01 · A different overlap
If the same marginals have overlap 0.12, are the events independent?
Hint
Compare 0.12 with 0.6×0.4.
Worked solution
No. 0.12≠0.24. Also P(B given A)=0.12/0.6=0.2, which differs from 0.4.
Understand · explore · practise
Independent events
Test independence using the intersection product, calculate probabilities with complements, and distinguish an assumed model from evidence in observed data.
Before you startVenn regions, the addition rule and complements.
01 / Compare the intersection with the product
An unchanged chance has a precise probability test.
A and B are independent when P(A ∩ B)=P(A)P(B).
When P(A)>0, this is equivalent to P(B given A)=P(B).
The “given A” proportion is P(A∩B)/P(A): restrict the eligible outcomes to A. Independence says that this restriction leaves the probability of B unchanged. If either event has probability 0, use the product definition rather than divide by 0.
Change the overlapExplore
Keep P(A)=0.6 and P(B)=0.4 fixed. Adjust P(A∩B) and compare it with the product 0.24.
At overlap 0, the four region probabilities are A only 0.6, both 0, B only 0.4, neither 0. These events are mutually exclusive and dependent because0≠0.24.
02 / Test before multiplying
Known marginal probabilities alone do not prove independence.
Watch: only one overlap gives independence
Pause, replay or seek freely. The notes explain the same idea and stay in view.
P(A)=0.6, P(B)=0.4 and P(A∩B)=0.24.Worked example
P(A)P(B)=0.6×0.4=0.24
This matches the known intersection, so the events are independent.
P(B given A)=0.24/0.6=0.4
The probability of B is unchanged within A.
The four regions are 0.36,0.24,0.16,0.24
They are nonnegative and sum to 1.
02 · Missing overlap
Only P(A)=0.6 and P(B)=0.4 are given. Can you calculate P(A∩B) as 0.24?
Hint
What additional information is required?
Worked solution
Only if independence is stated or justified. The marginals alone allow any overlap from 0 to 0.4.
03 / Independence also works with complements
Change an event without changing the logic.
If A and B are independent:
P(A ∩ B′)=P(A)(1−P(B))
P(A′ ∩ B′)=(1−P(A))(1−P(B)).
These identities follow by subtracting regions from the known totals.
03 · A only
Using the independent 0.6 and0.4 model, find P(A∩B′).
Hint
Subtract the overlap from P(A), or multiply by P(B′).
Worked solution
0.6−0.24=0.36. Equivalently 0.6×0.6=0.36.
04 · Neither
Find P(A′∩B′).
Hint
Both complementary events must occur.
Worked solution
0.4×0.6=0.24. The complements are independent too.
05 · Exactly one
Find P(exactly one of A and B).
Hint
Add A only and B only.
Worked solution
0.36+0.16=0.52. Equivalently 0.6×0.6+0.4×0.4.
04 / At least one is often a complement
Independent does not mean add without correction.
Two independent alarms sound with probabilities 0.7 and0.8.Worked example
Neither sounds: 0.3×0.2=0.06
Use the complementary probabilities.
At least one sounds: 1−0.06=0.94
This includes the possibility that both sound.
Both sound:0.7×0.8=0.56
The addition rule also gives 0.7+0.8−0.56=0.94.
06 · Exactly one alarm
Find the probability exactly one sounds.
Hint
One sounds and the other does not, in either order.
Worked solution
0.7×0.2+0.3×0.8=0.14+0.24=0.38.
07 · State the assumption
Why might independence be questionable for two alarms in the same room?
Hint
Could they share a cause or failure mechanism?
Worked solution
Shared power, environment or faults can link their outcomes. The calculations are valid under an independent-alarm model; physical separation or a verbal story alone does not prove the assumption.
05 / Find an unknown probability
Use a stated independence relationship.
08 · Missing marginal
Independent events satisfy P(A)=0.35 and P(A∩B)=0.14. Find P(B).
Hint
Solve 0.35×P(B)=0.14.
Worked solution
P(B)=0.14/0.35=0.4.
09 · Missing complement
Independent events satisfy P(A)=0.3 and P(A′∩B′)=0.42. Find P(B).
Hint
P(A′)=0.7.
Worked solution
0.7(1−P(B))=0.42, so 1−P(B)=0.6 and P(B)=0.4.
06 / A table can test a finite model
Observed association and population independence are distinct.
10 · Uniform record selection
Among 100 records, 40 have A,50 have B and20 have both. For uniform selection of one record, are A and B independent?
Hint
Compare 20/100 with(40/100)(50/100).
Worked solution
Yes for this finite selection model:0.2=0.4×0.5.
11 · Generalising the table
Does that table prove independence in the wider population that produced the records?
Hint
Observed frequencies vary from sample to sample.
Worked solution
No. The exact finite-record calculation does not prove a population property. Sampling variation, collection methods and other evidence matter.
07 / Know which independence is given
Two events and several events require careful wording.
12 · Exclusive events
Events with probabilities 0.2 and0.3 are mutually exclusive. Are they independent?
Hint
Compare their intersection with the product.
Worked solution
No. The intersection probability is 0 but the product is 0.06. Both probabilities are positive.
13 · Pairwise warning
For more than two events, is pairwise independence alone enough to multiply all their probabilities for a joint event?
Hint
Pairwise statements concern only pairs.
Worked solution
No. A model requiring several events together needs the appropriate mutual independence. Do not extend a two-event test beyond the information given.
08 / Use the product only with a justified model
Independent means the probability test holds.
Compare the intersection with the product when testing independence. When independence is given, calculate intersections and complementary intersections by multiplication, then use the addition or complement rule for unions. State assumptions and avoid treating sample proportions as proof of a wider property.
Section 1 of 8 · Compare the intersection with the product