The first job in any probability experiment is to record the results clearly. A frequency table does this: it lists every outcome alongside how many times it happened.
A die is rolled $40$ times. Complete the table and find the probability of rolling a $5$, based on these results.
| Score | $1$ | $2$ | $3$ | $4$ | $5$ | $6$ |
|---|---|---|---|---|---|---|
| Frequency | $6$ | $8$ | $5$ | $7$ | $?$ | $5$ |
This is an experimental probability, based on what actually happened.
A two-way table records data classified in two different ways at the same time — for example gender and choice of subject.
- Fill in every value the question gives you.
- Use the row totals and column totals to work out the missing cells.
- Start with any row or column that has only one gap.
- Keep going — each value you find often unlocks another.
- Check that the row totals and column totals both reach the grand total.
Complete this table showing how $80$ students travel to school.
| Walk | Bus | Car | Total | |
|---|---|---|---|---|
| Boys | $14$ | $?$ | $9$ | $42$ |
| Girls | $?$ | $13$ | $?$ | $?$ |
| Total | $31$ | $?$ | $17$ | $80$ |
| Walk | Bus | Car | Total | |
|---|---|---|---|---|
| Boys | $14$ | $19$ | $9$ | $42$ |
| Girls | $17$ | $13$ | $8$ | $38$ |
| Total | $31$ | $32$ | $17$ | $80$ |
Using the completed table above, a student is chosen at random. Find:
(a) $P(\text{travels by bus})$ (b) $P(\text{a girl who walks})$ (c) $P(\text{a boy, given they travel by car})$
A frequency tree looks like a tree diagram, but the branches carry numbers of people or things rather than probabilities.
$200$ people were tested for a condition. $30$ tested positive. Of those who tested positive, $24$ actually had the condition. Of those who tested negative, $6$ actually had the condition. Complete the frequency tree and find the probability that a randomly chosen person actually had the condition.
Check: $24 + 6 + 6 + 164 = 200$ ✓
The table shows the times taken by $60$ runners.
| Time $t$ (min) | Frequency |
|---|---|
| $20 \leq t \lt 25$ | $8$ |
| $25 \leq t \lt 30$ | $21$ |
| $30 \leq t \lt 35$ | $19$ |
| $35 \leq t \lt 40$ | $12$ |
A runner is chosen at random. Find (a) $P(t \lt 30)$, (b) $P(t \geq 30)$.
Check directly: $19 + 12 = 31$ ✓
Frequency table
Lists each outcome with how many times it occurred.
Always check
The frequencies must add to the number of trials.
Two-way table
Classifies data in two ways at once; use row and column totals.
Start where there is one gap
Fill the easiest cell first; it unlocks the rest.
Frequency tree
Branches carry counts, not probabilities.
Branch rule
Each pair of branches adds back to its parent.
Probability
$\dfrac{\text{frequency}}{\text{total}}$.
"Given that"
Use only one row or column as the denominator.
A coin is tossed $50$ times, landing heads $23$ times. Complete the frequency table and find the experimental probability of tails.
▶ Show solution
Tails frequency $= 50 - 23 = 27$.
$P(\text{tails}) = \dfrac{27}{50} = 0.54$
A spinner is spun $80$ times with these results: red $22$, blue $19$, green $26$, yellow $?$. Find the missing frequency and $P(\text{green})$.
▶ Show solution
$22 + 19 + 26 = 67$
Yellow $= 80 - 67 = 13$
$P(\text{green}) = \dfrac{26}{80} = \dfrac{13}{40} = 0.325$
Complete this two-way table for $50$ people.
| Tea | Coffee | Total | |
|---|---|---|---|
| Adults | $?$ | $18$ | $30$ |
| Children | $14$ | $?$ | $?$ |
| Total | $?$ | $?$ | $50$ |
▶ Show solution
Adults tea $= 30 - 18 = 12$
Children total $= 50 - 30 = 20$, so children coffee $= 20 - 14 = 6$
Tea total $= 12 + 14 = 26$; coffee total $= 18 + 6 = 24$
Check: $26 + 24 = 50$ ✓
Using the completed table from Question 3, a person is chosen at random. Find (a) $P(\text{drinks tea})$, (b) $P(\text{an adult who drinks coffee})$.
▶ Show solution
(a) $\dfrac{26}{50} = \dfrac{13}{25} = 0.52$
(b) $\dfrac{18}{50} = \dfrac{9}{25} = 0.36$
$150$ students sat an exam. $90$ were girls. $72$ of the girls passed and $45$ of the boys passed. Draw a frequency tree and find how many students failed altogether.
▶ Show solution
Boys $= 150 - 90 = 60$.
Girls who failed $= 90 - 72 = 18$.
Boys who failed $= 60 - 45 = 15$.
Total failing $= 18 + 15 = 33$ students.
Check: $72 + 18 + 45 + 15 = 150$ ✓
Using the data in Question 5, find (a) $P(\text{a student passed})$, (b) $P(\text{a girl, given the student passed})$.
▶ Show solution
(a) Passed $= 72 + 45 = 117$; $P = \dfrac{117}{150} = 0.78$
(b) Restrict to the $117$ who passed, of whom $72$ are girls.
$P = \dfrac{72}{117} = \dfrac{8}{13} = 0.615$ (3 d.p.)
The table shows the ages of $200$ visitors to a museum. Find $P(\text{aged under } 30)$.
| Age $a$ | $0 \leq a \lt 15$ | $15 \leq a \lt 30$ | $30 \leq a \lt 50$ | $50 \leq a \lt 80$ |
|---|---|---|---|---|
| Frequency | $42$ | $56$ | $61$ | $41$ |
▶ Show solution
Under $30$: $42 + 56 = 98$
$P = \dfrac{98}{200} = \dfrac{49}{100} = 0.49$
A frequency tree shows $240$ car journeys. $150$ were in the morning. $60$ of the morning journeys were delayed, and $27$ of the afternoon journeys were delayed. Find the number of journeys that were on time.
▶ Show solution
Afternoon journeys $= 240 - 150 = 90$.
Morning on time $= 150 - 60 = 90$.
Afternoon on time $= 90 - 27 = 63$.
Total on time $= 90 + 63 = 153$ journeys.
In a survey of $120$ people, $70$ owned a dog, $55$ owned a cat and $22$ owned both. Draw a two-way table and find how many owned neither.
▶ Show solution
Dog only $= 70 - 22 = 48$; cat only $= 55 - 22 = 33$.
| Cat | No cat | Total | |
|---|---|---|---|
| Dog | $22$ | $48$ | $70$ |
| No dog | $33$ | $17$ | $50$ |
| Total | $55$ | $65$ | $120$ |
Neither $= 120 - (48 + 22 + 33) = 120 - 103 = \mathbf{17}$ people.
A school of $400$ pupils is surveyed about school dinners. $240$ are in Key Stage 3. Of the Key Stage 3 pupils, $65\%$ have school dinners. Of the rest of the school, $45$ have school dinners.
(a) Complete a frequency tree. (b) Find $P(\text{a pupil has school dinners})$. (c) Given that a pupil has school dinners, find the probability they are in Key Stage 3. (d) A newspaper claims "most pupils have school dinners". Comment.
▶ Show solution
(a) Not KS3 $= 400 - 240 = 160$.
KS3 with dinners $= 65\%$ of $240 = 0.65 \times 240 = 156$; KS3 without $= 240 - 156 = 84$.
Not-KS3 with dinners $= 45$; without $= 160 - 45 = 115$.
Check: $156 + 84 + 45 + 115 = 400$ ✓
(b) Total with dinners $= 156 + 45 = 201$.
$P = \dfrac{201}{400} = 0.5025$
(c) Restrict to the $201$ dinner-eaters, of whom $156$ are KS3.
$P = \dfrac{156}{201} = 0.776$ (3 d.p.)
(d) $50.25\%$ is only just over half, so "most" is technically true but misleading — it is very close to an even split. The claim also hides the fact that the split is very uneven between key stages: $65\%$ of KS3 but only $28\%$ of the rest.