9.3 Probability: Compound, Independent & Conditional Events
Key Takeaways
- Basic probability P(A) = favorable outcomes / total outcomes, always in [0, 1]; P(impossible) = 0, P(certain) = 1.
- Complement rule: P(not A) = 1 − P(A) — the fastest path for 'at least one' problems.
- Mutually exclusive events: P(A or B) = P(A) + P(B). Independent events: P(A and B) = P(A) × P(B). Do not confuse the two.
- Conditional probability: P(A | B) = P(A and B) / P(B), defined only when P(B) > 0; the denominator is the reduced sample space.
- With replacement keeps probabilities constant across draws; without replacement changes the denominator on each draw — track the remaining count carefully.
Probability: Compound, Independent & Conditional Events
Quick Answer: GRE probability reduces to four rules: P(A) = favorable/total, the complement rule P(not A) = 1 − P(A), the OR rule (add when mutually exclusive), and the AND rule (multiply when independent). Conditional probability P(A | B) = P(A and B)/P(B) shrinks the sample space to event B. Master 'at least one' via the complement — it is the single highest-yield probability trick on the GRE.
Basic Probability
For an event A in a sample space S with equally likely outcomes:
P(A) = (number of outcomes in A) / (number of outcomes in S)
Probabilities always lie in [0, 1]. P = 0 means impossible; P = 1 means certain.
Worked Example — Single Die
P(roll a 4 on a fair six-sided die) = 1/6. P(roll even) = 3/6 = 1/2 (outcomes 2, 4, 6). P(roll greater than 4) = 2/6 = 1/3 (outcomes 5, 6).
The Complement Rule
P(not A) = 1 − P(A)
This is the fastest tool for "at least one" problems, because direct calculation requires summing many cases while the complement is a single subtraction.
Worked Example — At Least One via Complement
A fair coin is flipped 4 times. P(at least one head) = ?
- Direct: P(1 head) + P(2 heads) + P(3 heads) + P(4 heads) — tedious.
- Complement: P(no heads) = P(all tails) = (1/2)⁴ = 1/16.
- P(at least one head) = 1 − 1/16 = 15/16.
This pattern — "at least one X in n trials" — almost always rewards the complement.
Mutually Exclusive Events — The OR Rule
Two events are mutually exclusive (disjoint) if they cannot both occur: P(A and B) = 0.
P(A or B) = P(A) + P(B) (only when mutually exclusive)
If events are NOT mutually exclusive, use the general OR rule:
P(A or B) = P(A) + P(B) − P(A and B)
The subtraction avoids double-counting the overlap.
Worked Example — Mutually Exclusive vs. Overlap
Draw one card from a standard 52-card deck.
- P(King or Queen) = 4/52 + 4/52 = 8/52 = 2/13 (mutually exclusive — a card cannot be both).
- P(King or Heart) = 4/52 + 13/52 − 1/52 = 16/52 = 4/13 (overlap: the King of Hearts is both).
Independent Events — The AND Rule
Two events are independent if the occurrence of one does not affect the probability of the other:
P(A and B) = P(A) × P(B) (only when independent)
Worked Example — Independent Dice
Roll two fair six-sided dice. P(both show 6) = (1/6) × (1/6) = 1/36. P(first is even AND second is greater than 4) = (3/6) × (2/6) = 6/36 = 1/6.
Independent vs. mutually exclusive is a top GRE trap. Independent events can both occur (they are unrelated); mutually exclusive events cannot. If P(A) > 0 and P(B) > 0 and the events are mutually exclusive, they are NOT independent (knowing A occurred means B did not).
Conditional Probability
P(A | B) = P(A and B) / P(B), defined when P(B) > 0.
The intuition: knowing that B occurred shrinks the sample space to B. Within that reduced space, we ask what fraction also satisfies A.
Worked Example — Conditional with Cards
Draw one card. P(it is a King | it is a face card) = ?
- Face cards = Jack, Queen, King of each suit → 12 face cards.
- Kings among face cards = 4.
- P(King | face card) = 4/12 = 1/3.
Using the formula: P(King and face card) = 4/52; P(face card) = 12/52; ratio = (4/52)/(12/52) = 4/12 = 1/3. Same result.
With vs. Without Replacement
- With replacement: drawn item is returned; probabilities stay constant across draws. Events are independent.
- Without replacement: drawn item is removed; denominators shrink and probabilities change. Events are dependent.
Worked Example — Marbles Without Replacement
A bag has 3 red and 2 blue marbles. Draw two without replacement. P(both red) = ?
- First draw: 3/5 red.
- Second draw (given first was red): 2/4 red.
- P(both red) = (3/5) × (2/4) = 6/20 = 3/10.
With replacement: (3/5) × (3/5) = 9/25, a different answer.
Compound Events — Strategy Checklist
- Identify the events and the connectors (AND, OR, "at least one").
- Determine independence or mutual exclusivity.
- Decide on the rule: AND → multiply (if independent); OR → add (if mutually exclusive); at least one → complement.
- Track replacement status carefully.
- Reduce fractions; confirm the result lies in [0, 1].
Worked Example — Putting It All Together
A bag has 4 red, 3 blue, and 2 green marbles. Draw two without replacement. P(both same color) = ?
- P(both red) = (4/9) × (3/8) = 12/72.
- P(both blue) = (3/9) × (2/8) = 6/72.
- P(both green) = (2/9) × (1/8) = 2/72.
- These are mutually exclusive, so add: 12/72 + 6/72 + 2/72 = 20/72 = 5/18.
Common GRE Probability Traps
- Confusing P(A and B) with P(A or B). "And" requires overlap or independence; "or" requires addition.
- Forgetting to subtract the overlap in the general OR rule.
- Using P(A) + P(B) when events are not mutually exclusive.
- Treating without-replacement draws as independent.
- Computing 'at least one' directly instead of via the complement.
A fair six-sided die is rolled once. What is the probability of rolling a number that is either even or greater than 4?
A fair coin is flipped 5 times. What is the probability of getting at least one tail?
A bag contains 5 red marbles and 3 blue marbles. Two marbles are drawn without replacement. What is the probability that both are blue?
A standard 52-card deck is shuffled. One card is drawn. What is P(King | face card)?
Which of the following pairs of events are mutually exclusive but NOT independent?