What is the difference between independent and dependent events in Further Probability?

In Further Probability, independent events are those whose outcomes do not affect each other. For example, flipping a coin and rolling a die are independent events. Dependent events, on the other hand, are those where the outcome of one event affects the outcome of another. An example would be drawing two cards from a deck without replacement; the outcome of the first draw affects the probabilities of the second draw. Understanding these concepts is crucial for solving complex probability problems in the IB DP Mathematics curriculum.

How do you calculate conditional probability in Further Probability?

Conditional probability is the probability of an event occurring given that another event has already occurred. It is calculated using the formula P(A|B) = P(A and B) / P(B), where P(A|B) is the probability of event A occurring given that B has occurred. This concept is essential in Further Probability as it helps in analyzing situations where events are interdependent, a common scenario in IB DP Mathematics problems.

What is Bayes' Theorem and how is it applied in Further Probability?

Bayes' Theorem is a fundamental concept in Further Probability that allows us to update the probability of a hypothesis based on new evidence. The theorem is expressed as P(A|B) = [P(B|A) * P(A)] / P(B). It is particularly useful in situations where we need to revise probabilities with additional information, a skill often tested in the IB DP Mathematics curriculum.

How do you use probability distributions in Further Probability?

Probability distributions describe how probabilities are distributed over the values of a random variable. In Further Probability, common distributions include the binomial, normal, and Poisson distributions. Each distribution has its own set of parameters and applications, such as using the binomial distribution for binary outcomes or the normal distribution for continuous data. Mastery of these distributions is essential for solving complex probability problems in IB DP Mathematics.

What role does the Law of Total Probability play in Further Probability?

The Law of Total Probability is used to calculate the probability of an event by considering all possible ways that the event can occur. It is expressed as P(A) = Σ P(A|B_i) * P(B_i), where B_i represents all possible scenarios. This law is crucial in Further Probability for breaking down complex problems into simpler parts, a technique frequently employed in IB DP Mathematics to tackle challenging probability questions.

Why is "Computing $P(A \cap B)$ as $P(A) \cdot P(B)$ when events are NOT independent." a common mistake in Further Probability?

Why it happens: Defaulting to independence. How to avoid it: Use P(A B) = P(A) P(B | A) in general. The independence form is a SPECIAL CASE.

Why is "Computing a probability without naming the distribution." a common mistake in Further Probability?

Why it happens: Treating Paper 2 as 'plug-and-chug'. How to avoid it: Start every distribution problem with 'Let X B(n, p)' or 'Let X N(, ^2)'. AO3 marks reward this.

Why is "Stating an expected value without interpreting in context." a common mistake in Further Probability?

Why it happens: Treating the calculation as the whole answer. How to avoid it: Always finish with 'so the expected ... is ... (units)'. State whether the gain is positive or negative and what action is recommended.

ProbabilityIB Maths AA SL

Further Probability

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Further Probability — IB Maths AA SL: longer multi-stage problems, conditional probability and modelling with distributions

This note brings together the AA SL probability toolkit for the longer Paper-2 questions: chained conditional problems, distribution modelling, expectation in real contexts, and the most-tested decision-making scenarios.

What you’ll learn

Mapped to the IB DP Maths AA SL subject guide (2021 onwards (applies to 2026 exams)).

AO2 — Solve longer probability problems with chained conditionals.
AO2 — Apply the law of total probability and the partition formula.
AO2 — Determine expected gain / loss in financial / decision contexts.
AO3 — Communicate which distribution is being used and why.

Multiplication rule and the law of total probability

Two formulae handle most multi-stage problems.

Multiplication rule. Rearranging the conditional definition: $P(A \cap B) = P(A) \cdot P(B | A) = P(B) \cdot P(A | B).$

This is how to compute the probability of two events both happening when their occurrence is sequential or dependent.

Law of total probability. If $A$ and $A'$ partition the sample space: $P(B) = P(B | A) P(A) + P(B | A') P(A').$

For a finer partition $\{A_1, A_2, \ldots, A_k\}$ : $P(B) = \sum_{i = 1}^{k} P(B | A_i) P(A_i).$

Worked example. A factory has two machines: Machine 1 makes 60% of items with 2% defective; Machine 2 makes 40% with 5% defective. Find $P(\text{item is defective})$ .

Partition by machine. $P(D) = P(D | M_1) P(M_1) + P(D | M_2) P(M_2) = 0.02 \cdot 0.6 + 0.05 \cdot 0.4 = 0.012 + 0.020 = 0.032.$

Worked example (Bayes-style reverse). A defective item is selected. What's the probability it came from Machine 1?

$P(M_1 | D) = \dfrac{P(D \cap M_1)}{P(D)} = \dfrac{0.012}{0.032} = 0.375.$

(Although AA SL doesn't formally name Bayes' theorem, the calculation is exactly that.)

$P(A \cap B) = P(A) P(B | A)$ — works for dependent events.
Total probability: $P(B) = \sum P(B | A_i) P(A_i)$ .
'Reverse' conditionals via the conditional definition.

Modelling: when binomial, when normal?

Recognise the distribution from the language.

Binomial cues:

'How many of the $n$ defaults / successes / heads / customers ...'
Trials are SEPARATE and SAME probability.
Examples: number of correct answers in a multi-choice quiz, number of defective items in a sample.

Normal cues:

'Measurements of weight / height / time / amount that vary continuously.'
Often given $\mu$ and $\sigma$ directly.
Examples: bag weights, exam marks, daily sales.

Worked example (binomial in context). Past data shows that 12% of products fail a quality test. A box of 50 is sampled. Find: (a) Expected number of failures. (b) $P(\text{at most 8 fail})$ .

$X \sim B(50, 0.12)$ .

(a) $E(X) = np = 50 \cdot 0.12 = 6$ .

(b) $P(X \le 8) =$ binomCdf(50, 0.12, 8) $\approx 0.880$ .

Worked example (normal in context). A factory produces bags of sugar with mass $N(1000, 5^2)$ grams. The bag is sold as 1 kg. (a) Probability a bag weighs less than 990 g. (b) The lightest 1% are recycled — find the cut-off mass.

$X \sim N(1000, 25)$ , so $\mu = 1000$ , $\sigma = 5$ .

(a) normCdf $(-10^{99}, 990, 1000, 5) \approx 0.0228$ .

(b) invNorm $(0.01, 1000, 5) \approx 988.4$ g.

Distribution choice is an AO3 communication mark.
Binomial: discrete, fixed $n$ , constant $p$ .
Normal: continuous, $\mu$ and $\sigma$ given.
Always state the distribution + parameters.

Expected value in decisions

$E(X)$ in money, time, or any unit.

Expected value is the long-run average — useful for evaluating decisions like insurance, lottery purchases, and game strategies.

Worked example. A lottery costs $3 to enter. Possible prizes: $100 with probability 0.01, $10 with probability 0.05, nothing with probability 0.94. Should you play?

Let $X$ = net gain (prize minus cost): $E(X) = (100 - 3)(0.01) + (10 - 3)(0.05) + (-3)(0.94) = 0.97 + 0.35 - 2.82 = -1.50.$

Expected loss $1.50 per ticket — you should NOT play if you're risk-neutral.

Worked example (insurance). A 60-year-old has a 0.05 probability of needing $20,000 in health care this year, 0.20 probability of needing $1000, and 0.75 probability of needing nothing. What annual premium would be 'fair' (break-even for the insurer)?

$E(\text{cost}) = 20\,000 \cdot 0.05 + 1000 \cdot 0.20 + 0 \cdot 0.75 = 1000 + 200 = \$ 1200$.

A fair premium is $1200 — anything above this gives the insurer expected profit.

Pattern. Set up the random variable. Compute $E(X) = \sum x_i p_i$ . Interpret in context.

Examiner reports praise students who EXPLAIN the answer in context — 'expected loss of $1.50 per game' beats just '−1.50'.

$E(X) = \sum x_i p_i$ for a discrete r.v.
Used in decision-making contexts.
Always interpret in CONTEXT with units.

How it’s examined

Paper 2: 8–12-mark multi-part questions chaining conditional probability with a named distribution. Examiner reports stress AO3 marks for stating the distribution and AO3 for interpretation in context.

Sources: IB Diploma Programme Mathematics: Analysis and Approaches subject guide (IBO, official). Last reviewed 2026-05-25.

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Step-by-step worked examples — Further Probability

Step-by-step solutions to past-paper-style questions on further probability, written exactly the way a tutor would explain them at the board.

1Law of total probability
Building confidence• partition
Question
Bag A contains 4 red and 6 blue marbles; Bag B contains 7 red and 3 blue. A bag is chosen at random and one marble drawn. Find $P(\text{red})$ .
Step-by-step solution
1. Step 1
  $P(\text{red} | A) = 4/10$ , $P(\text{red} | B) = 7/10$ , $P(A) = P(B) = 1/2$ .
2. Step 2
  Total: $P(R) = (4/10)(1/2) + (7/10)(1/2) = 11/20$ .
Answer
$P(R) = 11/20$ .
2Reverse conditional (Bayes-style)
Stretch• conditional
Question
Same setup: a red was drawn. Find the probability it came from Bag A.
Step-by-step solution
1. Step 1
  $P(A | R) = \dfrac{P(R \cap A)}{P(R)} = \dfrac{(4/10)(1/2)}{11/20}$ .
2. Step 2
  $= \dfrac{2/10}{11/20} = \dfrac{2/10 \cdot 20}{11} = \dfrac{4}{11}$ .
Answer
$P(A | R) = \dfrac{4}{11}$ .
3Binomial in context
Building confidence• binomial
Question
A multiple-choice quiz has 20 questions, each with 4 options. A student guesses every question. Find the probability of getting at least 8 correct.
Step-by-step solution
1. Step 1
  $X \sim B(20, 0.25)$ . Want $P(X \ge 8)$ .
2. Step 2
  $P(X \ge 8) = 1 - P(X \le 7) = 1 -$ binomCdf(20, 0.25, 7) $\approx 1 - 0.898 = 0.102$ .
Answer
$\approx 0.102$ .
4Expected value — decision
Stretch• expected value
Question
A game costs $5 to play. With probability 0.1 you win $30, with probability 0.3 you win $5, otherwise you win nothing. Should you play?
Step-by-step solution
1. Step 1
  Let $X$ = net gain. Outcomes: $+25$ ( $p = 0.1$ ), $0$ ( $p = 0.3$ ), $-5$ ( $p = 0.6$ ).
2. Step 2
  $E(X) = 25(0.1) + 0(0.3) + (-5)(0.6) = 2.5 - 3 = -0.5$ .
3. Step 3
  Expected loss $0.50 per game — do NOT play if risk-neutral.
Answer
Expected loss $0.50 per game; do not play.

Key Definitions and Keywords — Further Probability

Definitions to memorise and the exact keywords mark schemes credit for further probability answers — sharpened from recent examiner reports for the 2026 IB DP Maths AA SL sitting.

Multiplication rule
Examiner keyword
$P(A \cap B) = P(A) \cdot P(B | A)$ .
Law of total probability
Examiner keyword
For a partition $\{A_1, \ldots, A_k\}$ : $P(B) = \sum_i P(B | A_i) P(A_i)$ .
Partition
A set of pairwise mutually exclusive events whose union is the whole sample space.

Common Mistakes and Misconceptions — Further Probability

The traps other students keep falling into on further probability questions — taken from recent IB DP Maths AA SL examiner reports and mark schemes — and how to avoid them.

✕Computing $P(A \cap B)$ as $P(A) \cdot P(B)$ when events are NOT independent.
Why it happens
Defaulting to independence.
How to avoid it
Use $P(A \cap B) = P(A) P(B | A)$ in general. The independence form is a SPECIAL CASE.
✕Computing a probability without naming the distribution.
Why it happens
Treating Paper 2 as 'plug-and-chug'.
How to avoid it
Start every distribution problem with 'Let $X \sim B(n, p)$ ' or 'Let $X \sim N(\mu, \sigma^2)$ '. AO3 marks reward this.
✕Stating an expected value without interpreting in context.
Why it happens
Treating the calculation as the whole answer.
How to avoid it
Always finish with 'so the expected ... is ... (units)'. State whether the gain is positive or negative and what action is recommended.

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