Probability: Venn Diagrams, Tree Diagrams, Conditional Probability and Expected Outcomes
Probability measures how likely an event is on a scale from 0 to 1. At Higher-GCSE level the TMUA tests Venn and tree diagrams, the addition and multiplication rules, conditional probability and expected outcomes - all by exact hand reasoning with no calculator, so keep fractions exact and eliminate impossible options fast.
Part of the TMUA syllabus — revision for the Test of Mathematics for University Admission (TMUA), the UAT-UK maths test used by Cambridge, Oxford, Imperial, UCL, LSE, Warwick and Durham. No calculator; multiple choice.
Key points
- For equally likely outcomes, P(event) = favourable/total; every probability lies between 0 and 1, and a complete set of outcomes sums to 1.
- Complement rule: P(not A) = 1 - P(A). For 'at least one' questions it is almost always quicker to compute 1 - P(none).
- Tree diagrams: multiply probabilities ALONG branches (this is the AND / 'both happen' case) and add ACROSS separate paths (the OR case). Each pair of branches from a point sums to 1.
- 'Without replacement' changes both the numerator and denominator on the second draw; 'with replacement' leaves them unchanged (the draws are then independent).
- Addition rule: P(A or B) = P(A) + P(B) - P(A and B). Subtracting the overlap avoids double counting; mutually exclusive events have overlap 0.
- Conditional probability restricts the sample space: P(A | B) = P(A and B) / P(B). On a Venn diagram, just divide the overlap count by the count in set B.
- Independence test: events are independent exactly when P(A and B) = P(A) × P(B), equivalently when P(A | B) = P(A).
- Expected number of occurrences in n trials = n × P(event); a frequency tree splits one total headcount by successive proportions.
Diagram
Formulae
P(not A) = 1 - P(A) Finding a complement; especially 'at least one' questions via 1 - P(none).
P(A or B) = P(A) + P(B) - P(A and B) Combining two events with the addition rule; drop the last term only if they are mutually exclusive.
P(A and B) = P(A) × P(B) Independent events - multiply probabilities along the branches of a tree diagram.
P(A | B) = P(A and B) / P(B) Conditional probability, or testing independence by checking if P(A | B) equals P(A).
E(count) = n × P(event) Expected number of successes when a trial with fixed probability is repeated n times.
Definitions
- Conditional probability P(A | B)
- The probability that A happens given that B has already happened; it is found by narrowing the sample space to only the outcomes in B, so P(A | B) = P(A and B) / P(B).
- Mutually exclusive events
- Events that cannot both occur in the same trial, so P(A and B) = 0 and the addition rule simplifies to P(A or B) = P(A) + P(B).
- Independent events
- Events where one occurring does not change the probability of the other; formally P(A and B) = P(A) × P(B), which is why you multiply along independent tree branches.
- Frequency tree
- A tree in which the branches carry counts (frequencies) rather than probabilities, splitting a fixed total group into subgroups by each successive condition.
Worked examples
In a year group of 30 students, 18 study French, 14 study Spanish and 5 study neither language. Find the probability that a randomly chosen student studies both languages, and the probability that a student studies Spanish given that they study French.
- 1
Remove the 'neither' group to count those in at least one set:
n(F or S) = 30 - 5 = 25 - 2
Apply the addition rule for counts to find the overlap:
n(F and S) = 18 + 14 - 25 = 7 - 3
Probability of studying both languages:
P(both) = 7/30 - 4
Restrict the sample space to the 18 French students:
P(S | F) = 7/18
Answer: P(both) = 7/30 and P(Spanish | French) = 7/18.
A bag contains 3 red and 2 blue counters. Two counters are drawn at random without replacement. Find the probability that both counters are the same colour, and hence the expected number of same-colour results if this two-draw experiment is performed 40 times.
- 1
Multiply along the red-then-red branch:
P(RR) = 3/5 × 2/4 = 3/10 - 2
Multiply along the blue-then-blue branch:
P(BB) = 2/5 × 1/4 = 1/10 - 3
Add the two exclusive same-colour paths:
P(same) = 3/10 + 1/10 = 2/5 - 4
Multiply trials by the probability:
E = 40 × 2/5 = 16
Answer: P(same colour) = 2/5, so the expected number of same-colour results in 40 experiments is 16.
Common mistakes
- ×Forgetting to subtract the overlap in P(A or B), which double-counts every element that lies in both sets.
- ×Reusing the first-draw denominator on the second branch of a 'without replacement' tree instead of reducing the total by 1.
- ×Confusing P(A | B) with P(B | A) - the event after the bar names the total you divide by, so the two are usually different.
- ×Treating dependent events as independent and multiplying unchanged probabilities when the first outcome has altered the setup.
- ×Misreading a Venn diagram: taking the 'both' overlap as a whole circle, or ignoring the 'neither' region that sits outside all the circles.
No-calculator tips
- ✓Keep fractions exact and cancel before multiplying; 3/5 × 2/4 cancels the 2 to give 3/10 in one line rather than reaching for 0.6 × 0.5.
- ✓Use the complement for 'at least one' - computing 1 - P(none) is usually a single branch instead of adding several separate paths.
- ✓Put tree paths over a common denominator (often /20 or /10) so you can add them mentally rather than using decimals.
- ✓Sanity-check every option: a probability above 1 or a branch pair not summing to 1 is instantly wrong, letting you eliminate distractors.
- ✓For expected value, multiply the trial count straight by the fraction; the intended MCQ answer is frequently the clean whole number this produces.
Test yourself
Original practice questions, no calculator. Work each out before revealing the answer.
Q1.A factory uses two machines. Machine A makes 60% of the items and Machine B makes the remaining 40%. Of Machine A's output, 5% are defective; of Machine B's output, 10% are defective. An item is picked at random and found to be defective. What is the probability it was made by Machine A?
- A. 4/7
- B. 3/5
- C. 3/100
- D. 3/7
Show answer
Answer: D — 3/7
P(A and defective) = 0.6*0.05 = 0.03 and P(B and defective) = 0.4*0.10 = 0.04, so P(defective) = 0.07 and P(A | defective) = 0.03/0.07 = 3/7. The 3/5 trap ignores the defect evidence (just the prior), 4/7 uses the wrong branch (B), and 3/100 forgets to divide by P(defective).
Q2.In a survey, 60% of people like tea, 50% like coffee, and 40% like both. What is the probability that a randomly chosen person likes neither drink?
- A. 0.2
- B. 0.3
- C. 0.4
- D. 0.7
Show answer
Answer: B — 0.3
P(tea or coffee) = 0.6 + 0.5 - 0.4 = 0.7, so P(neither) = 1 - 0.7 = 0.3. The 0.2 trap wrongly assumes independence ((1-0.6)(1-0.5)), 0.7 forgets to take the complement, and 0.4 just reports P(both).
Q3.A fair six-sided die is rolled once. You win 2 pounds if it shows a 6, you lose 1 pound if it shows a 1, and nothing happens otherwise. What is your expected gain, in pounds?
- A. 1/2
- B. 1/3
- C. 1/6
- D. -1/6
Show answer
Answer: C — 1/6
E = (1/6)(2) + (1/6)(-1) + (4/6)(0) = 2/6 - 1/6 = 1/6. The 1/2 trap averages the two nonzero outcomes without weighting by 1/6, 1/3 counts only the win, and -1/6 slips the sign by subtracting the win from the loss.
Read this topic in the official UAT-UK TMUA content specification →
Keep going
- See how often each topic appears across past TMUA papers → — this one is about 1% of all questions.