01 · A tempting bound
Why is k≤1/3, from P(A)≤1, insufficient?
Hint
Inspect the outside region.
Worked solution
For example k=1/3 gives neither=1−4/3=−1/3. The stronger bound is k≤1/4.
Understand · explore · practise
Probability parameters and proofs
Find valid parameter ranges, solve equations from independence and union information, reject impossible roots and prove identities using disjoint regions.
Before you startProbability rules, independent events, linear inequalities and quadratic equations.
01 / The algebra needs a probability domain
An equation may have roots that are not valid models.
Every disjoint region probability must be between 0 and 1, and their sum must be 1.
Checking only individual event probabilities can miss a negative outside region.
Write the regions first. They expose the constraints and make it easier to see which equation represents the stated relationship.
Test a parameterExplore
P(A)=3k, P(B)=2k, P(A∩B)=k. The regions must all be nonnegative.
At k=0, only the neither region has probability 1. The product definition of independence holds, although both events have probability 0.
02 / Find the feasible interval
Use nonnegative regions before solving.
Watch: the outside region sets the upper bound
Pause, replay or seek freely. The notes explain the same idea and stay in view.
P(A)=3k, P(B)=2k and P(A∩B)=k.Worked example
A only=2k; both=k; B only=k
Subtract the overlap from each marginal.
Neither=1−4k
The union is 2k+k+k=4k.
All regions are valid exactly when 0≤k≤1/4
Nonnegativity gives k≥0 and 1−4k≥0. Their sum is already 1.
02 · Boundary model
Give all four region probabilities when k=1/4.
Hint
Substitute into 2k,k,k,1−4k.
Worked solution
A only 1/2, both 1/4, B only 1/4, neither 0. The events are exhaustive but not mutually exclusive.
03 / Use the intersection equation
Keep zero solutions until the context rules them out.
Which valid k values make A and B independent?Worked example
k=(3k)(2k)=6k²
The intersection must equal the product.
k(6k−1)=0
Do not divide by k before considering k=0.
k=0 or k=1/6
Both lie in the feasible interval. If both event probabilities must be positive, only 1/6 remains.
03 · Nonzero solution
Find P(A),P(B) and P(A∪B) at k=1/6.
Hint
Use3k,2k,4k.
Worked solution
P(A)=1/2, P(B)=1/3, union 2/3. The intersection 1/6 equals(1/2)(1/3).
04 · Lost solution
What goes wrong if you immediately cancel k in k=6k²?
Hint
Cancellation assumes k is nonzero.
Worked solution
It loses the valid zero solution unless the problem explicitly states k>0 or positive event probabilities.
04 / Reject a root for a stated reason
Probability constraints are part of the solution.
Independent events have P(C)=p, P(D)=p+1/5 and P(C∪D)=19/25.Worked example
p+(p+1/5)−p(p+1/5)=19/25
Combine the union rule with independence.
p²−(9/5)p+14/25=0
Rearrange and factor as(p−2/5)(p−7/5)=0.
p=2/5; reject 7/5
Since0≤p≤4/5, the larger root is not a probability model. The valid D probability is 3/5.
05 · Verify the root
Check the union with p=2/5.
Hint
Calculate both marginals and their product.
Worked solution
2/5+3/5−(2/5)(3/5)=1−6/25=19/25.
06 · Why 4/5?
Explain the upper bound p≤4/5.
Hint
P(D)=p+1/5 cannot exceed1.
Worked solution
p+1/5≤1 gives p≤4/5. Together with p≥0 this defines the feasible interval.
05 / Bound an unknown intersection
The four-region ledger gives the general limits.
max(0, P(A)+P(B)−1) ≤ P(A∩B)
≤ min(P(A),P(B)).
The lower bound keeps neither nonnegative; the upper bound keeps both single-only regions nonnegative.
07 · Feasible overlap
If P(A)=0.7 and P(B)=0.5, find the possible range of overlap r.
Hint
Apply all four nonnegativity constraints.
Worked solution
0.2≤r≤0.5. The regions are 0.7−r,r,0.5−r,r−0.2.
08 · Impossible exclusivity
Can those events be mutually exclusive?
Hint
Zero would have to lie in the overlap interval.
Worked solution
No. The minimum overlap is 0.2, so they must share positive probability.
09 · Independence within the range
Is the independent overlap feasible for those marginals?
Hint
Compare 0.7×0.5 with the interval.
Worked solution
Yes. 0.35 lies in[0.2,0.5]; the regions are 0.35,0.35,0.15,0.15.
06 / Prove a complement result
Use the original assumption explicitly.
If A and B are independent, prove A and B′ are independent.Worked example
P(A∩B′)=P(A)−P(A∩B)
The two intersections split A into disjoint parts.
=P(A)−P(A)P(B)
Substitute the given independence assumption.
=P(A)(1−P(B))=P(A)P(B′)
This is exactly the independence criterion for A and B′.
10 · Both complements
Prove P(A′∩B′)=P(A′)P(B′) under the same assumption.
Hint
Complement the union and expand.
Worked solution
P(A′∩B′)=1−P(A∪B)=1−P(A)−P(B)+P(A)P(B)=(1−P(A))(1−P(B)).
07 / Distinguish identities from assumptions
An equality can describe a special case.
11 · Missing subtraction
In a finite model with positive probability for each outcome, P(A∪B)=P(A)+P(B). What does this imply?
Hint
Compare with the general addition rule.
Worked solution
P(A∩B)=0. Under the stated finite positive-outcome model this means the intersection is empty, so the events are mutually exclusive.
12 · Inequality proof
Prove P(A∪B)≥P(A).
Hint
Write the union as two disjoint pieces.
Worked solution
A∪B is the disjoint union of A and B∩A′. Thus P(A∪B)=P(A)+P(B∩A′)≥P(A), since probabilities are nonnegative.
13 · Validity is not independence
For k=1/12 in the first model, show it is valid but dependent.
Hint
Check the regions and the product separately.
Worked solution
Regions are 1/6,1/12,1/12,2/3, all valid. Intersection1/12 differs from(1/4)(1/6)=1/24, so the events are dependent.
08 / Sometimes both roots are valid
Do not discard a root just because there are two.
Independent events have probabilities p and1−p, with intersection3/16.Worked example
p(1−p)=3/16
Rearrange to p²−p+3/16=0.
(p−1/4)(p−3/4)=0
The two roots are1/4 and3/4.
Both roots lie between0 and1
Both give nonnegative regions: p−3/16,3/16,1−p−3/16,3/16. Nothing in the information chooses between them.
14 · Check both roots
List the four regions for p=1/4 and p=3/4.
Hint
The two single-only probabilities swap.
Worked solution
For1/4:1/16,3/16,9/16,3/16. For3/4:9/16,3/16,1/16,3/16. Both sets sum to1.
15 · Extra information
If the problem additionally says the first event is less likely than the second, which root remains?
Hint
Compare p with1−p.
Worked solution
p<1−p implies p<1/2, so p=1/4. The additional condition, not the quadratic alone, selects this root.
09 / Use parity to reduce integer outcomes
Classify outcomes before enumerating everything.
16 · Even sum and even product
Independently select two integers uniformly from1 to8, allowing repeats. Find the probability their sum and product are both even.
Hint
An even sum needs matching parity. Which matching case gives an even product?
Worked solution
Both odd gives an odd product, so both integers must be even. Each has probability1/2 of being even, giving(1/2)²=1/4. There are4×4=16 qualifying ordered pairs out of64.
17 · A certain integer event
Select an integer n uniformly from1 to8. Find the probability n(n+1) is even and justify it without eight separate products.
Hint
Consecutive integers have opposite parity.
Worked solution
One of n,n+1 is even, so their product is always even. Every possible n qualifies and the probability is1.
10 / Constrain, solve and verify
Probability conditions select valid algebraic solutions.
Build the disjoint regions, find the parameter domain, form the stated equation and solve without losing zero cases. Reject roots with an explicit probability constraint and substitute the valid result back. In proofs, state where the independence assumption is used.
Section 1 of 10 · The algebra needs a probability domain