Probability
Probability measures how likely an event is, as a number from 0 (impossible) to 1 (certain). When all outcomes are equally likely, it is the number of favourable outcomes divided by the total number of outcomes. Always list the sample space first and check that the outcomes you count are equally likely.
Formulas and when they apply
- P(event) = favourable outcomes ÷ total outcomesApplies when: all outcomes equally likely
- P(not A) = 1 − P(A)
- P(A or B) = P(A) + P(B) − P(A and B)Applies when: for events that cannot happen together, P(A and B) = 0
- P(A and B) = P(A) × P(B)Applies when: A and B are independent, for example separate tosses or draws with replacement
- P(A given B) = P(A and B) ÷ P(B)Applies when: P(B) is not zero
- Two dice have 36 outcomes. A standard pack has 52 cards: 26 red, 26 black, 4 of each rank
- Drawing r items together from n: total ways = nCr; favourable ways are counted with combinations in the same way
Worked examples
Example 1. Two fair dice are rolled. What is the probability that the sum is 7 or 11?
Solution. Sum 7 has 6 outcomes: (1,6), (2,5), (3,4), (4,3), (5,2), (6,1). Sum 11 has 2 outcomes: (5,6) and (6,5). Together 8 of 36 outcomes, so the probability is 8/36 = 2/9.
Example 2. A bag has 6 red and 4 white balls. Two balls are drawn one after the other without replacement. What is the probability that both are red?
Solution. First red: 6/10. Then 5 red remain among 9 balls: 5/9. Both red = 6/10 × 5/9 = 30/90 = 1/3.
Example 3. A fair coin is tossed 3 times. What is the probability of getting at least one head?
Solution. The opposite event is no heads at all (TTT), which has probability (1/2)³ = 1/8. So P(at least one head) = 1 − 1/8 = 7/8.
Shortcuts and their limits
- For 'at least one', work out the probability of 'none' and subtract it from 1.Does not apply when: Needs independent trials or a clearly countable 'none' case.
- For a sum on two dice, remember the counts of outcomes for sums 2 to 12: 1, 2, 3, 4, 5, 6, 5, 4, 3, 2, 1 out of 36.Does not apply when: Only for two ordinary six-sided dice.
- Draws made 'together' can be treated as draws one after another without replacement.Does not apply when: Valid when the order of the draws does not matter in the question.
Common mistakes
- Treating the 11 possible sums of two dice as equally likely. The 36 outcomes (pairs) are equally likely.
- Adding the probabilities of overlapping events without subtracting the overlap.
- Treating draws without replacement as independent.
- Dividing by the whole sample space when a condition has already shrunk it.
- Giving a probability greater than 1, or forgetting to simplify the fraction.
Independent practice made by EduVistaHQ with original questions. It is not an official examination paper, it is not affiliated with or endorsed by any examination body or company, and a practice score does not predict a real result.