P Probability — printable coverage pack

P1 · Record, describe and analyse the frequency of outcomes of probability experiments using tables and frequency trees

A frequency table records how often each outcome occurs; the frequencies should add to the total number of trials.
On a frequency tree, subtract from the parent frequency to complete a missing branch, then check that the two child branches add back to their parent.
For example, if $70$ trials split into frequencies $43$ and $27$ , and $18$ of the first group succeed, the other frequency in that group is $43-18=25$ .
Do not put probabilities on a frequency tree when frequencies are requested, and do not compare raw frequencies when the group totals are different.

Tier 1 · Easy

1. A spinner is used $30$ times. It lands on red $18$ times and on blue $12$ times. Complete a frequency table for the two outcomes and work out the relative frequency of red.[2 marks]
Answer
- Red: $18$ ; blue: $12$ .
- Relative frequency of red $=\dfrac{18}{30}=0.6$ .
Method: Record the observed counts in the table and check that $18+12=30$ . Divide the red frequency by the total number of spins: $18\div30=0.6$ .

Tier 2 · Standard

1. A frequency tree describes $80$ visits to a kiosk. On $52$ visits a hot drink is chosen and on the remaining visits a cold drink is chosen. A snack is also bought on $31$ hot-drink visits and on $7$ cold-drink visits. Complete all terminal frequencies and work out the percentage of visits on which a snack is bought.[3 marks]
Answer
- Cold drink: $28$ ; hot drink and no snack: $21$ ; cold drink and no snack: $21$ .
- Snack percentage $=47.5\%$ .
Method: The cold-drink branch has frequency $80-52=28$ . The no-snack frequencies are $52-31=21$ and $28-7=21$ . Snacks are bought on $31+7=38$ visits, so the percentage is $\dfrac{38}{80}\times100=47.5\%$ .

Tier 3 · Hard

1. Two weeks of trials are combined. In week 1, $72$ of $120$ trials succeed; $45$ successful trials and $18$ unsuccessful trials are fast. In week 2, $99$ of $180$ trials succeed; $66$ successful trials and $27$ unsuccessful trials are fast. Construct a combined frequency table for success or failure against fast or not fast. Compare the proportions that are fast in the two outcome groups.[5 marks]
Answer
- Success: $111$ fast and $60$ not fast; failure: $45$ fast and $84$ not fast.
- Proportion of successful trials that are fast $=\dfrac{111}{171}\approx64.9\%$ ; proportion of failed trials that are fast $=\dfrac{45}{129}\approx34.9\%$ .
- A successful trial is more likely to be fast.
Method: Combine like branches: successful trials total $72+99=171$ , of which $45+66=111$ are fast, so $60$ are not fast. Failures total $(120-72)+(180-99)=129$ , of which $18+27=45$ are fast, so $84$ are not fast. Compare within each outcome group: $111/171\approx0.649$ and $45/129\approx0.349$ .

P2 · Apply ideas of randomness, fairness and equally likely events to calculate expected outcomes of multiple future experiments

Randomness means that an individual result is uncertain even when the long-run pattern can be predicted.
For equally likely outcomes, calculate an event probability by dividing favourable outcomes by total outcomes, then use expected frequency $=n\times p$ for $n$ future trials.
A fair game gives participants equal chances or has zero expected gain after any cost; fairness must be justified from the probabilities and rewards.
An expected frequency is a long-run prediction, not a guarantee that exactly that many outcomes will occur.

Tier 1 · Easy

1. A fair six-sided die is rolled $240$ times. Work out the expected number of rolls on which the score is greater than $4$ .[2 marks]
Answer
- $80$ rolls
Method: The favourable scores are $5$ and $6$ , so the probability is $2/6=1/3$ . The expected frequency is $240\times\dfrac{1}{3}=80$ .

Tier 2 · Standard

1. A fair spinner has four equal sectors labelled $1$ , $1$ , $2$ and $3$ . It is spun $600$ times. Work out the expected total of all the scores.[3 marks]
Answer
- $1050$
Method: The expected score per spin is $\dfrac{1+1+2+3}{4}=\dfrac{7}{4}$ . Over $600$ spins, the expected total is $600\times\dfrac{7}{4}=1050$ .

Tier 3 · Hard

1. A box contains $3$ green, $2$ yellow and $1$ red token. A token is chosen at random and replaced. A player receives $\pounds4$ for green, $\pounds1$ for yellow and nothing for red. The entry fee is $\pounds2.20$ . Work out the organiser's expected profit from $900$ plays and decide whether the game is fair to the player.[5 marks]
Answer
- Expected payout per play $=\pounds\dfrac{7}{3}$ .
- Expected organiser profit from $900$ plays $=-\pounds120$ .
- The game is not fair; the player has an expected gain of about $13.3$ pence per play.
Method: The expected payout is $4\times\dfrac{3}{6}+1\times\dfrac{2}{6}=2+\dfrac{1}{3}=\dfrac{7}{3}$ pounds. The organiser receives $\pounds2.20=\pounds\dfrac{11}{5}$ per play, so expected profit per play is $\dfrac{11}{5}-\dfrac{7}{3}=-\dfrac{2}{15}$ pounds. Over $900$ plays this is $900\times(-2/15)=-120$ pounds. A fair entry fee would equal the expected payout, so this game favours the player.

P3 · Relate relative expected frequencies to theoretical probability, using appropriate language and the 0-1 probability scale

Probability lies from $0$ to $1$ : $0$ means impossible, $1$ means certain, and values nearer $1$ describe more likely events.
Relative frequency is observed frequency divided by number of trials; theoretical probability comes from a mathematical model of the outcomes.
If a theoretical probability is $p$ , the relative frequency from many unbiased trials is expected to be close to $p$ , although it will usually not equal it exactly.
Do not describe an event with probability $0.5$ as certain, and do not treat a small experimental difference as proof that a theoretical model is wrong.

Tier 1 · Easy

1. Place these events in order from least likely to most likely: event A has probability $0.72$ , event B has probability $\dfrac{1}{4}$ , and event C has probability $0.5$ .[1 mark]
Answer
- B, C, A
Method: Write every probability as a decimal: $\dfrac{1}{4}=0.25$ , then compare $0.25<0.5<0.72$ .

Tier 2 · Standard

1. An outcome occurs $84$ times in $140$ trials. Its theoretical probability is $\dfrac{5}{8}$ . Work out the relative frequency and the expected frequency in $560$ further trials. Comment on the experimental result.[3 marks]
Answer
- Relative frequency $=0.6$ .
- Expected frequency $=350$ .
- $0.6$ is reasonably close to the theoretical probability $0.625$ .
Method: The relative frequency is $84/140=0.6$ . The theoretical probability is $5/8=0.625$ , so the expected frequency is $560\times5/8=350$ . The observed value differs by only $0.025$ , which is plausible experimental variation.

Tier 3 · Hard

1. A model says that a device flashes with probability $0.5$ on each trial. In a short run it flashes $41$ times in $80$ trials. After $500$ trials in total it has flashed $247$ times. Calculate both relative frequencies and decide which result gives stronger evidence about the model.[4 marks]
Answer
- Short-run relative frequency $=0.5125$ .
- Overall relative frequency $=0.494$ .
- The $500$ -trial result gives stronger evidence and is very close to $0.5$ , so it supports the model.
Method: For the short run, calculate $41/80=0.5125$ . For all trials, calculate $247/500=0.494$ . Both are close to $0.5$ , but the larger sample is less affected by random variation, so the overall relative frequency is the stronger comparison.

P4 · Apply the property that the probabilities of an exhaustive set of outcomes sum to one; apply the property that the probabilities of an exhaustive set of mutually exclusive events sum to one

An exhaustive set covers every possible outcome, so its probabilities have total $1$ .
Mutually exclusive events cannot occur together; for such events, add their probabilities to find the probability that any one occurs.
The complement rule is $P(\text{not }A)=1-P(A)$ , which treats $A$ and not $A$ as an exhaustive, mutually exclusive pair.
Do not add probabilities directly when events can overlap; first separate the overlap or subtract it once using $P(A\cup B)=P(A)+P(B)-P(A\cap B)$ .

Tier 1 · Easy

1. The probability that a parcel arrives early is $0.37$ . Work out the probability that it does not arrive early.[1 mark]
Answer
- $0.63$
Method: Early and not early are exhaustive and mutually exclusive, so use the complement: $1-0.37=0.63$ .

Tier 2 · Standard

1. A trial has four mutually exclusive outcomes A, B, C and D. Their probabilities are $0.28$ , $0.41$ , $x$ and $0.19$ respectively. Work out $x$ .[2 marks]
Answer
- $x=0.12$
Method: The outcomes are exhaustive, so their probabilities sum to $1$ . Therefore $x=1-(0.28+0.41+0.19)=1-0.88=0.12$ .

Tier 3 · Hard

1. For two events A and B, $P(A)=0.58$ , $P(B)=0.47$ and $P(A\cap B)=0.21$ . Work out the probability that neither A nor B occurs.[4 marks]
Answer
- $0.16$
Method: Split the possibility space into mutually exclusive regions. The probability of $A\cup B$ is $0.58+0.47-0.21=0.84$ , because the intersection was counted twice. Neither event is the complement of the union, so its probability is $1-0.84=0.16$ .

P5 · Understand that empirical unbiased samples tend towards theoretical probability distributions, with increasing sample size

An empirical distribution is built from observed relative frequencies, whereas a theoretical distribution is predicted from a probability model.
With unbiased, independent trials, increasing the sample size usually makes empirical proportions settle nearer their theoretical probabilities.
For a fair die, a large sample should give each score a relative frequency near $1/6$ , but exact equality is not required.
A larger sample reduces random variation but does not repair a biased sampling method or guarantee a closer result on every single run.

Tier 1 · Easy

1. A fair coin gives a relative frequency of heads of $0.64$ after $25$ tosses. Explain what is likely to happen to the relative frequency as many more unbiased tosses are made.[1 mark]
Answer
- It is likely to move closer to the theoretical probability $0.5$ , although it need not do so after every extra toss.
Method: Use the long-run tendency: for unbiased coin tosses the empirical proportion tends towards the theoretical probability $1/2$ as the sample becomes large.

Tier 2 · Standard

1. A spinner is designed to land on green with probability $0.3$ . It lands on green $8$ times in $20$ spins and $63$ times in $200$ spins. Compare the two empirical probabilities with the theoretical probability.[3 marks]
Answer
- For $20$ spins, the relative frequency is $0.4$ .
- For $200$ spins, the relative frequency is $0.315$ .
- The larger sample gives the value closer to $0.3$ .
Method: Calculate $8/20=0.4$ and $63/200=0.315$ . Their distances from $0.3$ are $0.1$ and $0.015$ , so the empirical result from $200$ spins is closer to the theoretical probability.

Tier 3 · Hard

1. A fair die is tested. In the first $60$ rolls, a six appears $16$ times. After $600$ rolls in total, a six has appeared $108$ times. A student says the die must be biased because neither relative frequency equals $\dfrac{1}{6}$ . Evaluate the student's claim.[4 marks]
Answer
- The relative frequencies are $\dfrac{16}{60}\approx0.267$ and $\dfrac{108}{600}=0.18$ .
- The theoretical probability is $\dfrac{1}{6}\approx0.167$ .
- The larger-sample result is closer to the theoretical value, and an exact match is not expected, so the evidence does not establish bias.
Method: Compare each empirical probability with $1/6$ . The differences are about $0.100$ for the first $60$ rolls and $0.013$ after $600$ rolls. Random variation explains a non-zero difference, and the movement towards $1/6$ as the sample grows is consistent with a fair die.

P6 · Enumerate sets and combinations of sets systematically, using tables, grids, Venn diagrams and tree diagrams

Systematic enumeration lists every permitted outcome once, using a consistent order so omissions and duplicates are visible.
Use a table or grid for two varying quantities, a tree for successive choices, and a Venn diagram for membership of overlapping sets.
For two sets, place the intersection first, then the parts belonging only to each set, and finally the values outside both sets.
Do not assume all listed outcomes are equally likely; enumeration identifies possibilities, while probability also depends on the model.

Tier 1 · Easy

1. A uniform is made from one of two shirts, blue or white, and one of three ties, red, silver or green. List all possible shirt-and-tie combinations.[2 marks]
Answer
- Blue-red, blue-silver, blue-green, white-red, white-silver, white-green.
Method: Hold the shirt colour fixed and list every tie, then repeat for the other shirt. This gives $2\times3=6$ combinations, each appearing once.

Tier 2 · Standard

1. The universal set is the integers from $1$ to $20$ . Set A contains the multiples of $3$ and set B contains the factors of $18$ . Enumerate the four regions of a Venn diagram for A and B.[4 marks]
Answer
- $A\cap B=\{3,6,9,18\}$ .
- A only $=\{12,15\}$ ; B only $=\{1,2\}$ .
- Neither $=\{4,5,7,8,10,11,13,14,16,17,19,20\}$ .
Method: List $A=\{3,6,9,12,15,18\}$ and $B=\{1,2,3,6,9,18\}$ . Put their common values in the intersection, remove these to find each 'only' region, then place every unused integer from $1$ to $20$ outside both circles.

Tier 3 · Hard

1. A three-digit number is formed from three different digits chosen from $1$ , $2$ , $3$ and $4$ . Enumerate all the numbers that are greater than $230$ and even. How many of these numbers do not contain the digit $1$ ?[4 marks]
Answer
- $234,312,314,324,342,412,432$ .
- $4$ of these numbers do not contain the digit $1$ .
Method: Work systematically by hundreds digit. Starting with $2$ gives $234$ ; starting with $3$ gives $312,314,324,342$ ; starting with $4$ gives $412,432$ . The values without digit $1$ are $234,324,342,432$ , so there are $4$ .

P7 · Construct theoretical possibility spaces for single and combined experiments with equally likely outcomes and use these to calculate theoretical probabilities

A possibility space displays every combined outcome, often as ordered pairs in a grid.
When all elementary outcomes are equally likely, probability is favourable outcomes divided by the total number of outcomes in the space.
For two fair choices with $m$ and $n$ outcomes, a complete rectangular space has $mn$ ordered pairs before restrictions are applied.
Do not count unordered pairs when the experiment records a first and second result; $(a,b)$ and $(b,a)$ are different ordered outcomes unless the context says otherwise.

Tier 1 · Easy

1. A fair coin is tossed and a fair spinner labelled $1$ , $2$ , $3$ is spun. Write the six outcomes as ordered pairs and find the probability of getting a head and an even number.[2 marks]
Answer
- $(H,1),(H,2),(H,3),(T,1),(T,2),(T,3)$ .
- $P(\text{head and even})=\dfrac{1}{6}$ .
Method: Pair each coin result with each spinner result. Only $(H,2)$ satisfies both conditions, so one of the six equally likely outcomes is favourable.

Tier 2 · Standard

1. Spinner A has equal sectors labelled $1$ , $2$ , $3$ . Spinner B has equal sectors labelled $1$ , $2$ , $3$ , $4$ . Construct a possibility space and find the probability that the two scores have a sum greater than $5$ .[3 marks]
Answer
- Possibility space: $(1,1),(1,2),(1,3),(1,4),(2,1),(2,2),(2,3),(2,4),(3,1),(3,2),(3,3),(3,4)$ .
- The favourable ordered pairs are $(2,4),(3,3),(3,4)$ .
- $P(\text{sum greater than }5)=\dfrac{3}{12}=\dfrac{1}{4}$ .
Method: A $3$ by $4$ grid contains $12$ equally likely ordered pairs. Mark the entries whose sums exceed $5$ : $6,6,7$ at $(2,4),(3,3),(3,4)$ . Therefore the probability is $3/12=1/4$ .

Tier 3 · Hard

1. Two different cards are chosen in order from cards labelled $1$ , $2$ , $3$ , $4$ and $6$ . The first card is the tens digit and the second is the units digit. Construct the theoretical possibility space and find the probability that the two-digit number is divisible by $3$ .[4 marks]
Answer
- Possibility space: $12,13,14,16,21,23,24,26,31,32,34,36,41,42,43,46,61,62,63,64$ .
- There are $5\times4=20$ equally likely ordered outcomes.
- The favourable numbers are $12,21,24,42,36,63$ .
- Probability $=\dfrac{6}{20}=\dfrac{3}{10}$ .
Method: Make a $5$ by $5$ grid and exclude its diagonal because the cards must be different. A number is divisible by $3$ when its digit sum is divisible by $3$ . Testing the $20$ allowed ordered pairs gives the six listed numbers, so the probability is $6/20=3/10$ .

P8 · Calculate the probability of independent and dependent combined events, including using tree diagrams and other representations, and know the underlying assumptions

For events along one route through a tree, multiply branch probabilities; for alternative mutually exclusive routes, add their products.
Independent events leave later probabilities unchanged, while dependent events change later probabilities because earlier outcomes affect the situation.
Without replacement, update both numerator and denominator after each draw; with replacement, the original probabilities repeat.
Do not multiply probabilities without checking the assumptions: identical branch probabilities require independence or a replacement mechanism.

Tier 1 · Easy

1. A fair coin is tossed and an independent spinner lands on red with probability $\dfrac{3}{5}$ . Work out the probability of a head and red.[2 marks]
Answer
- $\dfrac{3}{10}$
Method: The events are independent, so multiply their probabilities: $\dfrac{1}{2}\times\dfrac{3}{5}=\dfrac{3}{10}$ .

Tier 2 · Standard

1. A bag contains $4$ blue counters and $3$ amber counters. Two counters are chosen at random without replacement. Work out the probability that both counters are blue.[3 marks]
Answer
- $\dfrac{2}{7}$
Method: The first blue probability is $4/7$ . After a blue is removed, $3$ of the $6$ remaining counters are blue. Multiply along the route: $\dfrac{4}{7}\times\dfrac{3}{6}=\dfrac{12}{42}=\dfrac{2}{7}$ .

Tier 3 · Hard

1. A bag contains $5$ black and $3$ white counters. Three counters are chosen at random without replacement. Work out the probability that exactly two are black. State why the branch probabilities change after each choice.[5 marks]
Answer
- $P(\text{exactly two black})=\dfrac{15}{28}$ .
- The probabilities change because counters are not replaced, so the composition and total in the bag change.
Method: There are three possible colour orders: BBW, BWB and WBB. Their probabilities are $\dfrac{5}{8}\times\dfrac{4}{7}\times\dfrac{3}{6}=\dfrac{5}{28}$ , $\dfrac{5}{8}\times\dfrac{3}{7}\times\dfrac{4}{6}=\dfrac{5}{28}$ and $\dfrac{3}{8}\times\dfrac{5}{7}\times\dfrac{4}{6}=\dfrac{5}{28}$ . Add the mutually exclusive routes to get $15/28$ .

P9 · Calculate and interpret conditional probabilities through representation using expected frequencies with two-way tables, tree diagrams and Venn diagrams [Higher only]

A conditional probability restricts the sample space to outcomes satisfying the stated condition.
In a two-way table or Venn diagram, calculate $P(A\mid B)$ by dividing the frequency in both $A$ and $B$ by the total frequency in $B$ .
Expected frequencies can replace probabilities on a tree; after multiplying each branch by a common starting total, condition using the relevant terminal frequencies.
Do not divide by the overall total when a condition is given: the denominator must be the frequency of the conditioning event.

Tier 1 · Easy

1. A two-way table records $60$ students. Of the $24$ who travel by bus, $9$ arrive late. Find the probability that a randomly chosen bus traveller arrives late.[2 marks]
Answer
- $P(\text{late}\mid\text{bus})=\dfrac{9}{24}=\dfrac{3}{8}$
Method: The condition restricts the group to the $24$ bus travellers. Of these, $9$ are late, so divide $9$ by $24$ and simplify.

Tier 2 · Standard

1. Of $500$ expected customers, $60\%$ are members. Of the members, $70\%$ order online. Of the non-members, $35\%$ order online. Use expected frequencies to find the probability that an online customer is a member.[4 marks]
Answer
- $P(\text{member}\mid\text{online})=\dfrac{3}{4}$
Method: There are $0.60\times500=300$ members and $200$ non-members. Expected online frequencies are $0.70\times300=210$ members and $0.35\times200=70$ non-members. Among $210+70=280$ online customers, $210$ are members, so the conditional probability is $210/280=3/4$ .

Tier 3 · Hard

1. In an expected-frequency Venn diagram for $240$ people, $138$ are in set A, $102$ are in set B and $54$ are in both sets. A person is chosen from those who are in exactly one of the sets. Find the conditional probability that the person is in A.[4 marks]
Answer
- $\dfrac{7}{11}$
Method: The A-only frequency is $138-54=84$ and the B-only frequency is $102-54=48$ . Exactly one set contains $84+48=132$ people. Restricting to this group, the required probability is $84/132=7/11$ .