Favourable outcomes over total outcomes: When outcomes are equally likely, the probability of event A is the number of favourable outcomes ÷ the total number of outcomes.
Every probability lies between 0 and 1.
The sample space of two dice (36 outcomes): count the favourable squares over 36 to get the probability.
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IB-style question — single event
A bag holds 5 red, 3 blue and 2 green counters.
One is taken at random.
Find the probability it is red.
Step by step
- Total counters.
- Favourable ÷ total.
Final answer
P(red) = 1/2.
Probabilities never exceed 1: If you get a probability above 1 or below 0, you've made an error — recheck your counts.
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P(not A) = 1 − P(A): The complement of A is 'A does not happen', with P(A′) = 1 − P(A).
This is the fast route to 'at least one' questions: P(at least one) = 1 − P(none).
IB-style question — at least one
Two fair dice are rolled.
Find the probability of getting at least one six.
Step by step
- P(no six on one die) = 5/6, so P(no six on both).
- Complement.
Final answer
P(at least one six) = 11/36.
'At least one' → do the opposite: Counting every 'at least one' case is slow; 1 − P(none) is almost always faster.
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List every equally likely outcome in a grid: A sample space diagram (grid or list) shows every equally likely outcome.
For two dice there are 36 outcomes; count the ones matching the event and divide by 36.
IB-style question — two dice
Two fair dice are rolled and the scores added.
Find the probability the total is 7.
Step by step
- Total outcomes in the grid.
- Outcomes giving 7: (1,6),(2,5),(3,4),(4,3),(5,2),(6,1).
Final answer
P(total = 7) = 1/6.
Order matters in the grid: (2,5) and (5,2) are different cells — count ordered outcomes so the total is 36.
Multiply along the chain — and update the totals: For events one after another, multiply the probabilities along the way.
Without replacement, the counts change after each draw (one fewer item, and one fewer of that type).
IB-style question — without replacement
A box has 4 red and 6 blue pens.
Two are drawn without replacement.
Find the probability both are red.
Step by step
- First red: 4 of 10.
- Second red: now 3 of 9. Multiply.
Final answer
P(both red) = 2/15.
Reduce the totals each time: After the first draw there are 9 pens left, not 10 — forgetting to reduce both totals is the classic slip.