200}$ , which simplifies to $\frac{17}{20} = 0.85$ .

Example: Guessing on two independent quiz questions

A student guesses randomly on two independent multiple-choice questions. Question 1 has $4$ options with exactly one correct answer, and Question 2 has $5$ options with exactly one correct answer. Find the probability that the student guesses both questions correctly.

Solution

Step 1: let $A$ be "guesses question 1 correctly" and $B$ be "guesses question 2 correctly"; since each option is equally likely to be picked, $P(A) = \frac{1}{4}$ and $P(B) = \frac{1}{5}$ .

Step 2: guessing on the two questions does not influence each other, so $A$ and $B$ are independent, and the multiplication rule applies: $P(A \cap B) = P(A) \cdot P(B)$ .

Step 3: substituting the values from Step 1, $P(A \cap B) = \frac{1}{4} \cdot \frac{1}{5} = \frac{1}{20}$ , so the chance of guessing both correctly is only $5\%$ .

A fair six-sided die is rolled once. What is $P(A)$ , the probability that the outcome is a prime number, where $A = \{2, 3, 5\}$ ?

Among $40$ students, $18$ play chess, $15$ play go, and $6$ play both. What is the probability that a randomly chosen student plays chess or go?

A fair coin is flipped twice. Let $A$ be "the first flip is heads" and $B$ be "the second flip is heads". Since the two flips do not influence each other, what is $P(A \cap B)$ ?

A factory has two machines that operate completely independently. Each machine has a $10\%$ chance of producing a defective item on a given run. What is the probability that neither machine produces a defective item?

References

Sheldon Ross (2019). A First Course in Probability
Joseph K. Blitzstein, Jessica Hwang (2019). Introduction to Probability