The AND rule of probability applies when you want two events to both happen; the OR rule applies when you want at least one of them to happen. Knowing which to use — and when to subtract the overlap — is the central challenge of combined-events questions at GCSE. Both rules follow directly from careful counting of outcomes.
What is the AND rule for independent events?
Two events are independent when the outcome of one does not affect the probability of the other. For independent events A and B:
P(A and B) = P(A) × P(B)
Worked example: A fair coin is flipped and a fair six-sided die is rolled. What is the probability of getting heads AND a 4?
- P(heads) = 1/2.
- P(4) = 1/6.
- P(heads and 4) = 1/2 × 1/6 = 1/12.
The multiplication reflects the fact that every outcome of the coin pairs with every outcome of the die: there are 2 × 6 = 12 equally likely outcomes, and only one is (heads, 4).
What is the AND rule for dependent events?
When events are dependent — the outcome of the first affects the probability of the second — you must use a conditional probability:
P(A and B) = P(A) × P(B | A)
where P(B | A) is the probability of B given that A has already happened.
Worked example: A bag contains 4 red and 3 blue counters. Two counters are drawn without replacement. Find the probability that both are red.
- P(first red) = 4/7.
- After one red is removed: 3 red and 3 blue remain → P(second red | first red) = 3/6 = 1/2.
- P(both red) = 4/7 × 1/2 = 2/7.
What is the OR rule for mutually exclusive events?
Two events are mutually exclusive when they cannot both happen at the same time. For mutually exclusive events A and B:
P(A or B) = P(A) + P(B)
Worked example: A standard six-sided die is rolled. What is the probability of getting a 2 or a 5?
- P(2) = 1/6, P(5) = 1/6.
- These are mutually exclusive (you cannot roll both at once).
- P(2 or 5) = 1/6 + 1/6 = 2/6 = 1/3.
The addition reflects the count of favourable outcomes: 2 is favourable, 5 is favourable, and they do not overlap.
What is the general OR rule when events are not mutually exclusive?
When events CAN occur together, simply adding their probabilities double-counts the overlap. The correct formula is:
P(A or B) = P(A) + P(B) − P(A and B)
Worked example: In a class of 30 students, 18 play football (F) and 12 play netball (N). 7 play both. A student is chosen at random. Find P(F or N).
- P(F) = 18/30, P(N) = 12/30, P(F and N) = 7/30.
- P(F or N) = 18/30 + 12/30 − 7/30 = 23/30.
Without subtracting the overlap, the 7 students who play both would be counted twice.
When do you add and when do you multiply?
This is the key decision in every combined-probability question:
| Signal in the question | Rule | Operation |
|---|---|---|
| "Both … and …" | AND rule | Multiply |
| "Neither … nor …" | AND rule (complements) | Multiply complements |
| "At least one of …" | OR rule or complement | Add (or 1 − P(none)) |
| "Either … or …" | OR rule | Add (and subtract overlap if not exclusive) |
The word "and" maps to multiplication; the word "or" maps to addition. When events are not mutually exclusive, do not forget to subtract the joint probability.
How do you combine AND and OR in a single question?
More complex GCSE questions may require both rules. Build the answer step by step:
Worked example: A bag has 5 red, 3 green, and 2 blue beads. Two beads are drawn with replacement. Find the probability of at least one red bead.
There are 10 beads in total, of which 5 are not red. So P(not red on one draw) = 5/10 = 1/2.
- P(at least one red) = 1 − P(no red beads in either draw).
- P(no red on first draw) = 5/10 = 1/2.
- P(no red on second draw | no red on first) = 5/10 = 1/2 (same, because draws are with replacement).
- P(no red in either draw) = 1/2 × 1/2 = 1/4.
- P(at least one red) = 1 − 1/4 = 3/4.
This complement approach — computing P(none) and subtracting from 1 — is almost always more efficient than listing every case where at least one red appears.
Frequently asked questions
How do I know if two events are independent?
Ask: does knowing the outcome of event A change the probability of event B? If the answer is no, they are independent. Coin flips, die rolls, and draws with replacement are all independent. Draws without replacement, drawing cards after not replacing them, and conditional scenarios are usually dependent.
Can I use tree diagrams for AND/OR problems?
Yes — tree diagrams are an excellent visual tool for these rules. For the AND rule, multiply along the branches; for the OR rule, add the probabilities of all the branch-ends (leaves) that satisfy the condition. Tree diagrams are particularly helpful for problems with two stages, such as drawing two items from a bag.
What if probabilities are given as decimals instead of fractions?
The AND and OR rules work identically with decimals. P(A and B) = P(A) × P(B) regardless of whether probabilities are expressed as fractions, decimals, or percentages. The only care needed is that all probabilities for mutually exclusive outcomes must add up to 1 (or 100%).
Does the order matter in the AND rule?
For independent events, no: P(A and B) = P(B and A) because multiplication is commutative. For dependent events, the order does matter in terms of which conditional probability you use. P(A first, then B) uses P(B | A); P(B first, then A) would use P(A | B). In "without replacement" questions, always track which event happens first in the sequence.
For guided probability practice at GCSE with Professor Pi, visit aitutors.me.