01 · Train journey
Find P(train).
Hint
Add both train cells.
Worked solution
(20+5)/50 = 1/2.
Understand · explore · practise
Probabilities from tables and histograms
Choose the correct population and denominator, calculate event probabilities from category tables and distinguish exact grouped counts from histogram estimates.
Before you startFractions, frequency density and reading grouped data.
01 / Define what is chosen
The denominator describes the selection.
For uniform random selection of one record: P(event) = qualifying records ÷ all eligible records.
State the population and the selection rule before calculating.
This can be exact for the recorded finite population. Using its proportion to predict a future wait is a separate modelling step: representativeness and stable conditions matter.
Choose a waiting-time thresholdExplore
Forty recorded waits: 0 ≤ t < 10: 8; 10 ≤ t < 20: 12; 20 ≤ t < 40: 20. Select one record uniformly at random.
At c = 0, no recorded wait is below 0: probability 0, exactly from these class boundaries.
Inside a class, the shaded proportion estimates the count by assuming constant density within that class.
02 / Count joint categories
A two-way table keeps combinations distinct.
Original records for 50 journeys: bus/on-time 18, bus/late 7, train/on-time 20, train/late 5. Select one of these 50 records uniformly.
Find the probability of a late bus journey.Worked example
The event specifies both bus and late
Only 7 records meet both conditions.
P(bus and late) = 7/50
All 50 records remain eligible for selection.
02 · On time
Find P(on time).
Hint
Include both types of transport.
Worked solution
(18+20)/50 = 38/50 = 19/25.
03 · Late bus
Why is 7/25 not the answer for a uniformly chosen journey being both bus and late?
Hint
Which records are eligible?
Worked solution
The experiment chooses from all 50 journeys, not only the 25 bus journeys. The joint probability is 7/50. Selecting only from bus records would define a different experiment.
03 / Avoid counting an overlap twice
Either includes the possibility of both.
04 · Bus or late
Find P(bus or late), including journeys that satisfy both.
Hint
Count bus/on-time, bus/late and train/late once each.
Worked solution
(18+7+5)/50 = 30/50 = 3/5. Equivalently (25+12−7)/50.
05 · Neither condition
Find P(neither bus nor late).
Hint
Which single cell remains?
Worked solution
Only train/on-time qualifies: 20/50 = 2/5. This complements the previous union.
04 / Use histogram area
Bar heights alone are not frequencies.
Watch: count the shaded area
Pause, replay or seek freely. The notes explain the same idea and stay in view.
The waiting-time classes have widths 10, 10, 20 and frequencies 8, 12, 20.Worked example
Frequency densities are 0.8, 1.2 and 1
Each density is frequency divided by class width.
Total histogram area is 8 + 12 + 20 = 40
The widest bar contains the most records despite not being the tallest.
P(t < 20) = 20/40 = 1/2
The first two complete classes determine this exactly for the recorded population.
06 · Last class
Find P(20 ≤ t < 40).
Hint
Use frequency, not bar height.
Worked solution
20/40 = 1/2. Dividing its height by the sum of the heights is invalid with unequal widths.
07 · Tallest bar
Which class has the highest frequency density, and which has the most records?
Hint
Compare density and frequency separately.
Worked solution
The 10–20 class has the highest density, 1.2. The 20–40 class has the greatest frequency, 20.
05 / A threshold inside a class
Grouped data do not reveal every individual value.
Estimate a partial count using the fraction of the class width.
Assume observations are spread uniformly within that class.
Estimate P(t < 15).Worked example
All 8 records in 0–10 qualify
The first class is complete.
Estimate half of the 12 records in 10–20 below 15
Estimated partial count = (5/10)×12 = 6.
Estimated probability = (8+6)/40 = 7/20
The exact count below 15 cannot be recovered from these groups.
08 · Thirty minutes
Estimate P(t < 30).
Hint
Use half of the final class.
Worked solution
Estimated count = 8+12+(10/20)×20 = 30, giving 30/40 = 3/4.
09 · Five minutes
Estimate P(t < 5).
Hint
Use half the first class.
Worked solution
Estimated count = (5/10)×8 = 4, giving 4/40 = 1/10. This depends on the within-class model.
06 / Say what is known exactly
A plausible estimate is not the missing raw data.
10 · Bounds without uniformity
What bounds on P(t < 15) follow from the groups alone?
Hint
All 8 first-class observations qualify; anywhere from 0 to 12 of the next class may qualify.
Worked solution
The count lies from 8 to 20 inclusive, so 1/5 ≤ P(t < 15) ≤ 1/2. The estimate 7/20 is within these bounds.
11 · Between thresholds
Estimate P(15 ≤ t < 30).
Hint
Take half the second class and half the third.
Worked solution
Estimated count = 6+10 = 16, giving 16/40 = 2/5.
12 · Future journeys
Does the exact proportion of on-time journeys in these 50 records prove the probability for next month is 19/25?
Hint
The population has changed from a fixed collection to future journeys.
Worked solution
No. It provides an estimate under assumptions about how representative these records are and whether conditions remain similar.
07 / Read the interval convention
Boundary values belong to the stated class.
13 · A wait of exactly 20
Does a recorded wait of exactly 20 minutes contribute to P(t < 20)?
Hint
The event uses a strict inequality.
Worked solution
No. It belongs to 20 ≤ t < 40 and is excluded from t < 20. For rounded records, the exact probability of t ≤ 20 cannot be found unless the count equal to 20 is known.
14 · Missing records
Five additional journeys have unknown punctuality. Can you silently count them as late?
Hint
Unknown is not a response category you observed.
Worked solution
No. Preserve the missing status. State whether the calculation concerns only the 50 complete records, and explain that extending it to all 55 requires more information or assumptions.
08 / Counts, areas and assumptions
Check the event, denominator and strength of the claim.
Identify who or what is selected. Count joint categories carefully, subtract overlap when needed and use histogram areas for unequal class widths. Distinguish exact probabilities for a recorded finite population, within-class estimates and predictions about future observations.
Section 1 of 8 · Define what is chosen