Learn in your language
Skip to content
A Level · Cambridge (CIE) · 9709

Reverse a tree condition using joint and total probabilities

Combine alternative starting sources, then divide the relevant joint path by the total probability of the observed result.

Statistics 1 · Probability · 5.3
Exam scope and assessment notes

Probability & Statistics 1 (Paper 5, AS & A Level) · Mathematics 9709, 2026–2027, section 5.3. Define the sample space and state its probability assumptions before counting. Fair independent dice give equiprobable ordered pairs; their sums are not equiprobable. Random selection means each specified individual is equally likely under the stated sampling model. Without replacement, later branch probabilities change; with replacement and independent mixing, they remain the same. A′ is a complement, A ∩ B means both, A ∪ B means either or both, and P(A|B) means A given B with P(B) > 0. Mutually exclusive and independent are different properties. The explicit general addition formula is not required by this section; explain overlap through disjoint regions. Conditional probabilities may use P(A|B) = P(A ∩ B)/P(B). Numerical models describe the stipulated experiment, not automatically a real population. Written practice is self-checked; human subject-teacher review remains pending.

Before you begin

  • Multiply down probability trees and add disjoint paths.
  • Calculate conditional probabilities with a positive denominator.

Work on paper. Try each question before revealing help, and explain why your method works.

Quick readiness check

Which paths contribute to total red probability?

What you will learn

  • Calculate an observed-result probability across disjoint source paths.
  • Recover a source probability given the observed result using a conditional ratio.
Choose box A with probability 1/3 or B with 2/3. A red and blue branches are 4/5 and 1/5; B branches are 1/5 and 4/5. Joint leaves A-red, A-blue, B-red, B-blue have probabilities 4/15, 1/15, 2/15, 8/15. Total red is 6/15 = 2/5; given red, box A has probability (4/15)/(6/15) = 2/3.Weight the source, then conditionStart1/3A4/5A-R 4/151/5A-B 1/152/3B1/5B-R 2/154/5B-B 8/15Total red: 6/15 = 2/5P(A | red) = (4/15)/(6/15) = 2/3
Choose box A with probability 1/3 or B with 2/3. A red and blue branches are 4/5 and 1/5; B branches are 1/5 and 4/5. Joint leaves A-red, A-blue, B-red, B-blue have probabilities 4/15, 1/15, 2/15, 8/15. Total red is 6/15 = 2/5; given red, box A has probability (4/15)/(6/15) = 2/3.

Weight each starting source

Choose box A with probability 1/3 or box B with probability 2/3. A contains four red and one blue; B contains one red and four blue. After choosing a box, choose a ball uniformly from it. The joint paths A-and-red and B-and-red have probabilities 4/15 and 2/15, so total red probability is 6/15 = 2/5.

The average (4/5 + 1/5)/2 = 1/2 would wrongly give equal starting weights to the boxes. Starting-source probabilities belong in the tree. Its four joint leaves are 4/15, 1/15, 2/15 and 8/15, summing to one. A red result can arise from either source, so include both source paths in its denominator.

Condition on what was observed

Given a red ball, P(A|red) = P(A and red)/P(red) = (4/15)/(6/15) = 2/3. This differs from P(red|A) = 4/5 and from the prior box-A probability 1/3. Given blue instead, P(B|blue) = (8/15)/(9/15) = 8/9.

The method is a simple conditional ratio; no general symbolic theorem needs to be memorised to solve this numerical tree. In a stipulated item-production model, source weights and mark rates likewise determine the chance of a source given a mark. Such a probability describes the supplied model and should not be claimed as verified real manufacturing evidence.

Pause and explain

The box model has A-and-red 4/15 and total red 6/15. What is P(A|red)?

Put the idea to work

Worked example

Choose A with probability 1/3, B with 2/3. Red probabilities within A and B are 4/5 and 1/5. Given red, find P(A).

Show the worked solution
  1. Joint A-and-red is (1/3)(4/5) = 4/15.
  2. Total red is 4/15 + (2/3)(1/5) = 6/15.
  3. Divide joint by observed total: (4/15)/(6/15) = 2/3.

Answer P(A|red) = 2/3.

Try a topic test

Choose Easy, Normal or Difficult. Save your answers and open worked review after finishing.

Test this topic →

From guided practice to a new situation

Write your answer and reasoning first. Use a hint only if you are stuck. The model solution is for self-checking; your written work is not automatically marked.

1 · Guided

Choose box A with probability 1/3 or B with 2/3. A has red probability 4/5; B has red probability 1/5. Find total P(red).

Give me a hint

Multiply each source weight by its red branch, then add.

Compare my reasoning
  1. A-and-red contributes 4/15.
  2. B-and-red contributes 2/15.
  3. Their sum is 6/15 = 2/5.

2/5.

Look for these in your work

  • Use unequal starting box weights.
  • Include both disjoint red paths.
2 · Independent

For the same box model, given that the ball is blue, find P(box B | blue).

Give me a hint

Use B-and-blue as numerator and both blue paths in the denominator.

Compare my reasoning
  1. B-and-blue has probability (2/3)(4/5) = 8/15.
  2. A-and-blue has probability (1/3)(1/5) = 1/15, so total blue is 9/15.
  3. The conditional ratio is (8/15)/(9/15) = 8/9.

8/9.

Look for these in your work

  • Keep the condition direction.
  • Use the total observed-blue probability as denominator.
3 · Transfer

In a stipulated model, 60% of items come from machine A and 40% from B. A marks 10% of its items; B marks 25%. Given a marked item, find P(it came from B).

Give me a hint

The source-and-mark contributions are 0.60 × 0.10 and 0.40 × 0.25.

Compare my reasoning
  1. A-and-mark is 0.06; B-and-mark is 0.10.
  2. Total marked probability is 0.16.
  3. P(B|marked) = 0.10/0.16 = 5/8.

5/8 in the stipulated model.

Look for these in your work

  • Weight each source before conditioning.
  • Limit the conclusion to the supplied model.

Common mistakes

  • Weight alternative sources by their actual starting probabilities.
  • Reverse conditioning by a joint-over-observed-total ratio.
  • Do not confuse prior source, source-given-result and result-given-source probabilities.
Recall without your notes

What can you explain now?

Why does a red observation change box-A probability from 1/3 to 2/3?

Compare with the explanation

A contributes four of the six fifteenths of total red probability.

A red result is more likely from A under this model, so the conditional source weights differ.

After trying it yourself, choose your next review. This is your self-assessment.

Your review choice appears on Today. Sign in to sync it across devices.

Make it stick

Multiply source paths, add all paths for the observation, then divide the requested joint path by that total.

If you needed the solution, close it and solve the problem again from a blank page. A correct answer is stronger when you can explain the reason for each step.

Keep learning Mathematics

Course scope and assessment

312 sequenced lessons provide 936 guided, independent and application problems covering the 160 objectives in sections 1.1–1.8, 2.1–2.6, 3.1–3.9, 4.1–4.5, 5.1–5.5 and 6.1–6.5. The map lists all 38 sections across six alternative components; all eight Pure Mathematics 1 sections and all six Pure Mathematics 2 sections, from Paper 2 Algebra through Numerical solution of equations, currently have full teaching sequences. Paper 3 Algebra, section 3.1, links the six shared foundation lessons plus six lessons on partial fractions and rational binomial expansions; Paper 3 logarithms, trigonometry and numerical equations use the shared lessons, with the Paper 1 graph foundation linked into trigonometry. Paper 3 differentiation combines the shared methods with its inverse-tangent derivative lesson. Paper 3 Integration links four shared standard and identity lessons plus seven advanced lessons, covering all six section 3.5 requirements. Eight Vectors lessons cover all six section 3.7 requirements, with 2D and 3D line and solid applications. Nine Differential equations lessons cover all four section 3.8 requirements, including Paper 3 integration techniques, initial conditions, equilibrium solutions and conditional model interpretations. Eleven Complex numbers lessons cover all eight section 3.9 requirements, with seven Argand diagrams and explicit argument and boundary checks. All nine Pure Mathematics 3 sections now have teaching linked to every mapped requirement. Nine Forces and equilibrium lessons cover all seven section 4.1 requirements, with six force diagrams, signed incline-friction intervals, reaction-dependent pull and push limits and separate-body third-law checks. Nine Kinematics lessons cover the four official section 4.2 requirements, with the graph requirement mapped to three relationships. Six motion diagrams support directed quantities, graph slopes and signed areas, Paper 1 time-calculus methods, reversal distances, constant-acceleration stages and shared-clock particle meetings. Six Momentum lessons cover both section 4.3 requirements, including signed system totals, conservation assumptions, outgoing velocities, coalescence, unknown masses and successive impacts. Four diagrams show system momentum through the stipulated events, and supplied kinetic-energy checks distinguish conservation of momentum from changes in kinetic energy. Twelve Newton’s laws lessons cover all four section 4.4 requirements, with vertical support and release stages, rough horizontal and incline motion, inferred parameters, hanging and surface pulley systems, rigid tow-bar thrust and slack-string transitions. Seven force diagrams support the separate-particle balances. Nine Energy, work and power lessons cover all five section 4.5 requirements, with angle-dependent work, kinetic and reference potential energy, signed net-work balances, smooth curved slides, rough and driven slopes, ideal connected-system energy and instantaneous power on hills. Five diagrams distinguish the energy inputs, losses and valid positive-speed force model. All five Mechanics sections now have teaching and staged practice linked to their mapped requirements. Thirteen Representation of data lessons cover all five section 5.1 requirements, including choice and limitations of displays, ordered and back-to-back stems, raw summaries and box plots, unequal-width density histograms, cumulative-frequency interpolation, descriptive moment calculations, grouped estimates, coding, combining two datasets and correcting totals. Six accessible diagrams retain repeated observations and distinguish numerical spread, frequency area and estimated readings. Ten Permutations and combinations lessons cover both official section 5.2 requirements, with selection and ordered roles, repeated objects, adjacency blocks, non-adjacency gaps, leading-zero cases, labelled multi-row seating and mixed restrictions. Six accessible diagrams show outcome trees, ordered versus unordered choices, repeated copies, blocks, gaps and distinct row seats. Eleven Probability lessons cover all four section 5.3 requirements, with equiprobable elementary outcomes, selection-based probabilities, complements and disjoint regions, independence tests, restricted sample spaces, two-way tables, replacement and removal trees, reversed conditions and unknown parameters. Seven accessible diagrams show dice outcomes, random subsets, event regions, a conditioned grid, a count table and two probability trees. Twelve Discrete random variables lessons cover all three official section 5.4 requirements, with complete probability tables, moment and parameter checks, binomial assumptions, exact probabilities and moments, geometric first-success probabilities, tails, expectation and integer reliability thresholds. Six accessible diagrams distinguish outcome grouping, weighted moments, fixed trials, discrete probability bars, waiting conventions and integer tail deadlines. Twelve Normal distribution lessons cover all three official section 5.5 requirements, with continuous-area interpretation, MF19 cumulative tables, full standardisation, tail and interval regions, conditional probabilities, percentiles, one- and two-parameter inference, binomial suitability, continuity corrections and approximate integer thresholds. Seven accessible diagrams show model areas, table reading, aligned scales, intervals, quantiles, matched binomial and normal shapes and whole integer cells. All five Statistics 1 sections now have teaching and staged practice linked to their mapped requirements. Fourteen Poisson distribution lessons cover all five official section 6.1 requirements, with homogeneous event-count assumptions, exposure scaling, exact masses and tails, conditional count events, moments, parameter inference, probability duration targets, disjoint and overlapping windows, rare-count binomial approximations, matched normal approximations and continuity corrections. Nine accessible diagrams show event exposures, unit conversions, discrete bars, complete tail complements, population moments, shared windows, approximate count models and integer cells. Exact checks distinguish approximate and exact close-target decisions. Fourteen Linear combinations lessons cover the six listed results in official section 6.2, with affine moments, signed scales and constant cases, expectation without independence, independent squared variance contributions, fixed and random fees, independent copies versus reused values, affine normal events, independent normal sums and differences, weighted and conditional areas, coefficient inference, exact independent Poisson sums and overlapping exposure limits. Nine accessible diagrams distinguish scale changes, variance contributions, component costs, copy mechanisms, aligned normal scales, summed areas, Poisson count pairs and unsupported distribution claims. Fourteen Continuous random variables lessons cover both official section 6.3 requirements, with density validity and units, support-aware intervals, complements and conditional regions, first and second moments, percentiles by direct density areas, symmetry checks, parameter inference, infinite-support power and exponential integrals, divergent-moment limits and combined applications. Nine accessible diagrams distinguish density height from probability, normalisation, negative-density failures, central and conditional areas, mean versus median, percentile locations and partial infinite tails. Explicit cumulative distribution function knowledge is not required. Eighteen Sampling and estimation lessons cover all eight official section 6.4 requirements, with population and frame definitions, random-number selection, sampling-quality limits, independent sample-mean moments, normal means, CLT conditions, SE scaling, raw, summarised and coded unbiased estimates, known-variance and large-sample mean intervals, repeated-sampling interpretation, precision planning, proportion intervals, sparse-outcome limits and contextual applications. Ten accessible diagrams distinguish selection frames, number rejection, repeated sample means, population versus mean density, CLT models, variance divisors, interval targets, illustrative coverage, integer size checks and proportion margins. Eighteen Hypothesis tests lessons cover all five official section 6.5 requirements, with pre-specified alternatives, inclusive count tails, explicit equal-tail conventions, integer critical regions and actual significance, exposure-aware Poisson tests, continuity-corrected count approximations, normal and large-sample mean tests, confidence-interval connections, error classifications, fixed-region binomial, Poisson and normal error probabilities and sample-size power comparisons. Nine accessible diagrams show the testing procedure, exact count regions, unequal Poisson tails, discrete significance jumps, corrected count boundaries, matching mean intervals and alternative error distributions. All shared lessons are counted once in the course total. All 38 sections now link teaching and practice. Eight earlier overview lessons remain available, but do not count as full section coverage. This is a content-availability measure, not full programme completion or exam readiness. Written practice is self-checked, not officially marked. Check the 2028–2030 syllabus separately if sitting an examination from 2028; this teaching map uses 2026–2027 requirements.

StudyForA · Learn with understanding

Original teaching content and links to official sources. Check your exam year and qualification before using a resource.