§-Syllabus dot point

NSWMaths AdvancedSyllabus dot point

How do we combine probabilities across several stages, and update a probability once we know something has happened?

Apply tree diagrams, conditional probability, independence and complementary events to solve multi-step probability problems

A focused answer to the HSC Maths Advanced probability dot point. Sample spaces and the basic rules, complementary 'at least one' events, the addition rule, independent versus dependent events, multiplying along tree branches, conditional probability and Bayes-style reverse reasoning, with worked examples drawn from recent HSC questions.

Generated by Claude Opus 4.818 min answerUpdated 2026-06-21

Reviewed by: AI editorial process; not yet individually human-reviewed

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What this dot point is asking

NESA wants you to handle probability as a process: set up the sample space, combine probabilities of compound events with the addition and multiplication rules, run those multiplications along the branches of a tree diagram, decide whether two events are independent, and update a probability once you are told that something has happened. Almost every Advanced paper opens with one of these (a sample-space count, a complementary "at least one", a tree, or a conditional probability), and the harder version near the end of the paper chains them together and asks you to "show that" a given value follows. The skill being marked is choosing the right rule for the wording and laying the working out so the method is visible.

The answer

Sample space and the basic rules

The sample space is the set of all equally detailed outcomes of an experiment. For a single fair action you find a probability by counting: $P(A) = \dfrac{\text{number of favourable outcomes}}{\text{total number of outcomes}}$ . Rolling a die and spinning a four-sector spinner, for instance, has $6 \times 4 = 24$ equally likely outcomes, and you count the ones that meet the condition. Every probability obeys $0 \le P(A) \le 1$ , the certain event has probability $1$ , and the impossible event has probability $0$ .

Two pieces of vocabulary do a lot of work in the exam. Events are mutually exclusive when they cannot both happen (rolling a $2$ and rolling a $5$ on one die), and complementary when one is exactly "the other does not happen". The complement of $A$ is written $\overline{A}$ or $A'$ , and because something must occur,

P(\overline{A}) = 1 - P(A).

This single identity is the most useful line in the topic: whenever the favourable cases are many but the unfavourable cases are few, count the few and subtract.

Complementary events and "at least one"

The phrase "at least one" is the standard trigger for the complement. Listing every way to get one or more successes is slow and error-prone; the opposite event, "none", is usually a single product. So

P(\text{at least one}) = 1 - P(\text{none}).

If a player scores on each shot with probability $0.15$ independently, the chance of "at least one goal in $n$ shots" is $1 - 0.85^{\,n}$ , because "none" means missing every time. The 2024 paper asked exactly this (find the smallest $n$ making the probability exceed $0.8$ ), and the 2023 paper asked for "at least one of four students not available" as $1 - P(\text{all four available})$ . Reach for the complement the moment you read "at least one".

The addition rule

For the probability that $A$ or $B$ happens (the union, meaning at least one of them),

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

You subtract $P(A \cap B)$ because the outcomes in the overlap have been counted once in $P(A)$ and again in $P(B)$ ; subtracting puts them back to a single count. A Venn diagram makes this concrete: the union is everything inside either circle, and the lens in the middle is the double-counted overlap. When $A$ and $B$ are mutually exclusive the overlap is empty, $P(A \cap B) = 0$ , and the rule collapses to the simple $P(A \cup B) = P(A) + P(B)$ . The 2024 multiple-choice item about $60$ students playing basketball or hockey is this rule rearranged: knowing the union and the two totals lets you solve for the overlap.

In the diagram, $P(\text{coffee}) = 0.45$ and $P(\text{pretzel}) = 0.30$ overlap in $P(\text{both}) = 0.20$ , so the union is $0.45 + 0.30 - 0.20 = 0.55$ and the outside region, $P(\text{neither}) = 1 - 0.55 = 0.45$ , falls straight out of the complement.

Independent versus dependent events

Two events are independent when knowing that one happened does not change the probability of the other. The test, and the working markers want to see, is the multiplication identity

A, B \text{ independent} \iff P(A \cap B) = P(A)\,P(B).

Tossing a coin and rolling a die are independent, so $P(\text{head and six}) = \frac12 \cdot \frac16 = \frac{1}{12}$ . Events are dependent when the first outcome changes the second, and the classic source of dependence is drawing without replacement: once a red marble is removed the bag has changed, so the second draw runs on different numbers. Drawing with replacement (or any genuinely repeated trial, like repeated shots at goal) puts the situation back as it was, so the trials are independent and you reuse the same probabilities each time. Deciding "with or without replacement" is the move that fixes every later number, so make it first.

The multiplication rule along a tree

A tree diagram lays out a multi-stage experiment one stage per column. The general multiplication rule is

P(A \cap B) = P(A)\,P(B \mid A),

read as " $A$ happens, and then $B$ happens given that $A$ has". On a tree this is literally "multiply along the branches": the probability of a complete path is the product of the branch probabilities you pass through. The conventions that earn marks are that the branches leaving any node sum to $1$ , that the second-stage probabilities are written conditional on the first stage (this is where without-replacement trees change their numbers), and that the path products for all outcomes sum to $1$ , which is your built-in check.

The tree above is the 2024 marbles question: a bag of $2$ red and $3$ white, two marbles drawn without replacement. The first draw is red with probability $\frac{2}{5}$ ; if it was red, only $1$ red remains among the $4$ left, so the second-stage red branch is $\frac{1}{4}$ , and the highlighted path gives $P(\text{RR}) = \frac{2}{5}\cdot\frac{1}{4} = \frac{1}{10}$ . The four path products $\frac{1}{10}, \frac{3}{10}, \frac{3}{10}, \frac{3}{10}$ sum to $1$ .

Conditional probability

The conditional probability of $A$ given $B$ , written $P(A \mid B)$ , is the probability of $A$ once you already know $B$ has occurred. Knowing $B$ shrinks the world to just the outcomes in $B$ , and you ask what fraction of that smaller world is also in $A$ :

P(A \mid B) = \frac{P(A \cap B)}{P(B)}, \qquad P(B) \neq 0.

This is the multiplication rule rearranged, and it is the engine behind every "given that" question. On the tree, the 2024 question continues: "given one marble is red, find the probability the other is also red" is $P(\text{RR} \mid \text{at least one red}) = \dfrac{P(\text{RR})}{P(\text{at least one red})}$ . The numerator is the path you want; the denominator is the total probability of the restricted world, not $1$ . Dividing by $P(B)$ instead of leaving it as $1$ is the whole point, and the most common place marks are lost.

Reverse conditional probability (Bayes-style)

A tree is naturally written "forwards": the first stage is a cause and the second is an effect, so the branch probabilities you are given are $P(\text{effect} \mid \text{cause})$ . Many questions then ask the reverse: given the effect, which cause was it? You cannot read $P(\text{cause} \mid \text{effect})$ straight off the tree, but you can build it from the definition. Find the one path you want (a single product), find the total probability of the observed effect by adding every path that produces it, and divide:

P(\text{cause}_i \mid \text{effect}) = \frac{P(\text{cause}_i)\,P(\text{effect} \mid \text{cause}_i)}{\displaystyle\sum_j P(\text{cause}_j)\,P(\text{effect} \mid \text{cause}_j)}.

The denominator is just "add up the branches that land on the effect" (a use of the total probability idea), and the numerator is the single branch that came from the cause you care about. The 2025 question, "given the team wins, find the probability Amara was selected", is exactly this shape, and so is any false-positive medical-test question. You do not need to memorise the formula as symbols if you can say the sentence: the favourable path over the sum of all paths that reach the same outcome.

How exam questions ask about probability

"What is the probability of a score of $\dots$ " with a die, spinner or table. Count the favourable outcomes over the total in the sample space.
"At least one $\dots$ " or "the least number of trials so the probability exceeds $\dots$ " Use $1 - P(\text{none})$ ; for repeated independent trials this is $1 - p^{\,n}$ .
"How many play both $\dots$ " or " $A$ or $B$ ". Apply the addition rule $P(A \cup B) = P(A) + P(B) - P(A \cap B)$ , often rearranged to find the overlap; a Venn diagram organises the regions.
"Are the events independent? Justify." Test whether $P(A \cap B) = P(A)\,P(B)$ , or equivalently whether $P(A \mid B) = P(A)$ ; state the comparison explicitly and conclude.
A tree diagram is drawn or implied (with / without replacement). Multiply along branches for a path; add the relevant path products for a combined event. Check all paths sum to $1$ .
"Show that $P(\dots) = \dots$ ". Write the intersection both ways, $P(A)\,P(B \mid A) = P(B)\,P(A \mid B)$ , and solve for the unknown probability.
"Given $B$ , find the probability of $A$ " where $B$ is the second-stage outcome. This is reverse-conditional: favourable path divided by the total probability of $B$ .

Worked examples

These examples move from the basic rules to a full Bayes-style tree, each in a context an Australian student would meet. Every numeric answer was verified by computation and kept as an exact fraction where natural.

At least one late parcel using the complement

Apex Couriers delivers each parcel on time with probability $0.92$ , independently of the others. A small business sends $5$ parcels. Find the probability that at least one arrives late.

Choose the rule. "At least one late" covers one, two, three, four or five late parcels, which is awkward to add. Its complement is the single clean event "all five on time", so use $P(\text{at least one late}) = 1 - P(\text{all on time})$ .

Compute the complement. The parcels are independent, so the chance all five are on time is the product of five equal factors:

P(\text{all on time}) = 0.92^{5} = 0.659082\ldots

Subtract. Therefore

P(\text{at least one late}) = 1 - 0.92^{5} = 0.3409 \;(\text{to } 4 \text{ d.p.}).

About a $34\%$ chance, which is far higher than the $8\%$ per-parcel late rate might suggest, and that gap is exactly why "at least one" questions are worth checking by the complement rather than by intuition.

Coffee or a pretzel: the addition rule

At a Westfield food court, $45\%$ of shoppers buy a coffee, $30\%$ buy a pretzel, and $20\%$ buy both. Find the probability that a random shopper buys a coffee or a pretzel, and the probability they buy neither.

Apply the addition rule. "Coffee or pretzel" is the union. Adding the two single probabilities double-counts the shoppers who buy both, so subtract that overlap once:

P(\text{coffee} \cup \text{pretzel}) = 0.45 + 0.30 - 0.20 = 0.55.

Use the complement for "neither". "Neither" is the complement of the union, so

P(\text{neither}) = 1 - 0.55 = 0.45.

Read off the parts if asked. Splitting the regions, "coffee only" is $0.45 - 0.20 = 0.25$ and "pretzel only" is $0.30 - 0.20 = 0.10$ ; these with the overlap $0.20$ and "neither" $0.45$ sum to $1$ , confirming the Venn diagram above.

Two marbles without replacement, and a conditional twist

A bag holds $2$ red and $3$ white marbles. Two are drawn without replacement. Find the probability of one of each colour, and then, given that at least one drawn marble is red, the probability that both are red.

Build the tree's path products. The first draw is red with probability $\frac{2}{5}$ or white with probability $\frac{3}{5}$ . Because there is no replacement, the second-stage probabilities run on the four remaining marbles. Multiplying along the branches gives

P(\text{RR}) = \tfrac{2}{5}\cdot\tfrac{1}{4} = \tfrac{1}{10}, \quad P(\text{RW}) = \tfrac{2}{5}\cdot\tfrac{3}{4} = \tfrac{3}{10}, \quad P(\text{WR}) = \tfrac{3}{5}\cdot\tfrac{2}{4} = \tfrac{3}{10}, \quad P(\text{WW}) = \tfrac{3}{5}\cdot\tfrac{2}{4} = \tfrac{3}{10}.

One of each colour. Two paths give one red and one white, so add them:

P(\text{one of each}) = P(\text{RW}) + P(\text{WR}) = \tfrac{3}{10} + \tfrac{3}{10} = \tfrac{3}{5}.

Condition on "at least one red". First, $P(\text{at least one red}) = 1 - P(\text{WW}) = 1 - \frac{3}{10} = \frac{7}{10}$ . The condition restricts the sample space to those outcomes, so divide the wanted path by this total:

P(\text{both red} \mid \text{at least one red}) = \frac{P(\text{RR})}{P(\text{at least one red})} = \frac{1/10}{7/10} = \frac{1}{7}.

This is the 2024 multiple-choice answer; the trap option $\frac{1}{4}$ comes from forgetting to divide by $\frac{7}{10}$ and just reading the second-stage red branch.

Conditional probability and independence from a two-way table

A Year 12 cohort of $200$ students was surveyed on how they travel to school and whether they were late on a given day. Of the $120$ who drive, $18$ were late; of the $80$ who catch public transport, $28$ were late. A student is chosen at random.

Set up the counts. Late students total $18 + 28 = 46$ . Probabilities are counts over $200$ , or over the relevant row for a conditional.

Find $P(\text{late} \mid \text{drives})$ . Conditioning on "drives" restricts attention to the $120$ drivers:

P(\text{late} \mid \text{drives}) = \frac{18}{120} = \frac{3}{20} = 0.15.

Reverse it to $P(\text{drives} \mid \text{late})$ . Now the world is the $46$ late students, of whom $18$ drove:

P(\text{drives} \mid \text{late}) = \frac{18}{46} = \frac{9}{23} \approx 0.391.

Test independence of "drives" and "late". Here $P(\text{drives}) = \frac{120}{200} = \frac{3}{5}$ , $P(\text{late}) = \frac{46}{200} = \frac{23}{100}$ , and $P(\text{drives} \cap \text{late}) = \frac{18}{200} = \frac{9}{100}$ . Compare:

P(\text{drives})\,P(\text{late}) = \tfrac{3}{5}\cdot\tfrac{23}{100} = \tfrac{69}{500} = 0.138, \qquad P(\text{drives} \cap \text{late}) = 0.09.

Since $0.09 \neq 0.138$ , the events are not independent: drivers are late less often than the cohort as a whole, so how a student travels does carry information about lateness.

Showing a probability and an "at least one", in the style of 2023 Q31

Four students each have probability $P(F)$ of being free on Friday and $P(S)$ of being free on Saturday, with $P(F) = \frac{3}{10}$ , $P(S \mid F) = \frac{1}{3}$ and $P(F \mid S) = \frac{1}{8}$ . Show that $P(S) = \frac{4}{5}$ , decide whether the two days are independent, and find the probability that at least one of the four students is not free on Saturday.

Write the intersection two ways. The probability of being free on both days can be built from either day first:

P(F \cap S) = P(F)\,P(S \mid F) = \tfrac{3}{10}\cdot\tfrac{1}{3} = \tfrac{1}{10}.

Solve for $P(S)$ . Using the other order, $P(F \cap S) = P(S)\,P(F \mid S)$ , so $\frac{1}{10} = P(S)\cdot\frac{1}{8}$ and

P(S) = \tfrac{1}{10}\times 8 = \tfrac{8}{10} = \tfrac{4}{5}.

Test independence. If Friday and Saturday were independent, $P(S \mid F)$ would equal $P(S)$ . But $P(S \mid F) = \frac{1}{3}$ while $P(S) = \frac{4}{5}$ , and these are different, so the two days are not independent.

At least one not free on Saturday. Across the four students availability is independent, so the complement "all four free on Saturday" is a single product:

P(\text{all four free}) = \left(\tfrac{4}{5}\right)^{4} = \tfrac{256}{625},

P(\text{at least one not free}) = 1 - \tfrac{256}{625} = \tfrac{369}{625} = 0.5904.

Reverse-conditional from a tree, in the style of 2025 Q19

Three players, Amara, Bala and Cassie, are in contention for one spot, with selection chances in the ratio $1 : 2 : 3$ . The team wins with probability $0.5$ if Amara plays, $0.4$ if Bala plays and $0.2$ if Cassie plays. Given that the team wins, find the probability that Amara was selected.

Turn the ratio into probabilities. The ratio $1 : 2 : 3$ sums to $6$ , so $P(A) = \frac{1}{6}$ , $P(B) = \frac{2}{6}$ , $P(C) = \frac{3}{6}$ . These are the first-stage branches; "wins" is the second stage.

Multiply along each winning branch. Each path to a win is selection times win-given-selection:

P(A \cap \text{win}) = \tfrac{1}{6}\cdot\tfrac{5}{10} = \tfrac{1}{12}, \quad P(B \cap \text{win}) = \tfrac{2}{6}\cdot\tfrac{4}{10} = \tfrac{2}{15}, \quad P(C \cap \text{win}) = \tfrac{3}{6}\cdot\tfrac{2}{10} = \tfrac{1}{10}.

Add for the total probability of a win. This is the denominator of the reverse conditional:

P(\text{win}) = \tfrac{1}{12} + \tfrac{2}{15} + \tfrac{1}{10} = \tfrac{19}{60}.

Divide the wanted path by the total. The reverse conditional asks for $P(A \mid \text{win})$ :

P(A \mid \text{win}) = \frac{P(A \cap \text{win})}{P(\text{win})} = \frac{1/12}{19/60} = \frac{5}{19} \approx 0.263.

So even though Amara is the least likely to be picked, she is the most likely winner-maker, and conditioning on a win lifts her probability from $\frac{1}{6} \approx 0.167$ up to $\frac{5}{19} \approx 0.263$ .

A false-positive medical test (why a positive can still be unlikely)

A condition affects $2\%$ of a population. A screening test returns positive for $95\%$ of people who have the condition (its sensitivity) and, separately, returns positive for $10\%$ of people who do not (its false-positive rate). A randomly chosen person tests positive. Find the probability they actually have the condition.

Lay out the tree. First stage: has the condition ( $0.02$ ) or not ( $0.98$ ). Second stage: tests positive, with probability $0.95$ down the "has" branch and $0.10$ down the "not" branch.

Probability of each positive path. Multiply along the branches:

P(\text{has} \cap +) = 0.02 \times 0.95 = 0.019, \qquad P(\text{not} \cap +) = 0.98 \times 0.10 = 0.098.

Total probability of a positive result. Add the two positive paths:

P(+) = 0.019 + 0.098 = 0.117.

Divide for the reverse conditional. The chance a positive result is a true positive is

P(\text{has} \mid +) = \frac{0.019}{0.117} = \frac{19}{117} \approx 0.162.

Only about $16\%$ , because the condition is rare: the many false positives from the healthy $98\%$ swamp the few true positives. This counterintuitive result is the whole reason reverse-conditional questions are asked.

Edge cases worth knowing

Selecting "at the same time" equals "without replacement". Picking two marbles together is not a new rule; it is two sequential draws with no replacement, and the order-free events combine the relevant ordered paths.
Mutually exclusive is not the same as independent. If $A$ and $B$ cannot both occur then $P(A \cap B) = 0$ , so (unless one has probability $0$ ) they are actually dependent: knowing $A$ happened forces $B$ to be impossible.
Conditioning can raise or lower a probability. $P(A \mid B)$ may be larger or smaller than $P(A)$ ; only when they are equal are the events independent.
The denominator in a conditional is the restricted world, not $1$ . Forgetting to divide by $P(B)$ is the single most common slip in "given that" questions.
A "show that" almost always wants both orderings. $P(A)\,P(B \mid A) = P(B)\,P(A \mid B)$ lets you solve for whichever single probability is unknown.

Exam technique

Decide first whether trials are independent or without replacement, because that fixes every later number. For "at least one", write $1 - P(\text{none})$ immediately rather than adding cases. On a tree, label the branches, check each fan of branches sums to $1$ , multiply along a path, and add path products for a combined event; quote the path products as exact fractions so a method mark survives an arithmetic slip. For "given that" questions, write the definition $P(A \mid B) = \frac{P(A \cap B)}{P(B)}$ and identify $B$ as the restricted sample space before substituting. For a reverse "given the effect, find the cause", state the denominator as the total probability of the effect (the sum of every branch reaching it). When asked to test independence, show the comparison $P(A \cap B)$ versus $P(A)\,P(B)$ explicitly and write a one-line conclusion; an unjustified "yes/no" earns nothing.

Common traps

Adding cases for "at least one": Use the complement $1 - P(\text{none})$ ; listing every success case is slow and usually drops a term.
Forgetting to subtract the overlap: $P(A \cup B) = P(A) + P(B) - P(A \cap B)$ ; leaving out $P(A \cap B)$ double-counts and can even give a probability above $1$ .
Reusing first-draw probabilities after no replacement: Once a marble is removed the totals change, so the second-stage branches must use the reduced counts.

Dividing a conditional by $1$ instead of $P(B)$ . $P(A \mid B)$ restricts the world to $B$ , so the denominator is $P(B)$ , not the whole sample space.

Confusing $P(A \mid B)$ with $P(B \mid A)$ . They are different in general (the test and the disease example differ a lot); reverse conditionals must be rebuilt, not flipped.

Calling mutually exclusive events independent. If they cannot co-occur, $P(A \cap B) = 0 \neq P(A)\,P(B)$ in general, so they are dependent.

Exam-style practice questions

Practice questions written in the style of NESA exam questions on this dot point, with worked answer explainers. The year tag is the paper they imitate, not the source.

2024 HSC Q91 marksA bag contains 2 red and 3 white marbles. Two marbles are selected at the same time. Given that one of the marbles selected is red, what is the probability that the other marble is also red?

Show worked answer →

Selecting two at once is the same as drawing without replacement. The tree gives $P(\text{RR}) = \frac{2}{5}\cdot\frac{1}{4} = \frac{1}{10}$ and $P(\text{WW}) = \frac{3}{5}\cdot\frac{2}{4} = \frac{3}{10}$ , so $P(\text{at least one red}) = 1 - \frac{3}{10} = \frac{7}{10}$ .

"Given one is red" restricts the sample space to the at-least-one-red outcomes, so $P(\text{both red} \mid \text{at least one red}) = \dfrac{P(\text{RR})}{P(\text{at least one red})} = \dfrac{1/10}{7/10} = \dfrac{1}{7}$ .

The answer is $\frac{1}{7}$ . Markers reward recognising that the condition shrinks the sample space and dividing by the conditioning probability, not by $1$ .

2023 HSC Q315 marksFour students each have probability

P(F)

of being available on Friday and

P(S)

of being available on Saturday, with

P(F) = \frac{3}{10}

P(S\mid F) = \frac{1}{3}

and

P(F\mid S) = \frac{1}{8}

. (a) Is availability on Friday independent of availability on Saturday? (b) Show that

P(S) = \frac{4}{5}

. (c) Find the probability that at least one of the four students is NOT available on Saturday.

Show worked answer →

(a) $P(F \cap S) = P(F)\,P(S\mid F) = \frac{3}{10}\cdot\frac{1}{3} = \frac{1}{10}$ . If the events were independent then $P(S\mid F)$ would equal $P(S)$ ; since $P(S\mid F) = \frac13 \neq \frac45 = P(S)$ (from part b), they are not independent.

(b) The same intersection can be written two ways: $P(F \cap S) = P(S)\,P(F\mid S)$ , so $\frac{1}{10} = P(S)\cdot\frac{1}{8}$ , giving $P(S) = \frac{8}{10} = \frac{4}{5}$ .

(c) "At least one not available" is the complement of "all four available". Availability is independent across students, so $P(\text{all four available}) = \left(\frac45\right)^4 = \frac{256}{625}$ and $P(\text{at least one not available}) = 1 - \frac{256}{625} = \frac{369}{625}$ .

Markers reward writing the intersection both ways to unlock $P(S)$ , and using $1 - P(\text{none})$ for the "at least one" part rather than adding four cases.

Practice questions

Original practice questions graded from foundation to exam level, each with a full worked solution. Try them before revealing the solution.

foundation2 marksA fair six-sided die is rolled and a fair four-sector spinner is spun. List how many outcomes are in the sample space, then find the probability of rolling a

6

and spinning sector

A

Show worked solution →

Count the sample space. The two actions are independent, so the total number of equally likely outcomes is the product

6 \times 4 = 24.

Find the single favourable outcome. Exactly one pair, (roll a $6$ , spin $A$ ), meets the condition, so

P(6 \text{ and } A) = \frac{1}{24}.

Equivalently, $P(6)\,P(A) = \tfrac{1}{6}\cdot\tfrac{1}{4} = \tfrac{1}{24}$ , confirming the count.

Marker's note: one mark for the sample space of $24$ outcomes, one for $P = \frac{1}{24}$ . Either the count method or the multiplication of independent probabilities earns full marks.

foundation3 marksA spinner lands on green with probability

0.7

each spin, independently of other spins. It is spun

3

times. Find the probability of at least one green.

Show worked solution →

Choose the complement. "At least one green" covers one, two or three greens, which is awkward to add, so use its complement "no greens at all":

P(\text{at least one green}) = 1 - P(\text{no green}).

Compute "no green". Each spin is not green with probability $1 - 0.7 = 0.3$ , and the spins are independent, so

P(\text{no green}) = 0.3^{3} = 0.027.

Subtract.

P(\text{at least one green}) = 1 - 0.027 = 0.973.

Marker's note: one mark for writing $1 - P(\text{none})$ , one for $0.3^3 = 0.027$ , one for $0.973$ . Adding the three separate green cases is the slow route that usually drops a term.

core3 marksA box holds

4

blue and

6

yellow pens. Two are drawn without replacement. Find the probability that both are blue, and the probability that the two pens are different colours.

Show worked solution →

First draw. The chance the first pen is blue is $\tfrac{4}{10} = \tfrac{2}{5}$ .

Second draw, without replacement. After one blue is removed, $3$ blue remain among $9$ pens, so

P(\text{both blue}) = \frac{4}{10}\cdot\frac{3}{9} = \frac{12}{90} = \frac{2}{15}.

Different colours. The two mixed paths are blue-then-yellow and yellow-then-blue:

P(\text{different}) = \frac{4}{10}\cdot\frac{6}{9} + \frac{6}{10}\cdot\frac{4}{9} = \frac{24}{90} + \frac{24}{90} = \frac{48}{90} = \frac{8}{15}.

Marker's note: one mark for the without-replacement second-stage fractions, one for $P(\text{both blue}) = \frac{2}{15}$ , one for adding both mixed paths to get $\frac{8}{15}$ . Reusing $\frac{4}{10}$ on the second draw is the classic replacement slip.

core4 marksA survey of

250

commuters records device use and lateness one morning. Of the

150

who used a phone on the trip,

45

arrived late; of the

100

who did not,

20

arrived late. Present these as a two-way table in words, then find

P(\text{late} \mid \text{used phone})

and test whether lateness is independent of phone use.

Show worked solution →

Read the table. The four cells are: used phone and late $45$ , used phone and on time $150 - 45 = 105$ , no phone and late $20$ , no phone and on time $100 - 20 = 80$ . Late commuters total $45 + 20 = 65$ ; the grand total is $250$ .

Conditional probability. Restrict to the $150$ phone users:

P(\text{late} \mid \text{phone}) = \frac{45}{150} = \frac{3}{10} = 0.3.

Test independence. Compute the marginals and the intersection over $250$ :

P(\text{phone}) = \frac{150}{250} = \frac{3}{5}, \qquad P(\text{late}) = \frac{65}{250} = \frac{13}{50},

P(\text{phone})\,P(\text{late}) = \frac{3}{5}\cdot\frac{13}{50} = \frac{39}{250} = 0.156, \qquad P(\text{phone} \cap \text{late}) = \frac{45}{250} = 0.18.

Since $0.18 \neq 0.156$ , the events are not independent: phone users are late more often than the cohort as a whole.

Marker's note: one mark for the completed cell counts, one for $P(\text{late} \mid \text{phone}) = 0.3$ , one for computing both $P(\text{phone})P(\text{late})$ and $P(\text{phone} \cap \text{late})$ , one for the comparison and conclusion. An unjustified "not independent" earns nothing.

core3 marksA factory makes bolts on two machines. Machine A makes

60\%

of the bolts, of which

3\%

are defective; Machine B makes the rest, of which

5\%

are defective. A bolt is chosen at random and found to be defective. Find the probability it came from Machine A.

Show worked solution →

Multiply along each defective path.

P(A \cap D) = 0.60 \times 0.03 = 0.018, \qquad P(B \cap D) = 0.40 \times 0.05 = 0.020.

Total probability of a defective bolt. Add the two paths that reach a defect:

P(D) = 0.018 + 0.020 = 0.038.

Divide for the reverse conditional.

P(A \mid D) = \frac{P(A \cap D)}{P(D)} = \frac{0.018}{0.038} = \frac{9}{19} \approx 0.474.

Marker's note: one mark for the two path products, one for $P(D) = 0.038$ , one for $\frac{9}{19} \approx 0.474$ . Answering $0.018$ (the numerator alone) forgets to divide by the total probability of the effect.

exam5 marksA biased coin shows heads with probability

p

on each toss, independently. Let

q = 1 - p

. (a) Show that the probability of getting at least one head in

n

tosses is

1 - q^{n}

. (b) The coin has

p = \frac{1}{4}

. Find the least number of tosses so that the probability of at least one head exceeds

0.9

. (c) For this coin, given that at least one head occurred in

3

tosses, find the probability that exactly one head occurred.

Show worked solution →

Part (a): show the formula. On each toss the coin is not a head with probability $q = 1 - p$ . The tosses are independent, so the probability of no heads across all $n$ tosses is the product of $n$ equal factors,

P(\text{no heads}) = q^{n}.

"At least one head" is the complement of "no heads", hence

P(\text{at least one head}) = 1 - P(\text{no heads}) = 1 - q^{n}. \qquad \blacksquare

Part (b): solve the inequality. With $p = \tfrac{1}{4}$ , $q = \tfrac{3}{4}$ , and we need

1 - \left(\tfrac{3}{4}\right)^{n} > 0.9 \;\Longleftrightarrow\; \left(\tfrac{3}{4}\right)^{n} < 0.1.

Take logarithms (dividing by $\ln\tfrac{3}{4} < 0$ reverses the inequality):

n \ln\tfrac{3}{4} < \ln 0.1 \;\Longrightarrow\; n > \frac{\ln 0.1}{\ln 0.75} = \frac{-2.302585}{-0.287682} = 8.004.

Since $n$ must be a whole number, the least value is $n = 9$ tosses. (Check: $(0.75)^{8} = 0.1001 > 0.1$ , but $(0.75)^{9} = 0.0751 < 0.1$ .)

Part (c): a conditional after "at least one". Restrict the world to "at least one head in $3$ tosses", which has probability

P(\text{at least one}) = 1 - \left(\tfrac{3}{4}\right)^{3} = 1 - \tfrac{27}{64} = \tfrac{37}{64}.

"Exactly one head in $3$ tosses" can fall in any of the $3$ positions, each of probability $p\,q^{2} = \tfrac{1}{4}\cdot\left(\tfrac{3}{4}\right)^{2}$ :

P(\text{exactly one}) = 3 \cdot \tfrac{1}{4}\cdot\tfrac{9}{16} = \tfrac{27}{64}.

Both events sit inside "at least one head", so exactly one head implies at least one, and

P(\text{exactly one} \mid \text{at least one}) = \frac{27/64}{37/64} = \frac{27}{37} \approx 0.730.

Marker's note: one mark for the complement argument in (a); one for reducing (b) to $(0.75)^n < 0.1$ and one for $n = 9$ with the log or trial check; one for the restricted total $\frac{37}{64}$ in (c) and one for $\frac{27}{37}$ . Forgetting the factor of $3$ for the position of the single head, or dividing by $1$ instead of $\frac{37}{64}$ , are the mark-losing slips.

exam4 marksTwo archers, Priya and Wei, shoot at a target once each. Priya hits with probability

\frac{3}{5}

and Wei hits with probability

\frac{1}{2}

, independently. (a) Find the probability that exactly one of them hits the target. (b) Given that the target is hit at least once, find the probability that Priya was the only one to hit it.

Show worked solution →

Set up the independent probabilities. Priya hits with $\tfrac{3}{5}$ (misses with $\tfrac{2}{5}$ ); Wei hits with $\tfrac{1}{2}$ (misses with $\tfrac{1}{2}$ ).

Part (a): exactly one hit. Two disjoint paths give exactly one hit, Priya-hit-Wei-miss and Priya-miss-Wei-hit:

P(\text{exactly one}) = \tfrac{3}{5}\cdot\tfrac{1}{2} + \tfrac{2}{5}\cdot\tfrac{1}{2} = \tfrac{3}{10} + \tfrac{2}{10} = \tfrac{5}{10} = \tfrac{1}{2}.

Part (b): condition on at least one hit. The complement of "at least one hit" is "both miss":

P(\text{both miss}) = \tfrac{2}{5}\cdot\tfrac{1}{2} = \tfrac{2}{10} = \tfrac{1}{5}, \qquad P(\text{at least one hit}) = 1 - \tfrac{1}{5} = \tfrac{4}{5}.

"Priya the only hitter" is the single path Priya-hit-Wei-miss, with probability $\tfrac{3}{5}\cdot\tfrac{1}{2} = \tfrac{3}{10}$ . Dividing by the restricted world,

P(\text{Priya only} \mid \text{at least one hit}) = \frac{3/10}{4/5} = \frac{3}{10}\cdot\frac{5}{4} = \frac{15}{40} = \frac{3}{8}.

Marker's note: one mark for each mixed path summing to $\frac{1}{2}$ in (a); one for $P(\text{at least one}) = \frac{4}{5}$ via the complement, one for dividing $\frac{3}{10}$ by $\frac{4}{5}$ to get $\frac{3}{8}$ in (b). Dividing by $1$ instead of $\frac{4}{5}$ is the usual conditional-probability error.