WAEC / WASSCE
WAEC Probability Practice Questions, Solved Step by Step
WAEC probability questions stay close to a few setups: picking balls from a bag with or without replacement, throwing two dice, independent events like two students passing an exam, and choosing numbers with a property such as prime. This page solves five original questions in that style, with answers left as fractions the way WAEC expects.
Probability on the WASSCE theory paper is narrower than the textbook chapter suggests. The same handful of situations rotate through the years: a bag of coloured balls, two dice, two independent people passing or failing, numbers picked from a range. The arithmetic is short, so the marks really test whether you can classify the situation, with replacement or without, independent or not, and or or. The five questions below are original, written in WASSCE style at the exam's difficulty. I have kept every answer as a fraction, which is the safest form in a WAEC paper unless the question asks for a decimal.
The probability setups WAEC repeats
Learn four templates and you have covered most papers. One: a single draw from a bag, probability is favourable outcomes over total outcomes. Two: two dice or a die thrown twice, where the 36-outcome sample space answers everything, and questions ask for a sum or at least conditions. Three: two draws without replacement, where the denominator drops by one on the second draw and you must add the branches that satisfy the event. Four: independent events, two candidates passing an exam, where you multiply along and add across, and the phrase exactly one means two products added. A fifth, smaller template hides algebra inside: a bag holds x red balls, the probability of red is given, find x.
Presenting answers so the marks stick
Write the sample space size before any fraction; for two dice, state that there are 36 equally likely outcomes, because that line often carries a mark on its own. For without replacement questions, a tree diagram is the cheapest insurance available: each correctly labelled branch pair earns method credit even if a final addition slips. Leave answers as fractions, simplified if easy, though WAEC generally accepts unsimplified correct fractions. Two checks cost nothing: every probability must sit between 0 and 1, and where your cases cover all possibilities, both pass, exactly one passes, neither passes, the total must be exactly 1. If it is not, one branch is wrong, and the check tells you before the examiner does.
Worked questions, step by step
A bag contains 5 red, 3 blue and 4 green balls. A ball is picked at random. Find the probability that it is (a) red, (b) not green.
- Total number of balls: 5 plus 3 plus 4 equals 12.
- P(red) equals 5/12.
- Balls that are not green: 5 plus 3 equals 8, so P(not green) equals 8/12 equals 2/3.
- Check: P(green) equals 4/12 equals 1/3, and 1 minus 1/3 equals 2/3. Consistent.
Answer: (a) 5/12 (b) 2/3
Where marks slip: B1 for the total of 12, A1 for each probability. The complement route, 1 minus P(green), earns the same marks and is faster; either way, show the total before the fraction.
Try one yourself: A bag has 6 red, 4 blue and 2 white balls. Find the probability of picking a blue ball. [Answer: 1/3]
Two fair dice are thrown once. Find the probability that (a) the sum of the scores is 8, (b) the sum is at least 10.
- There are 6 times 6 equals 36 equally likely outcomes.
- Sums of 8: (2,6), (3,5), (4,4), (5,3), (6,2), which is 5 outcomes, so P equals 5/36.
- Sums of at least 10: (4,6), (5,5), (6,4), (5,6), (6,5), (6,6), which is 6 outcomes.
- P(at least 10) equals 6/36 equals 1/6.
Answer: (a) 5/36 (b) 1/6
Where marks slip: B1 for the 36-outcome sample space, then M1 A1 per part for listing or counting outcomes. Forgetting that (3,5) and (5,3) are different outcomes is the error this question exists to punish.
Try one yourself: Two fair dice are thrown once. Find the probability that the sum is 7. [Answer: 1/6]
A box contains 4 white and 6 black identical balls. Two balls are drawn one after the other without replacement. Find the probability that (a) both are black, (b) they are of different colours.
- First draw black: 6/10. Second draw black, with one black gone: 5/9.
- P(both black) equals 6/10 times 5/9 equals 30/90 equals 1/3.
- Different colours happens two ways: white then black is 4/10 times 6/9 equals 24/90, and black then white is 6/10 times 4/9 equals 24/90.
- P(different colours) equals 24/90 plus 24/90 equals 48/90 equals 8/15.
Answer: (a) 1/3 (b) 8/15
Where marks slip: M1 for reducing both numerator and denominator on the second draw, A1 for 1/3; then M1 for adding both orders and A1 for 8/15. Counting white-black but not black-white halves the answer and loses the A mark.
Try one yourself: A bag has 3 white and 5 black balls. Two are drawn without replacement. Find the probability that both are black. [Answer: 5/14]
The probability that Kofi passes an examination is 3/4 and the probability that Ama passes the same examination is 2/3. Their results are independent. Find the probability that (a) both pass, (b) exactly one of them passes, (c) neither passes.
- P(both pass) equals 3/4 times 2/3 equals 6/12 equals 1/2.
- Exactly one passes: Kofi passes and Ama fails, 3/4 times 1/3 equals 3/12, or Kofi fails and Ama passes, 1/4 times 2/3 equals 2/12.
- P(exactly one) equals 3/12 plus 2/12 equals 5/12.
- P(neither) equals 1/4 times 1/3 equals 1/12.
- Check: 1/2 plus 5/12 plus 1/12 equals 12/12 equals 1. Consistent.
Answer: (a) 1/2 (b) 5/12 (c) 1/12
Where marks slip: M1 A1 for each part, with the failing probabilities 1/4 and 1/3 shown explicitly. The three cases must total 1, and writing that check line can rescue you if a branch is wrong.
Try one yourself: P(A passes) = 2/5 and P(B passes) = 1/2, independently. Find the probability that both pass. [Answer: 1/5]
A number is chosen at random from the integers 10 to 30 inclusive. Find the probability that the number is (a) a prime number, (b) a multiple of 4.
- Count the integers from 10 to 30 inclusive: 30 minus 10 plus 1 equals 21.
- Primes in the range: 11, 13, 17, 19, 23, 29, which is 6 numbers.
- P(prime) equals 6/21 equals 2/7.
- Multiples of 4 in the range: 12, 16, 20, 24, 28, which is 5 numbers, so P equals 5/21.
Answer: (a) 2/7 (b) 5/21
Where marks slip: B1 for the count of 21, then A1 per part with the lists shown. The inclusive count is the trap: writing 20 instead of 21 shifts every answer, and listing the primes visibly protects your method credit.
Try one yourself: A number is picked at random from the integers 1 to 20 inclusive. Find the probability that it is a multiple of 3. [Answer: 3/10]
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Solve my question →Quiz me on this topicQuestions students ask
Should probability answers be fractions or decimals?
Either is accepted unless the question specifies, but fractions are safer because they are exact. If you give a decimal, use at least the accuracy the question asks for elsewhere, and never round a probability to something above 1 or below 0.
What is the difference between with replacement and without replacement?
With replacement, the first ball goes back, so the second draw has the same probabilities. Without replacement, both the numerator for that colour and the total drop by one on the second draw. Misreading this one phrase is the single biggest source of lost marks on these questions.
Are these actual WAEC probability past questions?
No. They are original questions written in the WASSCE format at matching difficulty, because WAEC's own past papers are copyrighted. The setups, bags, dice, independent candidates, number ranges, are the same families the real paper draws from.