Probability

Board	Paper	Notes
AQA	Paper 1, 2	Basic probability in P1; conditional, Bayes in P2
Edexcel	P1, P2	Similar
OCR (A)	Paper 1, 2	Includes Venn diagrams and tree diagrams
CIE (9709)	P1, P6	Probability in P1; conditional in P6

---

Definition. A probability function $P$ on a sample space $\Omega$ satisfies:

1. Non-negativity: $P(A) \geq 0$ for all events $A \subseteq \Omega$.
2. Normalisation: $P(\Omega) = 1$.
3. Countable additivity: If $A_1, A_2, \ldots$ are mutually exclusive, then $P\!\left(\bigcup_{i}A_i\right) = \sum_i P(A_i)$.

These three axioms are the foundation of all probability theory. Every theorem in probability can be Derived from them.

---

Theorem. $P(A') = 1 - P(A)$.

Proof. $A$ and $A'$ are mutually exclusive and $A \cup A' = \Omega$.

$P(A \cup A') = P(A) + P(A') = P(\Omega) = 1 \implies P(A') = 1 - P(A). \quad \blacksquare$

Corollary. For any event $A$, $P(\emptyset) = 0$.

Proof. $P(\emptyset) = P(\Omega') = 1 - P(\Omega) = 1 - 1 = 0$. $\blacksquare$

Corollary. If $A \subseteq B$Then $P(A) \leq P(B)$.

Proof. Write $B = A \cup (B \cap A')$ where the two sets are disjoint. Then $P(B) = P(A) + P(B \cap A') \geq P(A)$ since $P(B \cap A') \geq 0$. $\blacksquare$

Theorem. $P(A \cup B) = P(A) + P(B) - P(A \cap B)$.

Proof. $A \cup B$ can be partitioned into three disjoint sets: $A \cap B'$, $A \cap B$And $A' \cap B$.

$P(A \cup B) = P(A \cap B') + P(A \cap B) + P(A' \cap B)$

$P(A) = P(A \cap B') + P(A \cap B) \implies P(A \cap B') = P(A) - P(A \cap B)$

$P(B) = P(A \cap B) + P(A' \cap B) \implies P(A' \cap B) = P(B) - P(A \cap B)$

$P(A \cup B) = [P(A) - P(A \cap B)] + P(A \cap B) + [P(B) - P(A \cap B)] = P(A) + P(B) - P(A \cap B). \quad \blacksquare$

For mutually exclusive events ($A \cap B = \emptyset$): $P(A \cup B) = P(A) + P(B)$.

Corollary (Boole’s inequality). For any events $A$ and $B$, $P(A \cup B) \leq P(A) + P(B)$.

Proof. Since $P(A \cap B) \geq 0$We have $P(A \cup B) = P(A) + P(B) - P(A \cap B) \leq P(A) + P(B)$. $\blacksquare$

Theorem. $P(A \cap B) = P(A) \cdot P(B|A)$.

Proof. This follows directly from the definition of conditional probability (Section 3.1). $\blacksquare$

General multiplication rule. For events $A_1, A_2, \ldots, A_n$:

$P\!\left(\bigcap_{i=1}^{n} A_i\right) = P(A_1) \cdot P(A_2|A_1) \cdot P(A_3|A_1 \cap A_2) \cdots P(A_n|A_1 \cap \cdots \cap A_{n-1})$

---

Definition. The conditional probability of $A$ given $B$ is

$P(A|B) = \frac{P(A \cap B)}{P(B)} \quad \mathrm{for } P(B) > 0$

Intuition. $P(A|B)$ is the probability of $A$ occurring given that we already know $B$ has Occurred. Knowing $B$ has happened changes our sample space from $\Omega$ to $B$And we measure What fraction of $B$ is also in $A$.

Theorem. Conditional probability satisfies the Kolmogorov axioms for a fixed conditioning event $B$ (with $P(B) > 0$).

Proof.

1. $P(A|B) = P(A \cap B)/P(B) \geq 0$ since $P(A \cap B) \geq 0$ and $P(B) > 0$.
2. $P(\Omega|B) = P(\Omega \cap B)/P(B) = P(B)/P(B) = 1$.
3. If $A_1, A_2, \ldots$ are mutually exclusive, then so are $A_1 \cap B, A_2 \cap B, \ldots$And

$P\!\left(\bigcup_i A_i \,\middle|\, B\right) = \frac{P\!\left(\left(\bigcup_i A_i\right) \cap B\right)}{P(B)} = \frac{\sum_i P(A_i \cap B)}{P(B)} = \sum_i P(A_i|B). \quad \blacksquare$

Corollary. The complement rule holds for conditional probability: $P(A'|B) = 1 - P(A|B)$.

Proof. This follows from applying the complement rule within the conditional probability Measure, which is justified by the theorem above. $\blacksquare$

---

Theorem. For events $A$ and $B$ with $P(B) \gt 0$:

$P(A|B) = \frac{P(B|A) \cdot P(A)}{P(B)}$

$P(A|B) = \frac{P(A \cap B)}{P(B)} = \frac{P(B|A) \cdot P(A)}{P(B)} \quad \blacksquare$

If $B_1, B_2, \ldots, B_n$ partition $\Omega$ (mutually exclusive and exhaustive):

$P(A) = \sum_{i=1}^{n}P(A|B_i)P(B_i)$

$P(B_k|A) = \frac{P(A|B_k)P(B_k)}{\sum_{i=1}^{n}P(A|B_i)P(B_i)}$

---

Definition. Events $A$ and $B$ are independent if and only if

$P(A \cap B) = P(A) \cdot P(B)$

Theorem. $A$ and $B$ are independent if and only if $P(A|B) = P(A)$ (provided $P(B) \gt 0$).

Proof.

($\Rightarrow$) If $P(A \cap B) = P(A)P(B)$Then $P(A|B) = \dfrac{P(A \cap B)}{P(B)} = \dfrac{P(A)P(B)}{P(B)} = P(A)$.

($\Leftarrow$) If $P(A|B) = P(A)$Then $\dfrac{P(A \cap B)}{P(B)} = P(A)$So $P(A \cap B) = P(A)P(B)$. $\blacksquare$

Intuition. Independence means knowing $B$ occurred gives you no information about $A$. The Probability of $A$ is the same whether or not $B$ has happened.

### 5.3 Pairwise and mutual independence

Definition. Events $A_1, A_2, \ldots, A_n$ are mutually independent if for every subset $\\{i_1, \ldots, i_k\\} \subseteq \\{1, 2, \ldots, n\\}$ with $k \geq 2$:

$P(A_{i_1} \cap A_{i_2} \cap \cdots \cap A_{i_k}) = P(A_{i_1}) \cdot P(A_{i_2}) \cdots P(A_{i_k})$

Definition. Events $A_1, A_2, \ldots, A_n$ are pairwise independent if every pair $(A_i, A_j)$ with $i \neq j$ is independent.

---

Venn diagrams represent events as regions. Useful for visualising:

• A \cup B$$A \cap B$$A'
• Relationships between events
• Applying the addition rule

Tree diagrams are useful for sequential experiments. Each branch represents a possible outcome with Its probability. The probability along any path is the product of the probabilities along its Branches (multiplication rule). The probability of any event is found by adding the probabilities of All paths leading to it (addition rule for mutually exclusive paths).

Example. A bag contains 3 red and 2 blue balls. Two balls are drawn without replacement.

$P(\mathrm{both red}) = \frac{3}{5} \times \frac{2}{4} = \frac{6}{20} = \frac{3}{10}$

$P(\mathrm{one of each}) = \frac{3}{5} \times \frac{2}{4} + \frac{2}{5} \times \frac{3}{4} = \frac{6}{20} + \frac{6}{20} = \frac{12}{20} = \frac{3}{5}$

---

$n! = n(n-1)(n-2)\cdots 1$With $0! = 1$.

• Permutations: $^n P_r = \dfrac{n!}{(n-r)!}$ (order matters)
• Combinations: $^n C_r = \binom{n}{r} = \dfrac{n!}{r!(n-r)!}$ (order does not matter)

When all outcomes are equally likely: $P(A) = \dfrac{|A|}{|\Omega|} = \dfrac{\mathrm{number of favourable outcomes}}{\mathrm{total number of outcomes}}$.

---

Theorem (Inclusion-Exclusion for three events). For events $A$, $B$, $C$:

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

Proof. Apply the two-event inclusion-exclusion rule twice:

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

Now $P(B \cup C) = P(B) + P(C) - P(B \cap C)$And by the distributive law of set theory $A \cap (B \cup C) = (A \cap B) \cup (A \cap C)$So:

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

Substituting:

$P(A \cup B \cup C) = P(A) + P(B) + P(C) - P(B \cap C) - P(A \cap B) - P(A \cap C) + P(A \cap B \cap C). \quad \blacksquare$

Theorem. For events $A$, $B$, $C$:

$(A \cup B \cup C)' = A' \cap B' \cap C'$

$(A \cap B \cap C)' = A' \cup B' \cup C'$

Proof. The three-event case follows by induction from the two-event case. For the first law:

$(A \cup B \cup C)' = ((A \cup B) \cup C)' = (A \cup B)' \cap C' = A' \cap B' \cap C'. \quad \blacksquare$

When solving problems with three events, the Venn diagram is divided into 8 regions (including The exterior). The fundamental approach is:

1. Start from the innermost region $A \cap B \cap C$ and work outward.
2. Use the given information to find the value of each region.
3. Each region represents a disjoint event, so probabilities add.

Example. In a class of 40 students, 18 study Maths, 15 study Physics, and 12 study Chemistry. 5 Study all three, 8 study Maths and Physics, 6 study Maths and Chemistry, and 7 study Physics and Chemistry.

The region for “Maths only” is: $18 - 8 - 6 + 5 = 9$ (subtract overlaps, add back the triple Overlap).

Region	Description	Calculation	Count
$A \cap B \cap C$	All three	Given	5
$A \cap B \cap C'$	Maths and Physics only	$8 - 5$	3
$A \cap C \cap B'$	Maths and Chemistry only	$6 - 5$	1
$B \cap C \cap A'$	Physics and Chemistry only	$7 - 5$	2
$A \cap B' \cap C'$	Maths only	$18 - 3 - 1 - 5$	9
$B \cap A' \cap C'$	Physics only	$15 - 3 - 2 - 5$	5
$C \cap A' \cap B'$	Chemistry only	$12 - 1 - 2 - 5$	4
$A' \cap B' \cap C'$	None	$40 - 29$	11

Check: $5 + 3 + 1 + 2 + 9 + 5 + 4 + 11 = 40$. $\checkmark$

---

A multi-stage experiment consists of a sequence of trials. A tree diagram represents this as:

• Levels correspond to stages (trials).
• Branches at each node represent possible outcomes at that stage.
• Branch probabilities are the conditional probabilities of each outcome given the path so far.
• Path probability is the product of all branch probabilities along the path.
• Event probability is the sum of all relevant path probabilities.

With replacement. At each stage, the sample space and probabilities reset. The trials are Independent.

Without replacement. At each stage, the sample space shrinks. The trials are not Independent; later probabilities depend on earlier outcomes.

Example. A bag contains 5 balls: 2 red and 3 blue. Three balls are drawn without replacement. Find the probability of drawing exactly 2 red balls.

There are $\binom{3}{2} = 3$ ways to arrange the two red draws among three positions: RRB, RBR, BRR.

$P(\mathrm{RRB}) = \frac{2}{5} \times \frac{1}{4} \times \frac{3}{3} = \frac{6}{60} = \frac{1}{10}$

$P(\mathrm{RBR}) = \frac{2}{5} \times \frac{3}{4} \times \frac{1}{3} = \frac{6}{60} = \frac{1}{10}$

$P(\mathrm{BRR}) = \frac{3}{5} \times \frac{2}{4} \times \frac{1}{3} = \frac{6}{60} = \frac{1}{10}$

$P(\mathrm{exactly 2 red}) = \frac{1}{10} + \frac{1}{10} + \frac{1}{10} = \frac{3}{10}$

For “at least $k$” problems, it is often easier to compute the complement: $P(\mathrm{at least } k) = 1 - P(\mathrm{at most } k-1)$.

Example. A fair coin is tossed 4 times. Find $P(\mathrm{at least 3 heads})$.

$P(\mathrm{at least 3 heads}) = P(\mathrm{exactly 3 heads}) + P(\mathrm{exactly 4 heads})$

$= \binom{4}{3}\left(\frac{1}{2}\right)^4 + \binom{4}{4}\left(\frac{1}{2}\right)^4 = \frac{4}{16} + \frac{1}{16} = \frac{5}{16}$

Alternatively: $P(\mathrm{at least 3 heads}) = 1 - P(\mathrm{at most 2 heads}) = 1 - \frac{11}{16} = \frac{5}{16}$.

To find a conditional probability $P(X|Y)$ from a tree diagram:

1. Identify all paths leading to $Y$ (the conditioning event).
2. Sum these path probabilities to get $P(Y)$.
3. Among those paths, identify which also satisfy $X$.
4. Sum the relevant path probabilities to get $P(X \cap Y)$.
5. $P(X|Y) = P(X \cap Y)/P(Y)$.

---

Definition. A random variable is a function $X \colon \Omega \to \mathbb{R}$ that assigns a Real number to each outcome in the sample space.

Definition. A random variable $X$ is discrete if its set of possible values is countable (i.e. Finite or countably infinite).

Example. If a fair die is rolled, define $X$ = “the number shown”. Then $X$ takes values in $\\{1, 2, 3, 4, 5, 6\\}$So $X$ is discrete.

Example. If a coin is tossed until the first head appears, define $X$ = “number of tosses”. Then $X$ takes values in $\\{1, 2, 3, \ldots\\}$Which is countably infinite.

Definition. The probability mass function (PMF) of a discrete random variable $X$ is the Function $p(x) = P(X = x)$Defined for all $x \in \mathbb{R}$.

Properties of a PMF. A function $p \colon \mathbb{R} \to [0, 1]$ is a valid PMF if and only if:

1. $p(x) \geq 0$ for all $x$.
2. $\displaystyle\sum_{\mathrm{all } x} p(x) = 1$.

Proof. Property 1 follows from non-negativity of probability. Property 2 follows because the Events $\\{X = x\\}$ for all possible values of $x$ form a partition of $\Omega$So their Probabilities sum to 1 by the normalisation axiom. $\blacksquare$

Definition. The cumulative distribution function (CDF) of a discrete random variable $X$ is

$F(x) = P(X \leq x) = \sum_{t \leq x} p(t)$

The CDF is a non-decreasing, right-continuous function with $\lim_{x \to -\infty} F(x) = 0$ and $\lim_{x \to +\infty} F(x) = 1$.

Definition. The expected value (mean) of a discrete random variable $X$ is

$E(X) = \mu = \sum_{\mathrm{all } x} x \cdot p(x)$

Definition. The variance of $X$ is

$\mathrm{Var}(X) = \sigma^2 = E\!\left[(X - \mu)^2\right] = \sum_{\mathrm{all } x} (x - \mu)^2 \cdot p(x)$

An equivalent computational formula is:

$\mathrm{Var}(X) = E(X^2) - [E(X)]^2$

Proof of the computational formula:

$\mathrm{Var}(X) = E\!\left[(X - \mu)^2\right] = E(X^2 - 2\mu X + \mu^2) = E(X^2) - 2\mu E(X) + \mu^2 = E(X^2) - \mu^2. \quad \blacksquare$

A biased die has PMF:

$x$	1	2	3	4	5	6
$p(x)$	$1/12$	$1/6$	$1/4$	$1/4$	$1/6$	$1/12$

Check: $1/12 + 1/6 + 1/4 + 1/4 + 1/6 + 1/12 = 1/12 + 2/12 + 3/12 + 3/12 + 2/12 + 1/12 = 12/12 = 1$. $\checkmark$

$E(X) = 1\!\cdot\!\tfrac{1}{12} + 2\!\cdot\!\tfrac{1}{6} + 3\!\cdot\!\tfrac{1}{4} + 4\!\cdot\!\tfrac{1}{4} + 5\!\cdot\!\tfrac{1}{6} + 6\!\cdot\!\tfrac{1}{12}$

$= \tfrac{1}{12} + \tfrac{2}{6} + \tfrac{3}{4} + \tfrac{4}{4} + \tfrac{5}{6} + \tfrac{6}{12} = \tfrac{1 + 4 + 9 + 12 + 10 + 6}{12} = \tfrac{42}{12} = 3.5$

$E(X^2) = 1\!\cdot\!\tfrac{1}{12} + 4\!\cdot\!\tfrac{1}{6} + 9\!\cdot\!\tfrac{1}{4} + 16\!\cdot\!\tfrac{1}{4} + 25\!\cdot\!\tfrac{1}{6} + 36\!\cdot\!\tfrac{1}{12}$

$= \tfrac{1 + 8 + 27 + 48 + 50 + 36}{12} = \tfrac{170}{12} = \tfrac{85}{6}$

$\mathrm{Var}(X) = E(X^2) - [E(X)]^2 = \tfrac{85}{6} - \tfrac{49}{4} = \tfrac{170 - 147}{12} = \tfrac{23}{12} \approx 1.917$

---

Problem 1

Events $A$ and $B$ are such that $P(A) = 0.4$, $P(B) = 0.5$And $P(A \cup B) = 0.7$. Find $P(A \cap B)$ and $P(A|B)$.

Solution 1

$P(A \cap B) = P(A) + P(B) - P(A \cup B) = 0.4 + 0.5 - 0.7 = 0.2$.

$P(A|B) = P(A \cap B)/P(B) = 0.2/0.5 = 0.4$.

If you get this wrong, revise: Addition Rule — Section 2.2.

Problem 2

A disease affects 1% of a population. A test is 99% accurate (both sensitivity and specificity). A person tests positive. What is the probability they actually have the disease?

Solution 2

Let $D$ = has disease, $T^+$ = tests positive.

P(D) = 0.01$$P(T^+|D) = 0.99$$P(T^+|D') = 0.01.

By the law of total probability: $P(T^+) = P(T^+|D)P(D) + P(T^+|D')P(D') = 0.99(0.01) + 0.01(0.99) = 0.0099 + 0.0099 = 0.0198$.

By Bayes’ theorem: $P(D|T^+) = \dfrac{P(T^+|D)P(D)}{P(T^+)} = \dfrac{0.0099}{0.0198} = 0.5$.

Even with a 99% accurate test, a positive result means only a 50% chance of actually having the Disease, because the disease is so rare.

If you get this wrong, revise: Bayes’ Theorem — Section 4.

Problem 3

Prove that if $A$ and $B$ are independent, then so are $A$ and $B'$.

Solution 3

$P(A \cap B') = P(A) - P(A \cap B) = P(A) - P(A)P(B)$ (by independence) $= P(A)[1 - P(B)] = P(A)P(B')$. $\blacksquare$

If you get this wrong, revise: Independence — Section 5.

Problem 4

A bag contains 4 red, 3 blue, and 2 green balls. Three balls are drawn without replacement. Find the probability that all three are different colours.

Solution 4

Total ways to choose 3 from 9: $\binom{9}{3} = 84$.

Ways to get one of each colour: $\binom{4}{1}\binom{3}{1}\binom{2}{1} = 4 \times 3 \times 2 = 24$.

$P = 24/84 = 2/7$.

If you get this wrong, revise: Counting Principles — Section 7.

Problem 5

Events $A$$B$$C$ are such that $P(A) = 0.3$$P(B) = 0.4$$P(C) = 0.5$$P(A \cap B) = 0.1$$P(A \cap C) = 0.15$$P(B \cap C) = 0.2$And $P(A \cap B \cap C) = 0.05$. Find $P(A \cup B \cup C)$.

Solution 5

By the inclusion-exclusion principle:

$P(A \cup B \cup C) = P(A) + P(B) + P(C) - P(A \cap B) - P(A \cap C) - P(B \cap C) + P(A \cap B \cap C)$ $= 0.3 + 0.4 + 0.5 - 0.1 - 0.15 - 0.2 + 0.05 = 0.8$

If you get this wrong, revise: Addition Rule — Section 2.2.

Problem 6

Two coins are tossed. Given that at least one is heads, find the probability that both are heads.

Solution 6

$\Omega = \{HH, HT, TH, TT\}$. $A = \{\mathrm{at least one heads}\} = \{HH, HT, TH\}$. $B = \{\mathrm{both heads}\} = \{HH\}$.

$P(B|A) = P(B \cap A)/P(A) = P(B)/P(A) = (1/4)/(3/4) = 1/3$.

If you get this wrong, revise: Conditional Probability — Section 3.

Problem 7

A fair die is rolled. Let $A$ = "even number" and $B$ = "number greater than 3". Are $A$ and $B$ independent?

Solution 7

$A = \{2, 4, 6\}$$B = \{4, 5, 6\}$$A \cap B = \{4, 6\}$.

P(A) = 3/6 = 1/2$$P(B) = 3/6 = 1/2$$P(A \cap B) = 2/6 = 1/3.

$P(A)P(B) = 1/4 \neq 1/3 = P(A \cap B)$. So $A$ and $B$ are not independent.

If you get this wrong, revise: Independence — Section 5.

Problem 8

In a school, 60% of students study Maths, 40% study Physics, and 25% study both. A student is chosen at random. Given that they study Physics, find the probability they study Maths.

Solution 8

$P(M) = 0.6$$P(P) = 0.4$$P(M \cap P) = 0.25$.

$P(M|P) = P(M \cap P)/P(P) = 0.25/0.4 = 0.625$.

If you get this wrong, revise: Conditional Probability — Section 3.

Problem 9

A box contains 10 items, 3 of which are defective. Items are inspected one by one without replacement. Find the probability that the first defective item is the third one inspected.

Solution 9

First two non-defective, third defective:

$P = \frac{7}{10} \times \frac{6}{9} \times \frac{3}{8} = \frac{7 \times 6 \times 3}{720} = \frac{126}{720} = \frac{7}{40}$

If you get this wrong, revise: Tree Diagrams — Section 6.2.

Problem 10

A machine produces components. 5% are defective. Components are packed in boxes of 20. Find the probability that a box contains exactly one defective component.

Solution 10

This is a binomial scenario: $X \sim B(20, 0.05)$.

$P(X=1) = \binom{20}{1}(0.05)^1(0.95)^{19} = 20 \times 0.05 \times 0.95^{19} \approx 0.3774$.

If you get this wrong, revise: Binomial Distribution — Statistical Distributions chapter.

Problem 11

Prove that $P(A \cup B') = 1 - P(A' \cap B)$.

Solution 11

By De Morgan's law: $(A \cup B')' = A' \cap B$.

So $P(A \cup B') = 1 - P((A \cup B')') = 1 - P(A' \cap B)$. $\blacksquare$

If you get this wrong, revise: Complement Rule — Section 2.1.

Problem 12

From a standard 52-card deck, 5 cards are dealt. Find the probability of getting a flush (all 5 cards of the same suit).

Solution 12

Total ways: $\binom{52}{5} = 2598960$.

Ways to get a flush: choose suit (4 ways), then 5 cards from that suit ($\binom{13}{5} = 1287$).

Total flushes: $4 \times 1287 = 5148$.

$P(\mathrm{flush}) = 5148/2598960 \approx 0.00198 \approx 0.2\%$.

If you get this wrong, revise: Counting Principles — Section 7.

Problem 13

A discrete random variable $X$ has PMF $p(x) = kx$ for $x \in \{1, 2, 3, 4, 5\}$ and $p(x) = 0$ otherwise. Find the constant $k$Then find $E(X)$ and $\mathrm{Var}(X)$.

Solution 13

For a valid PMF: $\sum_{x=1}^{5} kx = k(1 + 2 + 3 + 4 + 5) = 15k = 1$So $k = 1/15$.

$E(X) = \sum_{x=1}^{5} x \cdot \frac{x}{15} = \frac{1 + 4 + 9 + 16 + 25}{15} = \frac{55}{15} = \frac{11}{3}$

$E(X^2) = \sum_{x=1}^{5} x^2 \cdot \frac{x}{15} = \frac{1 + 8 + 27 + 64 + 125}{15} = \frac{225}{15} = 15$

$\mathrm{Var}(X) = E(X^2) - [E(X)]^2 = 15 - \frac{121}{9} = \frac{135 - 121}{9} = \frac{14}{9}$

If you get this wrong, revise: Discrete Random Variables — Section 10.

Problem 14

A bag contains 4 red and 6 blue balls. Balls are drawn one at a time without replacement until a red ball is drawn. Find the probability that exactly 3 draws are needed.

Solution 14

We need the first two draws to be blue and the third to be red:

$P = \frac{6}{10} \times \frac{5}{9} \times \frac{4}{8} = \frac{120}{720} = \frac{1}{6}$

If you get this wrong, revise: Multi-Stage Experiments — Section 9.

Problem 15

Prove Boole's inequality: for events $A_1, A_2, \ldots, A_n$

$P\!\left(\bigcup_{i=1}^{n} A_i\right) \leq \sum_{i=1}^{n} P(A_i)$

Solution 15

By induction on $n$.

Base case ($n = 2$): $P(A_1 \cup A_2) = P(A_1) + P(A_2) - P(A_1 \cap A_2) \leq P(A_1) + P(A_2)$. $\checkmark$

Inductive step: Assume $P\!\left(\bigcup_{i=1}^{k} A_i\right) \leq \sum_{i=1}^{k} P(A_i)$. Then

$P\!\left(\bigcup_{i=1}^{k+1} A_i\right) = P\!\left(\bigcup_{i=1}^{k} A_i\right) + P(A_{k+1}) - P\!\left(\bigcup_{i=1}^{k} A_i \cap A_{k+1}\right)$

$\leq \sum_{i=1}^{k} P(A_i) + P(A_{k+1}) = \sum_{i=1}^{k+1} P(A_i). \quad \blacksquare$

If you get this wrong, revise: Basic Probability Results — Section 2.

Problem 16

In a survey, 70% of people like tea, 50% like coffee, and 35% like both. A person is chosen at random. Given that they like at least one of the two drinks, find the probability that they like both.

Solution 16

$P(T) = 0.7$$P(C) = 0.5$$P(T \cap C) = 0.35$.

$P(T \cup C) = P(T) + P(C) - P(T \cap C) = 0.7 + 0.5 - 0.35 = 0.85$.

$P(T \cap C \mid T \cup C) = \frac{P(T \cap C)}{P(T \cup C)} = \frac{0.35}{0.85} = \frac{7}{17} \approx 0.412$

If you get this wrong, revise: Conditional Probability — Section 3.

Problem 17

A fair coin is tossed 5 times. Using the complement rule, find the probability of getting at least one head.

Solution 17

Let $A$ = "at least one head". Then $A'$ = "no heads" = "all tails".

$P(A) = 1 - P(A') = 1 - \left(\frac{1}{2}\right)^5 = 1 - \frac{1}{32} = \frac{31}{32}$

If you get this wrong, revise: Complement Rule — Section 2.1.

Problem 18

Two events $A$ and $B$ satisfy $P(A) = 0.6$$P(B|A) = 0.4$And $P(B|A') = 0.7$. Find $P(B)$$P(A|B)$And determine whether $A$ and $B$ are independent.

Solution 18

By the law of total probability:

$P(B) = P(B|A)P(A) + P(B|A')P(A') = 0.4 \times 0.6 + 0.7 \times 0.4 = 0.24 + 0.28 = 0.52$

$P(A \cap B) = P(B|A)P(A) = 0.4 \times 0.6 = 0.24$

$P(A|B) = \frac{P(A \cap B)}{P(B)} = \frac{0.24}{0.52} = \frac{6}{13} \approx 0.462$

Check independence: $P(A)P(B) = 0.6 \times 0.52 = 0.312 \neq 0.24 = P(A \cap B)$. So $A$ and $B$ are not independent.

If you get this wrong, revise: Bayes’ Theorem — Section 4, and Independence — Section 5.

Problem 19

A discrete random variable $X$ has CDF $F(x) = 0$ for $x \lt 0$$F(x) = x/4$ for $0 \leq x \lt 1$$F(x) = 1/2$ for $1 \leq x \lt 2$$F(x) = 3/4$ for $2 \leq x \lt 3$And $F(x) = 1$ for $x \geq 3$. Find the PMF of $X$ and verify it sums to 1.

Solution 19

The PMF is obtained from the jumps in the CDF:

• $p(0) = F(0) - F(0^-) = 0 - 0 = 0$. But from the formula $F(x) = x/4$ at $x = 0$: $p(0) = 0$. Actually, the jump occurs at the boundary. Since $F$ is continuous at $x = 0$There is no point mass at 0. The value $X = 0$ has probability 0; we look at where jumps occur.

More carefully, the jumps occur at:

• $x = 1$: $p(1) = F(1) - F(1^-) = 1/2 - 1/4 = 1/4$
• $x = 2$: $p(2) = F(2) - F(2^-) = 3/4 - 1/2 = 1/4$
• $x = 3$: $p(3) = F(3) - F(3^-) = 1 - 3/4 = 1/4$

There is also a continuous component on $[0, 1)$But since $X$ is discrete, the CDF must be a step Function. The given CDF has a linear portion, which indicates this CDF actually corresponds to a mixed distribution. For a purely discrete $X$The CDF should be piecewise constant with jumps.

Assuming the problem intended a discrete distribution, the PMF from the jumps is:

$p(1) = \frac{1}{4}, \quad p(2) = \frac{1}{4}, \quad p(3) = \frac{1}{4}, \quad p(x) = 0 \mathrm{ otherwise}$

Check: $1/4 + 1/4 + 1/4 = 3/4 \neq 1$. This indicates the continuous portion $F(x) = x/4$ on $[0,1)$ contributes probability $1/4$ spread over a continuum, confirming this is not a purely Discrete distribution.

If you get this wrong, revise: Discrete Random Variables — Section 10.

Problem 20

Three machines $M_1$, $M_2$, $M_3$ produce items with proportions 50%, 30%, 20%. Their defect rates are 2%, 3%, 5% respectively. An item is found to be defective. Find the probability it was produced by $M_3$.

Solution 20

Let $D$ = "defective". By the law of total probability:

$P(D) = P(D|M_1)P(M_1) + P(D|M_2)P(M_2) + P(D|M_3)P(M_3)$ $= 0.02 \times 0.5 + 0.03 \times 0.3 + 0.05 \times 0.2 = 0.01 + 0.009 + 0.01 = 0.029$

By Bayes’ theorem:

$P(M_3|D) = \frac{P(D|M_3)P(M_3)}{P(D)} = \frac{0.05 \times 0.2}{0.029} = \frac{0.01}{0.029} = \frac{10}{29} \approx 0.345$

If you get this wrong, revise: Extended Bayes’ Theorem — Section 4.4.

---

## Common Pitfalls

1. Incorrectly applying integration by parts by choosing $u$ and $\frac{dv}{dx}$ the wrong way around.

2. Dropping negative signs during algebraic manipulation. Substitute back to verify your answer.

3. Confusing the domain and range of functions, or not considering restrictions (e.g., denominator cannot be zero).

4. Forgetting the $+c$ constant of integration in indefinite integrals, or misusing boundary conditions in definite integrals.

• Data Representation — Frequency tables and grouped data from this topic form the empirical basis for probability models.
• Correlation and Regression — Conditional probability underpins the interpretation of regression and correlation analysis.
• Statistical Distributions — The binomial and Poisson distributions extend probability theory to counting and rare events.
• Hypothesis Testing — Probability theory provides the p-values and significance levels used in hypothesis testing.