The learning goal
Calculate P(A∩B) and P(A)P(B) separately: equality establishes independence. Mutually exclusive events share no outcomes; in this four-outcome positive-probability model, a zero intersection probability confirms that. Mutually exclusive events with positive individual probabilities are not independent because their intersection probability is zero while their product is positive. Retain that condition.
List the possible outcomes first
Assume two fair, independent coin flips, using H for heads and T for tails. Fairness and independence of the flips are model assumptions given by this exercise, not properties measured in a physical coin experiment. The four equally likely outcomes are HH, HT, TH and TT, each with probability 1/4.
Define A as first flip heads, B as second flip heads and C as first flip tails. A contains HH and HT; B contains HH and TH; C contains TH and TT. Write the sets before calculating probabilities so similar-sounding descriptions do not silently become the same event.
| Outcome | A: first H | B: second H | C: first T |
|---|---|---|---|
| HH | Yes | Yes | No |
| HT | Yes | No | No |
| TH | No | Yes | Yes |
| TT | No | No | Yes |
A and B can coincide and still be independent
Both A and B contain two outcomes, so each has probability 1/2. Their only shared outcome is HH, giving intersection probability 1/4. Their product is also (1/2)×(1/2)=1/4. Equality establishes independence for this pair in this model.
They are not mutually exclusive because HH allows both to occur. OpenStax’s definitions provide the two criteria. Apply them to the specified events rather than assuming that any events defined from two independent flips must themselves be independent.
Explain independent but not mutually exclusive
A and B share an outcome, so assume they influence one another and cannot be independent.
The shared HH outcome has probability 1/4, equal to P(A)P(B). Independence holds; mutual exclusivity does not.
A and C cannot coincide and are not independent here
The first flip cannot be both H and T, so A∩C is empty and its probability is zero. Yet P(A)P(C)=1/4. The unequal values make A and C mutually exclusive but not independent. Once you know the first flip is H, the probability that it is T becomes zero instead of 1/2.
Mutually exclusive events within an experiment can therefore be strongly dependent. Independent flips do not make every event combination defined from those flips independent. Evaluate the named pair. If conditional-probability direction is unclear, revisit the denominator and reverse-question example.
Change the question: at least one head versus exactly one
Now define D as at least one H and E as exactly one H. Write the sets without borrowing the previous answer: D={HH,HT,TH}; E={HT,TH}. Thus P(D)=3/4, P(E)=1/2, and the intersection is E with probability 1/2.
Their product is (3/4)×(1/2)=3/8, different from 1/2, so they are not independent. Their intersection is nonzero, so they are not mutually exclusive either. This pair satisfies neither property; the categories are not opposites that must divide every case.
| Pair | Intersection versus product | Conclusion |
|---|---|---|
| A and B | 1/4 = 1/4; nonzero intersection | Independent, not mutually exclusive |
| A and C | 0 ≠ 1/4; zero intersection | Mutually exclusive, not independent |
| D and E | 1/2 ≠ 3/8; nonzero intersection | Neither property holds |
Put the reason beside the calculation
Do not stop at a label. Record the intersection, the product and the equality decision beside each answer. When something is wrong, check the event sets first, the equal-likelihood assumption second, and fraction arithmetic afterwards.
Use the worked-example self-explanation method to explain why each set was chosen. The following record is an editorial study suggestion, not a claim that a fixed review duration guarantees better grades.
Experiment conditions and equal-likelihood evidence:
Outcomes in event A:
Outcomes in event B:
P(A), P(B):
P(A∩B):
P(A)P(B):
Independence decision and reason:
Mutual-exclusivity decision and reason:
What must be recalculated if an event description changes:Check that you learned more than the label
Without the table, explain why A and B can coincide yet be independent, and why replacing B with C changes the answer. Then redefine D as both flips heads and calculate its relationship with E again. A one-sentence change in an event requires a fresh calculation, rather than the previous label.
Understanding checks for this example
- List all four equally likely outcomes and identify the given fairness and independence assumptions.
- Calculate the intersection and product for all three event pairs.
- Explain independence and mutual exclusivity separately, without treating them as opposites.
- Retain the positive-probability condition and return to definitions for zero-probability events.
Common questions
Must independent events be mutually exclusive?
No. A and B are independent in this example, yet HH lets them occur together. Check the intersection and probability product separately.
Are mutually exclusive events always dependent?
When both have positive probability, yes. With a zero-probability event, the intersection and product may both be zero, so return to the definition rather than dropping that condition.
Can I count cells when outcomes are not equally likely?
Do not assume equal probability per cell. Sum the supplied outcome probabilities before comparing the intersection and product. The criteria must still be applied with their conditions.
