Papy

Conditional Probability - Edexcel GCSE Higher Maths

Typically 3-5 marks, usually the closing part of a longer probability question · Spec 1MA1

A conditional probability is the probability of one thing once you already know that something else has happened, and the knowledge changes the arithmetic. Being told the person chosen plays squash removes everybody who does not play squash from the pool you are choosing from, so the probability is counted out of the squash players alone. On Edexcel's 1MA1 specification this is spec ref P9: calculating and interpreting conditional probabilities through representation using expected frequencies with two-way tables, tree diagrams and Venn diagrams. P9 is Higher tier only. Foundation candidates complete two-way tables and tree diagrams under P6 and P8, but they are never asked for a probability that begins with the words given that.

Edexcel rarely gives the idea a question to itself. It turns up as the last part of something longer: a table you have just completed, a Venn diagram you have just filled in, or a bag of counters that has already produced two ordinary products. That part runs to 2-3 marks and the wording is almost always 'Given that ...'. The exception is the without-replacement question where the number of counters is unknown, which is worth around 5 marks on its own: the early parts build an equation from two conditional stages and the last part solves the quadratic it produces. Any of the three papers can carry it. A calculator buys you very little here, because most of the work is counting and then writing a single fraction; on Paper 1 you simply leave the fraction as it stands.

The Higher Exam Aid issued with every paper carries P(A and B)=P(A given B)×P(B)P(A \text{ and } B) = P(A \text{ given } B) \times P(B), and the Foundation sheet stops one line earlier at P(A or B)P(A \text{ or } B), which is a fair map of where the tier boundary falls. Rearranged, the Higher line reads P(A given B)=P(A and B)P(B)P(A \text{ given } B) = \frac{P(A \text{ and } B)}{P(B)}, and the division is where the marks sit. These parts hang on the denominator, so write 'out of 40' before you have decided on the numerator: a correct restricted total normally carries the method mark by itself. The losses are predictable. Dividing by the whole group, giving P(A and B)P(A \text{ and } B) when the question asked for P(A given B)P(A \text{ given } B), and reading the condition backwards so it ends up on the wrong event.

Worked example

A company asked all 200 of its employees how they travel to work. 120 of the employees are women. Of the employees who travel by car, 46 are women and 34 are men. Of the employees who travel by bus, 31 are women and 26 are men. All of the other employees walk or cycle. One of the 200 employees is chosen at random. Given that the employee chosen is a man, work out the probability that this employee walks or cycles.
[4 marks]
  1. The condition names the men, so start by counting them: 200−120=80200 - 120 = 80 men.
    M1: the restricted total identified, which is the denominator of the final fraction
  2. Men who travel by car or by bus: 34+26=6034 + 26 = 60.
    M1: reading only the men's figures, ignoring the 46 and the 31
  3. The rest of the men walk or cycle: 80−60=2080 - 60 = 20.
    M1: the numerator, found by subtracting from the men's total rather than from 200
  4. The choice is already restricted to the 80 men, so the probability is 2080=14\frac{20}{80} = \frac{1}{4}.
    A1: 14\frac{1}{4} or 0.25; a denominator of 200 here scores nothing on this mark

Practice questions

These are original questions in Edexcel 1MA1 Higher style. Nothing is printed as a picture, so each two-way table arrives as a few sentences of counts and every bag states whether the first item goes back before the second is taken. Probabilities can be typed as fractions, so 5/17 is a perfectly good answer; where the exact value does not terminate the stem also offers a decimal to 3 decimal places.

No account needed: the questions below are always free to check, plus 8 extra questions today. Your score so far updates as you go.

  1. Question 1 Warm-up [2 marks]
    80 people were asked whether they own a car. 45 of the people are aged under 40, and 28 of these 45 own a car. The other 35 people are aged 40 or over, and 24 of them own a car. One of the 80 people is chosen at random. Given that the person chosen is aged under 40, work out the probability that this person owns a car. Give your answer as a fraction or as a decimal to 3 decimal places.
  2. Question 2 Exam pace [4 marks]
    A club has 140 members. 84 of the members are adults and the rest are juniors. 63 of the adults play in a team. 28 of the juniors play in a team. One of the 140 members is chosen at random. Work out (i) the probability that the member plays in a team, and (ii) the probability that the member is a junior, given that the member plays in a team. Give each answer as a fraction or as a decimal to 3 decimal places.
  3. Question 3 Exam pace [2 marks]
    A pack contains 40 cards numbered 1 to 40, one number on each card. One card is taken at random. Given that the number on the card is a multiple of 5, work out the probability that the number is also a multiple of 4.
  4. Question 4 Exam pace [2 marks]
    A bag contains 4 green counters, 5 blue counters and 3 red counters. Two counters are taken at random, one after the other, without replacement. Given that the first counter is blue, work out the probability that the second counter is also blue. Give your answer as a fraction or as a decimal to 3 decimal places.
  5. Question 5 Stretch [5 marks]
    A school has 180 students in Year 11. 108 of them study Spanish and the rest do not. Given that a student studies Spanish, the probability that the student also studies Music is 14\frac{1}{4}. Given that a student does not study Spanish, the probability that the student studies Music is 16\frac{1}{6}. One of the 180 students is chosen at random and this student studies Music. Work out the probability that this student studies Spanish. Give your answer as a fraction or as a decimal to 3 decimal places.
  6. Question 6 Stretch [5 marks]
    A bag contains 2 red counters, 5 blue counters and 4 green counters. Two counters are taken at random, one after the other, without replacement. Given that the two counters are the same colour, work out the probability that they are both blue. Give your answer as a fraction or as a decimal to 3 decimal places.

Common mistakes examiners see

  • Keeping the whole group as the denominator, so 20 men out of 200 employees is written as 20200=0.1\frac{20}{200} = 0.1 when the answer is 2080=0.25\frac{20}{80} = 0.25.

    Read the condition first and write its total down before anything else. Cover every row or region that fails the condition with your hand; what is still visible is both the pool you count from and the number you divide by.

  • Answering P(A and B)P(A \text{ and } B) when the question asked for P(A given B)P(A \text{ given } B). In a group of 30 where 12 study History and 7 study both History and Geography, the response is 730\frac{7}{30} rather than 712\frac{7}{12}.

    The words 'and' and 'given that' point at the same numerator but different denominators. 'And' divides by everybody; 'given that' divides by the named event. Underline whichever phrase the stem uses before you calculate.

  • Swapping the two events over. In a club of 50 women and 40 squash players, 15 of whom are women, P(woman given squash)=1540P(\text{woman given squash}) = \frac{15}{40} but P(squash given woman)=1550P(\text{squash given woman}) = \frac{15}{50}. Same numerator, different question.

    The event that comes after 'given that' is always the denominator. Say the sentence back as 'out of the squash players, how many are women' and the fraction writes itself.

  • Leaving the second-stage denominator alone when nothing is replaced, so P(second red given first red)P(\text{second red given first red}) from a bag of 5 red and 7 blue is given as 412\frac{4}{12} instead of 411\frac{4}{11}.

    Being told the first counter was red changes two numbers, not one: a red counter has gone and so has a counter. Write out what is left in the bag before you write the fraction.

  • Getting to n(n−1)=210n(n-1) = 210 and then dividing, writing n=210÷1n = 210 \div 1 or n=210n = \sqrt{210}, or solving n2−n−210=0n^2 - n - 210 = 0 correctly and then offering the negative root as the number of counters.

    Multiply out to a quadratic in the form n2−n−210=0n^2 - n - 210 = 0, factorise or use the formula, then reject any negative or fractional root with a one-line reason: a bag cannot hold −14-14 counters. Check your value back in the original probability before moving on.

Frequently asked questions

What does P(A given B) mean?
It is the probability that A happens when you already know B has happened. Because B is known, the outcomes outside B are no longer available, so the probability is counted out of B alone rather than out of everything. Edexcel writes it in words as P(A given B) on the Higher formulae sheet rather than with a vertical bar, and the calculation is P(A and B) divided by P(B).
Is conditional probability on the Foundation paper?
No. Spec point P9 is Higher tier only, so a Foundation candidate is never asked a 'given that' probability. Foundation does cover two-way tables, tree diagrams and Venn diagrams under P6 and P8, which is why the diagrams look familiar to both tiers even though the final part of the question does not.
Do I get the conditional probability formula in the exam?
Partly. The Higher Exam Aid that comes with each 1MA1 paper prints P(A and B) = P(A given B) P(B), so the relationship is there, but you have to rearrange it yourself to get P(A given B) = P(A and B) divided by P(B). The Foundation version of the sheet carries only P(A or B) = P(A) + P(B) - P(A and B).
How many marks is conditional probability worth on Edexcel Higher?
A 'given that' part tacked onto a table, a Venn diagram or a tree is usually 2-3 marks. The without-replacement question that ends in a quadratic is longer, around 5 marks across its parts. Over a full series of three 80-mark papers, expect roughly 3-5 marks that depend on it.

Related topics