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Independent events probability practice
For independent events, learning that one occurred does not change the probability of the other. Independence is a condition to justify, not a rule to assume for every problem.
By ExamCat · Original study lesson · Published September 24, 2026
Understand the method
If A and B are independent, P(A and B) = P(A)P(B). With sampling without replacement, probabilities often change after the first draw. Distinguish an intersection, where both happen, from a union, where at least one happens.
Worked example, step by step
Flip a fair coin and roll a fair six-sided die independently. Find the probability of heads and an even number.
- P(heads) = 1/2.
- Three of the six die faces are even, so P(even) = 3/6 = 1/2.
- Multiply for the independent intersection: 1/2 × 1/2 = 1/4.
- There are twelve equally likely coin-die outcomes and three favorable ones, providing a second check.
A mistake to watch for
Mutually exclusive events are not the same as independent events. Two positive-probability events that cannot happen together are dependent: knowing one occurred rules out the other.
Try it yourself: three questions
Write an answer before opening the explanation. For calculations, keep your units and intermediate steps; for concepts, state the reason for your choice.
1. Find the probability of two heads in two independent fair flips.
Show answer and explanation
1/2 × 1/2 = 1/4.
2. A bag has 3 red and 2 blue balls. Draw two without replacement. Find P(both red).
Show answer and explanation
3/5 × 2/4 = 3/10. The second probability changes because the first red ball is no longer in the bag.
3. On one fair die roll, can rolling 1 and rolling 2 be independent?
Show answer and explanation
No. They are mutually exclusive, and knowing the roll is 1 makes the probability of 2 zero.
Build your next practice session
Mark the first step where your reasoning differed from the explanation. Turn that step into a question, answer it without looking, then change one condition in the example and explain what would change. A correct guess is a reason to revisit the method.
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