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Module 10 · L01 of 20 ~40 min ⚡ +90 XP available

Probability Review for Statistics

A quality engineer inspects a batch of LEDs and needs the chance that exactly 3 of the next 20 are defective. Before you can model that with a binomial distribution, you need to be fluent in the underlying probability machinery — sample spaces, conditional probability, independence, and expected value. This lesson reactivates those tools so the binomial work in the rest of Module 10 lands cleanly.

Today's hook — Two fair dice are rolled. Before reading on, write down (a) the probability that the sum is 7, and (b) the probability that the sum is 7 given that the first die shows 4. Are those two probabilities equal? Compare your answer after card 05.

0/5QUESTS

Recall — your gut answer first

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A bag contains 3 red and 5 blue balls. You draw two balls without replacement. Before any formula — what is the probability that both are red? Sketch the tree or list the cases below.

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The two moves for any probability question

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Every probability problem rewards two habits: define the sample space (or use a tree / Venn diagram), then decide if events are dependent or independent before multiplying. Skipping the diagram is the single biggest cause of error.

The list-and-classify strategy: (1) list (or describe) the sample space, (2) identify the events of interest, (3) decide if they are mutually exclusive, independent, or conditional, and (4) apply the matching rule.

$P(A|B) = \dfrac{P(A\cap B)}{P(B)}$ · Independent: $P(A\cap B) = P(A)P(B)$

$P(A|B) = \dfrac{P(A\cap B)}{P(B)}$

Tree first, formula second

Drawing a probability tree forces you to track conditional probabilities branch by branch and stops you from blindly multiplying $P(A)P(B)$ when the events are not independent.

"Given that" means condition

Any time you read "given that", "if we know", or "of the X who …", the problem is conditional. Use $P(A|B) = P(A\cap B)/P(B)$ — not just $P(A)$.

$E(X) = \sum x \cdot P(X=x)$

The expected value of a discrete random variable is the probability-weighted average of its outcomes — not the most likely outcome. It need not be a value $X$ can actually take.

What you'll master

Know

Key facts

Addition rule: $P(A\cup B) = P(A) + P(B) - P(A\cap B)$
Conditional probability: $P(A|B) = \dfrac{P(A\cap B)}{P(B)}$
Independence test: $P(A\cap B) = P(A)P(B)$
Expected value: $E(X) = \displaystyle\sum_x x\,P(X=x)$

Understand

Concepts

Why "with/without replacement" changes whether events are independent
How conditional probability differs from joint probability
Why expected value is a long-run average, not a single-trial prediction

Can do

Skills

Build probability trees for multi-stage experiments
Compute conditional probabilities from a two-way table or Venn diagram
Calculate $E(X)$ for a discrete random variable from its distribution table

Key terms

Sample spaceThe set $\Omega$ of all possible outcomes of an experiment. For one die: $\Omega = \{1,2,3,4,5,6\}$; for two dice: 36 ordered pairs.

EventA subset of the sample space. $P(A) = \dfrac{|A|}{|\Omega|}$ when outcomes are equally likely.

Mutually exclusive$A\cap B = \varnothing$, so $P(A\cap B) = 0$ and $P(A\cup B) = P(A)+P(B)$.

Independent eventsKnowing $B$ happened does not change $P(A)$. Test: $P(A\cap B) = P(A)P(B)$.

Conditional probability$P(A|B) = \dfrac{P(A\cap B)}{P(B)}$ — the probability of $A$ given that $B$ has occurred.

Discrete random variableA variable $X$ taking countable values $x_1, x_2, \ldots$ with probabilities summing to 1. $E(X) = \sum x_i P(X=x_i)$.

ME12-5NESA outcome: solves problems using appropriate statistical processes — the umbrella outcome for the whole of Module 10.

The three probability rules you'll re-use

core concept

Almost every probability question in Module 10 reduces to one of three rules. Memorise them in this exact form:

Addition rule. $P(A\cup B) = P(A) + P(B) - P(A\cap B)$. If $A,B$ mutually exclusive, the last term is 0.
Multiplication rule. $P(A\cap B) = P(A)\,P(B|A)$. If $A,B$ independent, then $P(B|A) = P(B)$ and we get $P(A)P(B)$.
Conditional rule. $P(A|B) = \dfrac{P(A\cap B)}{P(B)}$ — a rearrangement of the multiplication rule.

Worked through the hook: Two fair dice are rolled.

Sample space: 36 equally-likely ordered pairs.
Sum 7 outcomes: $(1,6),(2,5),(3,4),(4,3),(5,2),(6,1)$ — 6 pairs. So $P(\text{sum}=7) = \dfrac{6}{36} = \dfrac{1}{6}$.
Given the first die is 4: only 6 outcomes possible, and only $(4,3)$ gives sum 7. So $P(\text{sum}=7\,|\,\text{first}=4) = \dfrac{1}{6}$.
The two are equal, which tells us "sum = 7" and "first die = 4" are independent — a special feature of sum 7 (any first-die value still leaves exactly one complement).

Connecting to binomial. The binomial setup later in Module 10 assumes independent trials with a fixed success probability. Every binomial calculation rests on the independence and multiplication rules above.

The three probability rules: (1) Addition: $P(A\cup B)=P(A)+P(B)-P(A\cap B)$; (2) Multiplication: $P(A\cap B)=P(A)P(B|A)$; (3) Conditional: $P(A|B)=P(A\cap B)/P(B)$.

Pause — copy all three probability rules in their exact HSC forms: addition rule, multiplication rule, and conditional rule into your book.

Quick check: A bag contains 4 red and 6 blue marbles. Two are drawn without replacement. What is $P(\text{both red})$?

Independence and expected value of a discrete RV

core concept

We just saw the three probability rules: $P(A\cup B)=P(A)+P(B)-P(A\cap B)$; $P(A\cap B)=P(A)P(B|A)$; and $P(A|B)=P(A\cap B)/P(B)$. That raises a question: the multiplication rule for independent events collapses to $P(A\cap B)=P(A)P(B)$ — but how do you test whether independence actually holds, and how do you compute the long-run average of a discrete random variable? This card answers it → test independence with $P(A\cap B)=P(A)P(B)$; compute $E(X)=\sum x_i p_i$.

Two events $A$ and $B$ are independent iff $P(A\cap B) = P(A)P(B)$. Equivalently, $P(A|B) = P(A)$ — knowing $B$ tells you nothing extra about $A$. "Without replacement" almost always destroys independence; "with replacement" preserves it.

For a discrete random variable $X$ with values $x_1, x_2, \ldots$ and probabilities $p_1, p_2, \ldots$ (summing to 1), the expected value is:

$$E(X) = \sum_i x_i\,P(X=x_i) = \sum_i x_i p_i$$

Example. $X$ is the score on a single fair die. Then $E(X) = \dfrac{1+2+3+4+5+6}{6} = \dfrac{21}{6} = 3.5$. The expected score is 3.5 even though $X$ never equals 3.5 — that is the point of "long-run average".

Common mistake. Students confuse $E(X)$ with the most likely value (the mode). For a fair die every outcome has $p = 1/6$, so the mode is undefined, yet $E(X) = 3.5$ is still meaningful. Always weight by probability, not by likelihood-of-occurring-most.

Two events $A$ and $B$ are independent iff $P(A\cap B) = P(A)P(B)$. Equivalently, $P(A|B) = P(A)$ — knowing $B$ tells you nothing extra about $A$. "Without replacement" almost always destroys independence; "with replacement" preserves it.

Pause — copy the independence test $P(A\cap B)=P(A)P(B)$ and the expected value formula $E(X)=\sum x_i p_i$ with a one-line worked example into your book.

Did you get this? True or false: if $X$ is the number of heads when a fair coin is tossed twice, then $E(X) = 1$.

Worked examples · 3 in a row, reveal as you go

PROBLEM 1 · CONDITIONAL PROBABILITY

A school survey found $60\%$ of students play sport and $40\%$ play a musical instrument. $25\%$ of students do both. (a) Find the probability that a randomly chosen student plays sport or a musical instrument. (b) Given a student plays sport, find the probability they also play a musical instrument.

Let $S$ = plays sport, $M$ = plays music. We are told $P(S) = 0.6$, $P(M) = 0.4$, $P(S\cap M) = 0.25$.

List the given probabilities first. A Venn diagram with $S\cap M = 0.25$, $S$-only $= 0.35$, $M$-only $= 0.15$ is a useful sketch.

PROBLEM 2 · INDEPENDENCE AND TREES

A traffic light is green $40\%$ of the time and red $60\%$ of the time. A driver passes through it on two independent occasions. Find the probability that the light is green on exactly one of the two occasions.

Let $G_i$ = green on visit $i$. Since the visits are independent, $P(G_1) = P(G_2) = 0.4$ and $P(\bar G_i) = 0.6$.

"Exactly one green" splits into two mutually exclusive cases: $G_1\bar G_2$ or $\bar G_1 G_2$.

PROBLEM 3 · EXPECTED VALUE OF A DISCRETE RV

A discrete random variable $X$ has the following distribution: $P(X=0) = 0.1$, $P(X=1) = 0.3$, $P(X=2) = 0.4$, $P(X=3) = 0.2$. Find $E(X)$.

Check the probabilities sum to 1: $0.1 + 0.3 + 0.4 + 0.2 = 1.0$ ✓.

Always confirm a valid distribution before computing $E(X)$. If the sum isn't 1, the table is wrong.

Fill the gap: A discrete random variable $X$ has $P(X=0)=0.2$, $P(X=1)=0.5$, $P(X=2)=0.3$. Then $E(X) = 0(0.2) + 1(0.5) + 2(0.3) = $.

Misconceptions to fix · the 3 traps that cost marks

Trap 01

Multiplying when events are NOT independent

"Without replacement" almost always means dependent. Drawing two balls from a bag changes the probability of the second draw. Always use $P(A\cap B) = P(A)P(B|A)$ and update the second probability — never blindly write $P(A)P(B)$ unless you have verified independence.

Trap 02

Confusing $P(A|B)$ with $P(A\cap B)$

$P(A\cap B)$ is the joint probability (both happen, out of the whole sample space). $P(A|B)$ is the conditional probability (A happens, restricted to outcomes where B has already occurred). They differ by a factor of $P(B)$ — never interchange them.

Trap 03

Reporting the mode instead of $E(X)$

$E(X)$ is the probability-weighted average of all values, not the value with the highest probability. For $\{0,1,2,3\}$ with probs $\{0.1,0.3,0.4,0.2\}$ the mode is 2 but $E(X) = 1.7$. Always sum $\sum x_i p_i$, never just pick the most likely outcome.

Did you get this? True or false: if two events $A$ and $B$ are independent, then $P(A|B) = P(A)$.

Work mode · how are you completing this lesson?

Activities · practice with the ideas

A card is drawn from a standard 52-card deck. Find the probability that it is a face card (J, Q, K) or a heart.

A bag has 5 red and 3 green marbles. Two are drawn without replacement. Find the probability that the second is green given the first was red.

A coin is biased so that $P(\text{head}) = 0.7$. It is tossed twice independently. Find the probability of exactly one head.

A discrete random variable $X$ has $P(X=1) = 0.2$, $P(X=2) = 0.5$, $P(X=3) = 0.3$. Find $E(X)$ and check the probabilities sum to 1.

In a class of 30 students, 18 study Chemistry, 14 study Physics, and 8 study both. Are "studies Chemistry" and "studies Physics" independent? Justify with $P(A)P(B)$ vs $P(A\cap B)$.

Odd one out: Three of these statements about a random variable $X$ with distribution $P(X=0)=0.1, P(X=1)=0.3, P(X=2)=0.4, P(X=3)=0.2$ are correct. Which one is NOT?

Revisit your thinking

Earlier you considered two dice and the probability of sum 7, unconditionally and given the first die is 4.

Both probabilities equal $\dfrac{1}{6}$. This special case shows the two events are independent — but only because every value of the first die leaves exactly one second-die value that completes sum 7. For sum 2 or sum 12 the equivalent conditional probability would be very different, breaking independence. The takeaway: never assume independence — verify it with the formal test $P(A\cap B) = P(A)P(B)$.

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Multiple choice

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Pick your answer, then rate your confidence — that tells the system what to drill next. Each retry pulls a fresh mix from the bank.

Short answer

ApplyBand 32 marks

Q1. A bag contains 5 red and 4 blue marbles. Two are drawn without replacement. Find the probability that both are red. (2 marks)

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ApplyBand 43 marks

Q2. Events $A$ and $B$ satisfy $P(A) = 0.5$, $P(B) = 0.4$, $P(A\cup B) = 0.7$. (i) Find $P(A\cap B)$. (ii) Find $P(A|B)$. (iii) Are $A$ and $B$ independent? Justify. (3 marks)

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AnalyseBand 53 marks

Q3. A discrete random variable $X$ has the distribution: $P(X=1) = 0.1$, $P(X=2) = a$, $P(X=3) = 0.4$, $P(X=4) = 0.2$. Find $a$, then compute $E(X)$. (3 marks)

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Comprehensive answers (click to reveal)

Activity answers:

1. $P(\text{face}) = 12/52$, $P(\text{heart}) = 13/52$, $P(\text{face} \cap \text{heart}) = 3/52$ (J, Q, K of hearts). $P(\text{face} \cup \text{heart}) = 12/52 + 13/52 - 3/52 = 22/52 = 11/26$.

2. $P(G_2 | R_1) = 3/7$ (after one red removed, 3 green out of 7 remaining).

3. $P(\text{exactly one head}) = 2(0.7)(0.3) = 0.42$.

4. Sum = 1 ✓. $E(X) = 1(0.2)+2(0.5)+3(0.3) = 0.2 + 1.0 + 0.9 = 2.1$.

5. $P(C) = 0.6$, $P(P) = 14/30 \approx 0.467$, so $P(C)P(P) \approx 0.28$. But $P(C\cap P) = 8/30 \approx 0.267$. These differ, so the events are not independent (they are close to independent but not exactly).

Q1 (2 marks): $P(R_1) = 5/9$ [1]. $P(R_2|R_1) = 4/8 = 1/2$, so $P(\text{both red}) = 5/9 \times 1/2 = 5/18$ [1].

Q2 (3 marks): (i) $P(A\cap B) = 0.5 + 0.4 - 0.7 = 0.2$ [1]. (ii) $P(A|B) = 0.2/0.4 = 0.5$ [1]. (iii) $P(A)P(B) = 0.5 \times 0.4 = 0.2 = P(A\cap B)$, so $A$ and $B$ are independent. (Equivalently: $P(A|B) = 0.5 = P(A)$.) [1]

Q3 (3 marks): $0.1 + a + 0.4 + 0.2 = 1 \Rightarrow a = 0.3$ [1]. $E(X) = 1(0.1) + 2(0.3) + 3(0.4) + 4(0.2)$ [1] $= 0.1 + 0.6 + 1.2 + 0.8 = 2.7$ [1].

Boss battle · The Probability Refresher

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Five timed questions on sample spaces, conditional probability, independence, and expected value. Beat the boss to bank a tier — gold (90% + speed), silver (75%), or bronze (50%). Replays welcome.

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Science Jump · platform challenge

Climb platforms by answering probability review questions. Lighter alternative to the boss.

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