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Conditional probability questions
Every probability lives in a sample space, and new information shrinks it. Conditional probability is the arithmetic of that shrinkage. Given that B happened, the world is B now, and everything gets remeasured against it.
6 original questions · 20 marks · the conditional probability notes · Statistics
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P(A ∩ B) = 0.18 and P(B) = 0.6. Find P(A|B), and state what P(A) would have to equal for A and B to be independent.
Worked answer
P(A|B) = P(A ∩ B)/P(B) = 0.18/0.6 = 0.3, the overlap measured as a share of B's world rather than of the whole. Independence means conditioning on B changes nothing, so P(A) would have to be 0.3 as well. M1 for the conditional formula, A1 for 0.3, B1 for the value P(A) would need under independence. Dividing P(A ∩ B) = P(A)P(B) through by P(B) turns the product form of independence into this conditional form, so the two definitions say one thing.P(B) = 0.4 and P(A|B) = 0.25. Find P(A ∩ B).
Worked answer
Multiply along the branch. P(A ∩ B) = P(B) × P(A|B) = 0.4 × 0.25 = 0.1. M1 for the product along the branch, A1 for 0.1. This is the conditional formula rearranged, and it is what every tree diagram does. The second-stage probabilities on a tree are already conditional on the first stage, so multiplying along a path gives the probability of the whole path.A bag holds 4 red and 3 blue counters. Two are drawn without replacement. Find the probability that the second counter is blue, and the probability that both are red.
Worked answer
Two routes end in a blue second draw. Red then blue is 4/7 × 3/6 = 12/42, and blue then blue is 3/7 × 2/6 = 6/42. Adding them gives 18/42 = 3/7. Both red is a single route, 4/7 × 3/6 = 12/42 = 2/7. The second branch carries a denominator of 6 because a counter has already gone, and that shift is what 'without replacement' means. M1 for the products along the branches, M1 for adding the two routes, A1 for the first probability, A1 for the second.A bag holds 4 red and 3 blue counters, and two are drawn without replacement. Find the probability that the first counter was red, given that the second is blue.
Worked answer
P(first red | second blue) = P(red then blue)/P(second blue) = (12/42)/(18/42) = 12/18 = 2/3. Conditioning on the later event to ask about the earlier one feels backwards, but the formula has no sense of time. It compares an overlap with a whole and nothing more. M1 for the conditional formula, A1 for the correct pair of probabilities, A1 for 2/3. The denominator has to be P(second blue). Putting P(first red) underneath instead is the standard slip and answers a different question.Of 100 gym members, 60 are adults and 40 are juniors. 45 adults and 10 juniors attend classes. Find P(attends | adult) and P(adult | attends).
Worked answer
Reading along the adult row, P(attends | adult) = 45/60 = 3/4. The attenders number 45 + 10 = 55, so P(adult | attends) = 45/55 = 9/11. B1 for the first probability, M1 for the 55 attenders, A1 for the second. Same overlap of 45 members, two different denominators. The event written after the bar decides which world the fraction lives in, and swapping the two earns no marks even though the numerator is identical.A factory has two machines. Machine A makes 60% of the components and 3% of them are faulty. Machine B makes the rest, and 8% of those are faulty. A component is chosen at random and found to be faulty. Find the probability that it came from machine B.
Worked answer
Draw the tree with the machine first, then condition. P(A and faulty) = 0.6 × 0.03 = 0.018 and P(B and faulty) = 0.4 × 0.08 = 0.032, so P(faulty) = 0.018 + 0.032 = 0.05. Then P(B | faulty) = 0.032/0.05 = 0.64. M1 A1 for the two products, M1 for adding them to reach the total probability of a fault, M1 for dividing the right product by that total, A1 for 0.64. Look at what the answer says. Machine B makes only 40% of the components yet accounts for 64% of the faulty ones, because the evidence has reweighted the branches. A patient who reads a 99% accurate positive test as a 99% chance of illness has run the same calculation backwards, quoting P(positive | ill) when the question asks for P(ill | positive). The direction of the bar fixes the denominator, and the denominator fixes the answer.
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