01 · Spinner sum
Independent fair spinners labelled 1–4 and 1–5 are used. Find P(sum=6).
Hint
List the qualifying ordered pairs.
Worked solution
There are 20 equally likely pairs. Sum6 occurs at(1,5),(2,4),(3,3),(4,2), giving4/20=1/5.
Understand · explore · practise
Probability: mixed practice
Test your choice of probability method with original sample-space, Venn, histogram, independence, parameter, tree and multi-stage problems with worked solutions.
Before you startThe preceding probability lessons; revisit a method when a question exposes a gap.
01 / Decide before calculating
Name the experiment, event and assumptions.
For each question: define the eligible outcomes, choose a rule, then check the result.
Try the question before opening a hint or worked solution.
These questions use fresh numbers. A correct numerical answer needs a valid probability model and a clear interpretation. There is no timer; move between sections as needed.
Choose a method, then checkExplore
Two independent fair dice are rolled. How would you find the probability of sum 8?
Check method and assumption
Count the five qualifying ordered pairs among36 equally likely pairs. Probability5/36. Independence and fairness justify equal joint probabilities.
02 / Sample-space choices
Derived totals need not be equally likely.
02 · Equal totals?
A learner assigns probability 1/8 to each possible sum 2–9. Explain the error using two sum values.
Hint
Compare the number of underlying pairs.
Worked solution
Sum2 has only(1,1), probability 1/20, while sum 6 has four pairs, probability 1/5. The eight sums are not equally likely.
03 / Two-event regions
The inclusive union counts the overlap once.
03 · At least one club
In 80 pupils, club A only has 17, both clubs 9, club B only 24 and neither 30. Find P(at least one club).
Hint
Add the three inside regions.
Worked solution
(17+9+24)/80=50/80=5/8.
04 · Exactly one club
For the same population, find P(exactly one club).
Hint
Exclude the overlap.
Worked solution
(17+24)/80=41/80.
05 · Not both
Find the probability a pupil is not in both clubs.
Hint
Complement the intersection, not the union.
Worked solution
1−9/80=71/80. This includes the pupils in neither club.
04 / Three-event regions
Keep pairwise totals separate from pairwise-only counts.
A population of 60 has disjoint regions: A only 12, B only 15, C only 8, AB only 5, AC only 4, BC only 6, ABC 3 and none 7.
06 · At least two sets
Find P(at least two sets).
Hint
Include the triple region.
Worked solution
(5+4+6+3)/60=18/60=3/10.
07 · Pair intersection
Find P(A∩B).
Hint
The pairwise intersection includes all three.
Worked solution
(5+3)/60=8/60=2/15.
08 · Exactly one set
Find P(exactly one set).
Hint
Use only the three single-only regions.
Worked solution
(12+15+8)/60=35/60=7/12.
05 / Exact counts or estimates?
State what the grouped records can determine.
09 · Threshold inside a class
Fifty recorded waits have classes 0≤t<5:10,5≤t<15:30,15≤t<25:10. Estimate P(t<10) for uniform random selection of one record.
Hint
Use half the middle class and state the assumption.
Worked solution
Assuming uniform spread in 5–15, estimated qualifying count=10+(5/10)×30=25, so estimated probability 1/2.
10 · Threshold at a boundary
Find P(t<15) from the same records. Is this an interpolation estimate?
Hint
Two whole classes qualify.
Worked solution
(10+30)/50=4/5, exact for uniform selection from these 50 records. No within-class assumption is needed; future prediction would be a separate modelling step.
06 / Test the relationship
Neither overlapping nor disjoint is a substitute for the product test.
11 · Known intersection
P(A)=0.45,P(B)=0.4,P(A∩B)=0.18. Test independence and find P(A∪B).
Hint
Compare the intersection with the product first.
Worked solution
0.45×0.4=0.18, so independent. Union=0.45+0.4−0.18=0.67.
12 · Impossible exclusivity
Can events with probabilities 0.6 and 0.5 be mutually exclusive?
Hint
The union cannot exceed1.
Worked solution
No. Their intersection is at least0.6+0.5−1=0.1. A zero overlap would produce union1.1.
07 / Probability algebra
Constraints come from all four regions.
13 · Parameter domain
P(A)=4k,P(B)=2k,P(A∩B)=k. Find the valid range of k.
Hint
Write A only,both,B only,neither.
Worked solution
The regions are3k,k,k,1−5k. All are nonnegative exactly when0≤k≤1/5.
14 · Independent case
For that model, find the positive k making the events independent.
Hint
Use k=(4k)(2k).
Worked solution
k=8k² gives k=0 or 1/8. The requested positive solution is 1/8, which lies within[0,1/5].
08 / With and without replacement
Explain each second-stage probability.
15 · Replacement
A bag has 2 red and 3 blue counters. Draw uniformly twice with replacement and independent mixing. Find P(different colours).
Hint
Add RB and BR.
Worked solution
(2/5)(3/5)+(3/5)(2/5)=12/25.
16 · No replacement
Repeat the preceding experiment without replacement. Find P(different colours) and compare.
Hint
Use denominator4 on each second branch.
Worked solution
(2/5)(3/4)+(3/5)(2/4)=12/20=3/5. This exceeds 12/25 by 3/25. The second-stage composition depends on the first colour.
09 / Weight the groups
A group rate alone is not a joint probability.
Watch: add the weighted reject paths
Pause, replay or seek freely. The notes explain the same idea and stay in view.
17 · Two suppliers
A supplier A provides 70% of parts with reject rate 2%; supplier B provides 30% with reject rate 8%. Using these model rates, find the overall reject probability and the fraction of rejected parts from B.
Hint
Multiply each supplier share by its reject rate, then condition on rejection.
Worked solution
Reject probability = 0.7 × 0.02 + 0.3 × 0.08 = 0.014 + 0.024 = 0.038. Among rejects, B fraction=0.024/0.038=12/19. This is not the same as B’s 8% reject rate.
10 / A first success needs earlier failures
Specify the stopping event, not just a later result.
18 · First on the third
Independent identical attempts succeed with probability 0.3. Find the probability the first success is on attempt 3, and the probability of at least one success within 3 attempts.
Hint
First-third is FFS; within-three complements FFF.
Worked solution
First-third=0.7²×0.3=0.147. Within-three=1−0.7³=0.657. These answer different questions.
11 / Use mistakes to choose the next revision
Locate the step that caused the difficulty.
Revisit sample spaces for equal-likelihood errors, Venn diagrams for overlapping events, independence for unjustified products, and trees for changing conditional probabilities. Keep exact finite-record answers separate from estimates and future-model claims.
Section 1 of 11 · Decide before calculating