Probability basics (expected #, complementary events, probability of event): notes and practice questions
- **Expected Value ():** The mean of a random variable .
- Expected Number of Occurrences: For trials with success probability , expected occurrences are .
- **Complementary Events ():** The event where does not happen; exactly one of or must occur.
- Sum of Probabilities: Probabilities of all possible outcomes sum to 1.
- Independent Events: .
- Conditional Probability: Probability of occurring given has occurred: .
- Venn Diagrams:
- Intersection (): "A AND B" (overlapping section).
- Union (): "A OR B OR BOTH" (all parts of A and B).
- Complement (): "NOT A" (everything outside A).
- Mutually Exclusive: Events A and B do not overlap ().
- Tree Diagrams: Map sequential events; branch order depends on given conditional probabilities.
- Solving Venn Diagram Problems:
- Work from the centre outwards, setting unknown intersection to .
- Form an equation where all separate sections sum to the total population/probability.
- For from Venn: (intersection of A and B) / (total of B).
- Conditional Probability Denominator: For "given that" scenarios, the total denominator shrinks to only include the population of the given condition.
- Diagrams: Drawing Venn or Tree diagrams aids visualization and problem-solving.
- SL/HL Distinction: Fundamental probability principles are identical across both syllabi.
How it is examined
Early, cheap marks. The expected number of occurrences is the part that surprises students, because the answer is normally not a whole number and rounding it costs the mark. Keep the distinction between relative frequency and theoretical probability available, since a question can give experimental data and ask which is which.
and .
- The concepts of trial, outcome, equally likely outcomes, relative frequency, sample space () and event.
- The probability of an event as .
- The complementary events and (not ).
- The expected number of occurrences.
Linking questions
- Other contexts: actuarial studies and the link between probability of life spans and insurance premiums, government planning based on projected figures, Monte Carlo methods.
- Links to other subjects: theoretical genetics and Punnett squares (biology); the position of a particle (physics).
- Aim 8: the ethics of gambling.
- International-mindedness: the St Petersburg paradox; Chebyshev and Pavlovsky (Russian).
- TOK: to what extent are theoretical and experimental probabilities linked? What is the role of emotion in our perception of risk, for example in business, medicine and travel safety?
- Use of technology: computer simulations may be useful here.
Practice questions
37 questions · 30 medium · 7 hardQuestion 1
MediumPaper 1 · calculator11 marksIn a security system, two independent sensors, System A and System B, report a threat level for an incident. System A reports a level with probabilities , , . System B reports a level with probabilities , , , . The overall threat assessment, , for an incident is defined as the higher of the two reported threat levels (i.e., ).
Complete the following table to show the probability distribution of .
| 1 | 2 | 3 | 4 | |
|---|---|---|---|---|
Find the probability that an incident has an overall threat assessment of at least 3.
Given that the overall threat assessment is at least 3, find the probability that System A reported a level of 2.
Calculate the expected overall threat assessment, .
Consider all possible pairs of and their joint probabilities. For each pair, determine and sum the probabilities for each unique value of .
Recall how to sum probabilities from a discrete probability distribution for a given range of values.
Remember the formula for conditional probability: . Identify events A and B correctly.
The expected value of a discrete random variable is the sum of each possible outcome multiplied by its probability.
Question 2
HardPaper 1 · calculator12 marksTwo unbiased dice, each with faces numbered from 1 to 6 inclusive, are rolled. The numbers on the uppermost faces of the dice are noted.
Let the random variable be the sum of the numbers on the dice.
(a)(i) Find .
(a)(ii) Find .
(b) Complete the table to show the probability distribution of .
The probability that is shown.
| 2 | 3 | 4 | 5 | 6 | 7 | 8 | 9 | 10 | 11 | 12 | |
|---|---|---|---|---|---|---|---|---|---|---|---|
(c) Calculate .
(d) Given that the sum of the numbers on the dice is a prime number, find the probability that .
List all possible outcomes when rolling two dice and identify those where the sum is 7. Remember there are total possible outcomes.
Similar to part (a.i), identify the outcomes where the sum is 10.
Systematically list all possible sums and the number of ways each sum can be achieved. Remember that the sum of all probabilities must equal 1.
The expected value is calculated as . Use the probabilities from your completed table.
Recall the definition of conditional probability: . First, identify all prime sums and calculate the probability of getting a prime sum.
Question 3
MediumPaper 1 · calculator6 marksA company conducted performance reviews for 120 employees across two departments: Marketing and Engineering. The outcomes are summarized in the following table.
| Department | Excellent | Needs Improvement |
|---|---|---|
| Marketing | 28 | 32 |
| Engineering | 45 | 15 |
An employee is chosen at random from this group.
(a) Find the probability that the randomly chosen employee received an 'Excellent' rating.
(b) Given that the chosen employee received an 'Excellent' rating, find the probability that they work in the Marketing department.
(c) Two different employees are chosen at random from the original group. Find the probability that both employees work in the Marketing department.
To find the probability of an event, divide the number of favorable outcomes by the total number of possible outcomes. In this case, count the total number of 'Excellent' ratings and the total number of employees.
This is a conditional probability question. You are given that the employee received an 'Excellent' rating, so your sample space is reduced to only those employees with an 'Excellent' rating. Then, find how many of those work in Marketing.
This involves probability without replacement. Consider the probability of the first employee being from Marketing, and then the probability of the second employee also being from Marketing, given that the first one was already chosen.
Question 4
HardPaper 2 · calculator14 marksThe diameter of a certain type of industrial component is modelled by a normal distribution with a mean of mm and a standard deviation of mm.
Find the probability that a randomly selected component has a diameter less than mm.
Find the probability that a randomly selected component has a diameter greater than mm.
Assuming the diameters of components are independent, find the probability that two consecutive components both have a diameter greater than mm.
A component is classified as 'premium' if its diameter is between mm and mm. A batch of three components is considered 'high quality' if all three components in the batch are premium. Find the probability that a randomly selected batch is NOT high quality.
In a production run, batches of three components are produced. Find the probability that at least of these batches are high quality.
Find the probability that between and (exclusive of ) of these batches are high quality.
Given that at least batches are high quality, find the probability that less than batches are high quality.
Use the normal cumulative distribution function (CDF) on your GDC. Remember that is directly calculated by the CDF.
Use the normal cumulative distribution function (CDF) or the normal survival function (SF) on your GDC. Remember that .
For independent events and , the probability of both occurring is .
First, find the probability that a single component is 'premium'. Then, use this to find the probability that a batch of three is 'high quality'. Finally, calculate the complementary probability.
This scenario involves a fixed number of trials (batches), each with two possible outcomes (high quality or not), and the trials are independent. This suggests a binomial distribution. Remember .
This means finding , which is equivalent to .
This is a conditional probability problem: . Here, is 'less than batches are high quality' and is 'at least batches are high quality'. The intersection is 'between and (exclusive of ) batches are high quality'.
Question 5
MediumPaper 1 · calculator6 marksA bakery produced a batch of 200 loaves of bread, consisting of sourdough and rye. The sales outcomes for these loaves are shown in the following table.
| Sold | Unsold | |
|---|---|---|
| Sourdough | 45 | 30 |
| Rye | 60 | 65 |
(a) Find the probability that a randomly chosen loaf from this batch was sold by the bakery.
A loaf is chosen at random from this batch. It is found that this loaf was sold.
(b) Find the probability that the loaf was a sourdough loaf.
Two different loaves are chosen at random from the original batch of 200 loaves.
(c) Find the probability that both loaves were rye.
To find the probability of an event, divide the number of favourable outcomes by the total number of possible outcomes. First, determine the total number of loaves sold.
This is a conditional probability problem. You are given that the loaf was sold, so your sample space is reduced to only the sold loaves. Then, find the number of sourdough loaves among those sold.
This involves probability without replacement. Calculate the probability of the first loaf being rye, then consider how the total number of loaves and the number of rye loaves changes for the second selection.
Question 6
HardPaper 2 · calculator14 marks(a) The number of customer service emails received by 'TechSolutions' per hour follows a Poisson distribution with a mean of . Using this model, find the probability that TechSolutions receives exactly emails in a given hour.
(b) Over a period of consecutive hours, find the probability that TechSolutions receives:
(i) exactly emails.
(ii) emails during the first and third hour only (i.e., at least one email in the first hour, no emails in the second hour, and at least one email in the third hour).
(c) Over a working day of hours, find the probability that there are exactly hours during which TechSolutions receives no emails.
(d) TechSolutions expands its operations and opens a new department, 'SupportPlus', which also receives emails. The number of emails received by each 'SupportPlus' agent per hour follows a Poisson distribution with a mean of . Assuming these are independent of the main TechSolutions department and each other, determine the least number of SupportPlus agents required so that the total probability of receiving at least emails across all departments (main TechSolutions and the SupportPlus agents) in an hour is greater than .
Recall the probability mass function for a Poisson distribution: . Identify the mean and the specific number of events from the question.
When combining independent Poisson processes, their means add up. Calculate the new mean for the -hour period and then apply the Poisson probability mass function.
This involves combining probabilities of independent events. Calculate and for a single hour, then multiply these probabilities for the specific sequence of events.
This scenario can be modeled by a binomial distribution. First, find the probability of 'no emails' in a single hour using the Poisson distribution. This will be the 'success' probability for your binomial model.
Let be the number of SupportPlus agents. The total mean number of emails per hour will be the sum of the mean from TechSolutions and times the mean from each SupportPlus agent. You need to find the smallest integer such that . You can do this by iterating through values of or by using a GDC's table/graphing function.
Question 7
MediumPaper 1 · calculator7 marksA player participates in a dart game where they throw one dart at a target with three distinct regions: an inner ring, a middle ring, and an outer ring. The probability of hitting each region in any given throw is shown in the following table.
| Region | Probability |
|---|---|
| Inner Ring | 0.2 |
| Middle Ring | 0.3 |
| Outer Ring | 0.5 |
The score awarded for hitting each region is:
- Inner Ring: 2 points
- Middle Ring: 1 point
- Outer Ring: 0 points
The player throws two darts independently. Find the probability that the player achieves a total score of exactly 2 points from the two throws.
In a different version of the game, the player wins points if they hit the Inner Ring, wins 4 points if they hit the Middle Ring, and loses 8 points if they hit the Outer Ring.
Find the value of such that the game is fair.
Consider all the combinations of two throws that result in a total score of 2 points. Remember that the throws are independent.
A game is considered fair if the expected value of the points won or lost is zero. Set up an equation for the expected value and solve for .
Question 8
HardPaper 2 · calculator15 marksThe 'SoundWave' concert hall has a capacity of seats. Historical data shows that of ticket holders attend the concert. The management decides to sell tickets, hoping that no more than concertgoers will arrive.
The number of concertgoers who arrive is assumed to follow a binomial distribution with a probability of .
(a) Calculate the probability that more than concertgoers arrive for the performance.
(b) (i) Write down the expected number of concertgoers who will arrive if tickets are sold.
(ii) Find the maximum number of tickets that could be sold if the expected number of concertgoers who arrive must be less than or equal to .
Each ticket costs 200 in compensation to each concertgoer who cannot be seated.
(c) Find, to the nearest integer, the expected increase or decrease in the money made by the venue if they decide to sell tickets rather than .
For a binomial distribution , the probability of successes is given by . To find the probability that more than concertgoers arrive, consider the probabilities for concertgoers, or use the complement rule . You will need a GDC for this calculation.
The expected value of a binomial distribution is given by .
Let be the number of tickets sold. The expected number of attendees is . Set up an inequality with this expected value and the capacity, then solve for . Remember that the number of tickets must be an integer.
Calculate the total expected money made for each scenario (selling tickets and selling tickets). For the -ticket scenario, you need to consider the income from all tickets sold and subtract the expected compensation. The expected compensation is calculated by summing the products of the probability of a specific number of overbooked attendees and the corresponding compensation amount.
Question 9
MediumPaper 1 · calculator7 marksA game involves two stages. First, a player draws a marble from a bag containing 3 red marbles and 2 blue marbles. After drawing a marble, the player spins a fair three-sided spinner with sections labeled 1, 2, and 3.
If a red marble is drawn, the player's final score is the number shown on the spinner.
If a blue marble is drawn, the player's final score is two more than the number shown on the spinner.
Find the probability that a player's final score is 4.
Complete the following table, showing the probability distribution of the final score.
| Final score () | 1 | 2 | 3 | 4 | 5 |
|---|---|---|---|---|---|
| Probability P() |
Calculate the expected value of the player's final score.
Consider the specific marble color and spinner outcome that would lead to a final score of 4. Calculate the probability of each event and then combine them.
List all possible combinations of marble drawn and spinner outcome. For each final score, identify all combinations that lead to it and sum their probabilities. Remember that the sum of all probabilities must be 1.
The expected value E(X) is calculated by summing the product of each possible score and its corresponding probability: E(X) = .
Question 10
HardPaper 2 · calculator18 marks(a) The manager of "The Daily Grind" coffee shop suggested that the number of customers arriving at the shop during a 5-minute interval can be modelled by a Poisson distribution.
Suggest two observations that the manager may have made that led him to suggest this model.
(b) Now assume that the model is valid and that the mean number of customers arriving at the shop during a 5-minute interval is .
The manager observes customer arrivals during a 15-minute interval.
Calculate the probability that exactly 6 customers arrive during this 15-minute interval.
(c) Using the same model as in part (b), find the probability that fewer than 4 customers arrive during a 15-minute interval.
(d) Find the probability that in four consecutive 5-minute intervals, at least one customer arrives in each interval.
(e) Following a new marketing campaign, the manager wished to determine whether the mean number of customers arriving during a 5-minute interval had increased.
State the hypotheses for the test.
(f) Find the critical region for the test at the 5% significance level.
(g) Given that the mean number of customers per 5-minute interval has actually risen to , find the probability that the manager makes a Type II error.
Recall the key assumptions of a Poisson distribution regarding the nature of events.
First, adjust the mean () for the new time interval. Then, use the Poisson probability mass function .
Remember that 'fewer than 4' means , which is equivalent to . Use the Poisson cumulative distribution function.
First, calculate the probability of at least one customer arriving in a single 5-minute interval. Then, consider how to combine probabilities for independent events.
Formulate the null hypothesis () as no change, and the alternative hypothesis () reflecting the manager's suspicion of an increase. Use the notation for the mean of a Poisson distribution.
For a one-tailed test for an increase, you need to find a value such that the probability of observing or more customers (under the null hypothesis) is less than or equal to the significance level.
A Type II error occurs when you fail to reject the null hypothesis when it is false. This means observing a value outside the critical region, given the true mean is .
Question 11
MediumPaper 1 · calculator8 marksIn a survey of students at a local college, it was found that the probability a student studies Art (event ) is . The probability a student studies Biology (event ) is . The probability that a student studies either Art or Biology or both is .
(a) Calculate the probability that a randomly selected student studies Biology but not Art.
(b) Determine the probability that a randomly selected student studies Art or does not study Biology.
(c) Find the probability that a randomly selected student does not study both Art and Biology.
Start by finding the probability that a student studies both Art and Biology using the formula for the union of two events.
Consider the complement of event B, . Then use the formula for the union of and , or consider regions in a Venn diagram.
This involves the complement of the intersection of A and B.
Question 12
HardPaper 2 · calculator27 marks(a) A single data packet can be transferred between three servers, S1, S2, and S3, in a network. The possible direct transfers are:
- From S1, the packet can be sent to S2 or stay in S1.
- From S2, the packet can be sent to S1 or S3.
- From S3, the packet can be sent to S2 or stay in S3.
Write down the adjacency matrix for the directed graph representing these possible direct data transfers, ordering the servers S1, S2, S3.
(b) Find the total number of distinct data paths of length 5 from server S1 to server S2.
(c.i) Every possible sequence of 5 data transfers has the same probability of occurring. State this probability.
(c.ii) Use your answer to part (b) to find the probability that if a data packet was initially on server S1, it will be on server S2 after 5 transfers.
(d.i) A network administrator monitors the movement of two data packets. The possible combined states of the two packets (assuming they occupy distinct servers) are (packets on S1 and S2), (packets on S1 and S3), and (packets on S2 and S3).
The transitions between these states are modelled by the following transition matrix , where rows represent the current state and columns represent the next state, in the order :
State the probability that if the packets are currently in state , they will be in state after one transfer.
(d.ii) Using the transition matrix from part (d.i), state the probability that if the packets are currently in state , they will be in state after one transfer.
(d.iii) Using the transition matrix from part (d.i), state the probability that if the packets are currently in state , they will be in state after one transfer.
(e) Given that the two data packets are initially in state (on servers S1 and S2), find the probability that they will be in state (on servers S2 and S3) after 5 transfers.
(f) The data packets continue this pattern of transfers for a long period. Find the server that is occupied least and the proportion of the time it is free.
An adjacency matrix for a directed graph has if there is an edge from node to node , and otherwise. Remember to consider self-loops.
The number of walks of length between two nodes can be found by raising the adjacency matrix to the power of . The entry in the resulting matrix gives the number of walks from node to node .
Consider the number of choices for each individual transfer and how probabilities combine for a sequence of independent events.
The probability is the number of successful paths divided by the total number of possible paths of that length.
Identify the correct entry in the transition matrix based on the given order of states.
Identify the correct entry in the transition matrix based on the given order of states.
Identify the correct entry in the transition matrix based on the given order of states.
To find the probabilities after multiple steps, you need to raise the transition matrix to the power corresponding to the number of transfers. Then, select the entry that corresponds to the initial and final states.
For a long period, the system reaches a steady state. Calculate the steady-state probability vector for the combined states. Then, use these probabilities to determine the proportion of time each individual server is occupied.
Question 13
MediumPaper 1 · calculator8 marksIn a survey of students at a high school, data was collected on their subject choices. It was found that students were studying Mathematics, and students were studying Physics.
Every student surveyed was studying at least one of these two subjects.
(a) Determine how many students were studying both Mathematics and Physics.
(b) Find the probability that a randomly selected student studies Mathematics, but not Physics.
(c) Explain why the events "studying Mathematics" and "studying Physics" are not independent events.
Recall the formula for the union of two sets: .
First, find the number of students who study Mathematics only. Then divide by the total number of students.
For two events and to be independent, must equal . Calculate these values and compare them.
Question 14
HardPaper 2 · calculator15 marksTech Solutions Inc. is a customer support company. They are analyzing their call center efficiency. Over a long period, they collect data on the number of calls, , already in the queue when a new customer's call arrives. The probability distribution of is shown in the following table.
| Number of calls in queue, | 0 | 1 | 2 | 3 | |
|---|---|---|---|---|---|
| P() | 0.15 | 0.30 | 0.35 | 0.20 | 0 |
Find the probability that there are at least two calls in the queue when a new customer's call arrives.
Find E().
The time in seconds, , taken to resolve a single customer's issue can be modelled by the normal distribution .
The company's management estimates that the expected total time a new customer will wait before their issue is resolved can be found by calculating E() E().
Find the value of E() E().
The company considers a service time to be 'long' if it takes more than three minutes to resolve a single customer's issue.
Using the distribution of given above, find the probability that it takes more than three minutes to resolve a randomly selected customer's issue.
Find the probability it takes more than four minutes in total to resolve two randomly selected customers' issues. You may assume all service times are independent of all other service times.
The company assumes that when a new customer's call arrives, the person at the front of the queue has only just reached a support agent. They also assume that if there are three or more customers already in the queue, the new customer will definitely wait more than three minutes before being served.
Using these assumptions and the probabilities for given in the table above,
find the probability a customer just arriving at the call center will wait more than three minutes before being served.
The company has a policy to employ more staff if the probability that a customer has to wait more than three minutes before being served is greater than .
Hence state whether Tech Solutions Inc. will decide to employ more staff.
To find the probability of 'at least two calls', sum the probabilities for and .
The expected value E() is calculated as the sum of (each value of its corresponding probability ).
Recall that for a normal distribution , the expected value E() is simply . Then multiply this by E() found in part (a.ii).
Convert three minutes to seconds. Then use your GDC to find for the given normal distribution.
If and are independent normal random variables, then their sum is also a normal random variable. The mean of the sum is the sum of the means, and the variance of the sum is the sum of the variances.
Consider the different scenarios for the number of customers in the queue ().
If , the wait time is 0.
If , the wait time is .
If , the wait time is .
If , the wait time is assumed to be minutes.
Compare the probability calculated in part (e.i) with the threshold of .
Question 15
MediumPaper 2 · calculator12 marksA group of students were surveyed about their participation in three extracurricular clubs: Environmental Club (E), Music Club (M), and Theater Club (T). The probabilities of their participation are represented in the Venn diagram below.
| Region | Probability |
|---|---|
| P(E only) | 0.18 |
| P(M only) | 0.23 |
| P(T only) | 0.15 |
| P(E M only) | 0.12 |
| P(E T only) | 0 |
| P(M T only) | 0.05 |
| P(E M T) | 0 |
| P(Neither E, M, nor T) | 0.27 |
Justify that events M and T are not independent.
Explain why events E and T are mutually exclusive.
Determine whether events E and M are independent.
Determine whether events E' and M' are mutually exclusive.
Find P(T E').
For two events to be independent, the probability of their intersection must be equal to the product of their individual probabilities. Calculate both sides and compare.
Recall the definition of mutually exclusive events in terms of their intersection.
Calculate P(E), P(M), and P(E M) from the Venn diagram. Then check the condition for independence.
Events E' and M' are mutually exclusive if P(E' M') = 0. Recall that E' M' is equivalent to (E M)'.
P(T E') means the probability of a student being in the Theater Club but NOT in the Environmental Club. Identify the regions that satisfy this condition.
Question 16
MediumPaper 1 · calculator7 marksA quality control team at a manufacturing plant inspects newly produced widgets. The probability that a single widget has a detectable defect is . The inspections are independent events.
(a) Find an expression for the probability that at least one defect is detected in inspections.
(b) Hence, determine the least number of inspections required for the probability of detecting at least one defect to be greater than .
Consider the complementary event: what is the probability that no defects are detected in inspections?
Set up an inequality using your expression from part (a) and solve for . Remember to use logarithms and consider how the inequality sign changes when dividing by a negative number.
Question 17
MediumPaper 1 · calculator7 marksA set of numbered cards is created, where each card has a number that is a multiple of . The numbers start from and go up to , where is an even positive integer.
Calculate the total number of cards in the set, in terms of .
Find the probability, in terms of , that a card selected at random from the set shows a number that is divisible by .
The numbers on the cards form an arithmetic sequence. Identify the first term, common difference, and the last term. Use the formula for the -th term of an arithmetic sequence to find .
For a number to be divisible by both and , it must be a multiple of their least common multiple (LCM). Determine the LCM of and , then count how many such multiples exist within the given range.
Question 18
MediumPaper 1 · calculator6 marksA collector has a box containing 20 vintage stamps. Of these, 4 are considered rare editions.
Three stamps are randomly selected from the box.
(a) Find the probability that exactly one of the selected stamps is a rare edition.
(b) Find the probability that at least one of the selected stamps is a rare edition.
Consider using combinations to find the number of ways to select the rare stamp and the common stamps, and the total number of ways to select any three stamps.
Consider calculating the complementary event: the probability that none of the selected stamps are rare, and subtract this from 1.
Question 19
MediumPaper 2 · calculator10 marksClara is taking three independent online quizzes: Mathematics, Physics, and Chemistry. The probability that she passes the Mathematics quiz is . The probability that she passes the Physics quiz is . The probability that she passes the Chemistry quiz is .
(a) Find the probability that Clara passes all three quizzes.
(b) Find the probability that Clara passes exactly one of the three quizzes.
(c) Given that Clara fails the Mathematics quiz, find the probability that she passes the Physics and Chemistry quizzes.
(d) Find the probability that Clara passes at least one of the three quizzes.
Since the quizzes are independent, the probability of passing all three is the product of their individual passing probabilities.
Consider the three mutually exclusive cases where only one quiz is passed. Calculate the probability for each case and then sum them up.
Remember that the outcomes of the quizzes are independent. If an event is independent, the conditional probability simplifies.
It's often easier to calculate the probability of the complementary event (failing all quizzes) and subtract it from 1.
Question 20
MediumPaper 1 · calculator6 marksA museum curator is selecting items for a new exhibit. From a collection of artifacts, are classified as ancient, and the remaining are modern. The curator randomly selects artifacts from the collection.
(a) Find the probability that exactly one of the selected artifacts is ancient.
(b) Find the probability that at least one of the selected artifacts is ancient.
To find the probability of selecting exactly one ancient artifact, consider the number of ways to choose one ancient artifact and one modern artifact, and divide this by the total number of ways to choose two artifacts from the collection.
The event 'at least one ancient artifact' can be calculated by summing the probabilities of 'exactly one ancient' and 'exactly two ancient', or by subtracting the probability of 'no ancient artifacts' from 1.
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