Printable · GCSE Higher · ages 14-16
Conditional probability worksheet — GCSE Higher
Fifteen questions on "conditional probability" — DfE statement P9. Print it, or print three versions so neighbours cannot copy by letter; the key gives the letter for each version.
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Conditional probability worksheet — GCSE Higher
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- 1.At a sports club, the numbers of members who play tennis and badminton are: 50 members play tennis, 40 members play badminton, and 18 members play both tennis and badminton. A member who plays tennis is chosen at random. Work out the probability that this member also plays badminton.
- 2.Two fair six-sided dice are rolled and the two scores are added together. Given that at least one of the dice shows a 5, work out the probability that the total is 8.
- 3.240 people were asked whether they had been to the cinema in the last month. 100 of the people are under 30 years old and 140 are aged 30 or over. 65 of the under 30s and 35 of those aged 30 or over had been to the cinema. One of the people aged 30 or over is picked at random. Work out the probability that this person had been to the cinema.
- 4.At a sports centre, 45% of the members are aged under 18. 60% of the members aged under 18 swim each week. 20% of the members aged 18 or over swim each week. Work out the percentage of all the members who swim each week.
- 5.A factory makes bolts on two machines. Machine A makes 60% of the bolts and machine B makes the other 40%. 5% of the bolts made by machine A are faulty and 10% of the bolts made by machine B are faulty. A bolt is picked at random from one day's production. Work out the probability that it was made by machine B and is faulty. Give your answer as a decimal.
- 6.A test for a medical condition is given to 1000 people. 50 of the people have the condition and 950 do not. The test is positive for 45 of the 50 people who have the condition, and it is also positive for 95 of the 950 people who do not have the condition. One of the people whose test is positive is picked at random. Work out the probability that this person has the condition.
- 7.For two events A and B, P(A) = 0.6 and P(A and B) = 0.15. Work out P(B | A).
- 8.A call centre finds that 40% of its calls are about billing. 35% of the calls about billing are dealt with in under five minutes. The centre takes 500 calls on Monday. Work out how many of Monday's calls you would expect to be about billing and dealt with in under five minutes.
- 9.A doctors' surgery has 400 patients. 3 in every 10 of the patients are over 65 years old. 90 of the patients over 65 and 70 of the patients aged 65 or under had a flu jab. One of the patients who had a flu jab is picked at random. Work out the probability that this patient is over 65.
- 10.250 people took a theory test at one test centre. 150 of them had taken a preparation course and the rest had not. 120 of those who had taken the course passed and 50 of those who had not taken the course passed. One of the people who passed is picked at random. Work out the probability that this person had taken the preparation course.
- 11.In a survey of 200 households, 120 have a garden and 80 own a dog. 54 of the households have a garden and own a dog. Work out the probability that a household owns a dog given that it has a garden, and compare it with the probability that a household picked from the whole survey owns a dog.
- 12.A bag contains 4 red sweets and 6 yellow sweets. Two sweets are taken at random, one after the other, and are not put back. The first sweet taken is red. Work out the probability that the second sweet taken is also red.
- 13.In a group of 50 people, 32 own a car, 20 own a bicycle and 9 own both a car and a bicycle. A person who owns a bicycle is picked at random. Work out the probability that this person does not own a car.
- 14.A group of 80 pupils was asked whether they walk to school. 45 of the pupils are girls and 35 are boys. 30 of the girls walk to school and 14 of the boys walk to school. One of the girls is picked at random. Work out the probability that she walks to school. Give your answer in its simplest form.
- 15.In a certain town, P(rain on Monday) = 0.3. If it rains on Monday, the probability that it also rains on Tuesday is 0.6. If it does not rain on Monday, the probability that it rains on Tuesday is 0.25. Work out the probability that it rains on Tuesday.
Answer key
- (a) 9/25 — Method: P(badminton | tennis) = n(tennis and badminton) ÷ n(tennis) — restrict to the tennis-players, then find what fraction of them also play badminton. Working: n(tennis and badminton) = 18, n(tennis) = 50, so P(badminton | tennis) = 18/50 = 9/25. Answer: 9/25. Watch out: dividing by 40 (the badminton total) finds P(tennis | badminton) instead of P(badminton | tennis) — the wrong direction. Dividing by 90 (all the members named in the question) ignores that you already know the member plays tennis. And dividing by 72 (50 + 40 − 18, the number who play at least one of the two sports) answers a question about the union, not the condition you were given.
- (b) 2/11 — Method: restrict the 36 equally likely outcomes to those where at least one die shows a 5, then find what fraction of THOSE give a total of 8. Working: outcomes with at least one 5: (5, 1), (5, 2), (5, 3), (5, 4), (5, 5), (5, 6), (1, 5), (2, 5), (3, 5), (4, 5), (6, 5) — 11 outcomes. Among these, the total is 8 for (5, 3) and (3, 5) — 2 outcomes. P(total 8 | at least one 5) = 2/11. Answer: 2/11. Watch out: 5/36 is P(total 8) over the full 36 outcomes — it ignores that you already know one die shows a 5. Treating the condition as 'the first die shows a 5' instead of 'at least one die shows a 5' restricts you to only 6 outcomes and misses the (3, 5) case, giving 1/6. And counting only (5, 3) but not its reverse (3, 5) inside the correct 11-outcome list gives 1/11 instead of 2/11.
- (b) 1/4 — Method: the person picked is known to be aged 30 or over, so the sample space is those 140 people; divide the number of them who had been to the cinema by 140. Working: 35 of the 140 people aged 30 or over had been to the cinema, giving 35/140. Dividing the numerator and the denominator by 35 gives 1/4. Answer: the probability is 1/4. The distractors: 7/20 is 35/100, taking the count from the older group but the total from the under 30s, which is reading across the wrong row; 7/48 is 35/240, dividing by everyone surveyed instead of by the age group named; 3/4 is 105/140, the probability that someone aged 30 or over had NOT been to the cinema, the opposite event inside the correct group.
- (b) 38% — Method: the two swimming percentages are quoted inside different age groups, so weight each one by the size of its group and add the two results. Working: the under 18s are 45% of the members and 60% of them swim, giving 0.45 × 60 = 27% of all the members. The members aged 18 or over are 55% of the members and 20% of them swim, giving 0.55 × 20 = 11% of all the members. Adding these gives 38%. Answer: 38% of the members swim each week. The distractors: 80% comes from adding 60% and 20% straight off, treating two rates quoted inside different groups as though they could be added; 40% is the mean of 60% and 20%, which would be right only if the two age groups were the same size, and they are not; 42% comes from pairing each swimming rate with the wrong age group, working out 0.45 × 20 added to 0.55 × 60.
- (c) 0.04 — Method: 'made by machine B and faulty' is the second branch of a tree followed after the first, so multiply the probability of machine B by the probability of a fault given machine B. Working: machine B makes 0.4 of the bolts, and 0.1 of those bolts are faulty, so the probability is 0.4 × 0.1 = 0.04. Answer: the probability is 0.04. The distractors: 0.5 comes from adding 0.4 and 0.1 instead of multiplying, treating two stages of one journey as two separate outcomes; 0.1 gives the fault rate for machine B on its own, as though every bolt in the factory came from machine B, so the 40% share is never used; 0.07 is 0.6 × 0.05 added to 0.4 × 0.1, the probability that a bolt is faulty whichever machine made it, which answers a question about all the production rather than about machine B.
- (c) 9/28 — Method: two linked steps. Total everyone whose test is positive, since the person picked is known to be one of them, then divide the positive tests that belong to people with the condition by that total. Working: 45 positive tests come from people who have the condition and 95 come from people who do not, so 140 tests are positive. The people with the condition give 45/140, and dividing the numerator and the denominator by 5 gives 9/28. Answer: the probability is 9/28. The distractors: 9/10 is 45/50, the probability of a positive test given that the person has the condition, which is the condition and the event the wrong way round and is the figure a candidate quotes when the two are confused; 9/200 is 45/1000, dividing by everyone tested rather than by the 140 who tested positive; 1/20 is 50/1000, the probability that a person has the condition before the test result is used at all.
- (c) 0.25 — Method: P(B | A) = P(A and B) ÷ P(A). Working: P(B | A) = 0.15 ÷ 0.6 = 0.25. Answer: 0.25. Watch out: multiplying 0.6 by 0.15 instead of dividing gives 0.09, and subtracting 0.15 from 0.6 gives 0.45 — neither uses the conditional probability formula. Leaving the answer as 0.15 mistakes the probability of A and B happening together for the probability of B once you already know A has happened — those are different quantities.
- (a) 70 — Method: two linked steps. Find the expected number of billing calls first, then take the 35% of those, because the 35% is quoted for billing calls only. Working: 40% of 500 is 200 billing calls. 35% of 200 is 70 calls. Answer: you would expect 70 calls. The distractors: 200 stops after the first step and gives the billing calls, forgetting that only some of them are dealt with quickly; 175 is 35% of 500, applying the quick response rate to every call the centre takes rather than to the billing calls only; 375 comes from adding 40% and 35% to get 75% and taking 75% of 500, which treats two stages of one journey as separate outcomes to be added.
- (c) 9/16 — Method: two steps. Total the patients who had a flu jab, since the patient picked is known to be one of them, then divide the over 65s who had a jab by that total. Working: 90 patients over 65 and 70 patients aged 65 or under had a jab, so 160 patients had one. The over 65s give 90/160, and dividing the numerator and the denominator by 10 gives 9/16. Answer: the probability is 9/16. The distractors: 7/16 is 70/160, the probability that the patient picked is aged 65 or under, which is the other part of the same restricted group; 3/4 is 90/120, the probability that a patient had a jab given that they are over 65, which is the condition and the event the wrong way round and needs the 120 patients over 65; 9/40 is 90/400, dividing by every patient on the list instead of by the 160 who had a jab.
- (d) 12/17 — Method: the person picked is known to have passed, so the sample space is everyone who passed; divide the course takers who passed by that total. Working: 120 course takers and 50 others passed, so 170 people passed. The course takers who passed give 120/170, and dividing the numerator and the denominator by 10 gives 12/17. Answer: the probability is 12/17. The distractors: 4/5 is 120/150, the probability that someone passed given that they took the course, which reverses the condition and the event; 12/25 is 120/250, dividing by everyone who sat the test rather than by the 170 who passed; 17/25 is 170/250, the probability that a person picked from everyone sitting the test passed, which answers a different question altogether.
- (a) 0.45, different from 0.4 for all the households — Method: work out the probability inside the restricted group of garden owners, then work out the probability across the whole survey, and compare the two. Working: 54 of the 120 households with a garden own a dog, so the conditional probability is 54 divided by 120, which is 0.45. Across the whole survey 80 of the 200 households own a dog, which is 0.4. Since 0.45 is not 0.4, having a garden changes the chance of owning a dog and the two events are not independent. Answer: 0.45, different from 0.4 for all the households. The distractors: 0.27 is 54/200, dividing the households with both by the whole survey instead of by the 120 with a garden; 0.675 is 54/80, the probability that a household has a garden given that it owns a dog, which is the condition and the event the wrong way round; 0.4 is 80/200, the probability of owning a dog with the garden information never used, which is why that route also reports no difference.
- (b) 1/3 — Method: the first sweet has already been taken and it was red, so work out the second probability from what is actually left in the bag. Working: one red sweet has gone, so 3 red sweets remain out of 9 sweets altogether, giving 3/9. Dividing the numerator and the denominator by 3 gives 1/3. Answer: the probability is 1/3. The distractors: 2/5 is 4/10, the probability for the first draw used again, which is only right if the first sweet is put back; 3/10 takes one off the red count but leaves the total at 10, updating half of the fraction; 4/9 takes one off the total but leaves the red count at 4, updating the other half of the fraction.
- (b) 11/20 — Method: the person picked is known to own a bicycle, so work inside the 20 bicycle owners and count how many of them do not own a car. Working: 9 of the 20 bicycle owners also own a car, so 11 of them do not. The probability is 11/20, which will not cancel. Answer: the probability is 11/20. The distractors: 9/20 gives the bicycle owners who DO own a car, answering the opposite event inside the correct group; 11/50 divides by the whole group of 50, keeping the restricted numerator but the full denominator; 11/32 puts the count over the number of car owners, conditioning on the wrong group entirely.
- (d) 2/3 — Method: the pupil picked is known to be a girl, so the sample space is the 45 girls and not all 80 pupils; divide the number of girls who walk by the number of girls. Working: 30 of the 45 girls walk to school, which gives 30/45. Dividing the numerator and the denominator by 15 gives 2/3. Answer: the probability is 2/3. The distractors: 3/8 is 30/80, dividing the girls who walk by every pupil in the group, which is the commonest slip on a conditional probability because it keeps the restricted numerator but the whole denominator; 1/3 is 15/45, counting the 15 girls who do not walk to school and so answering the opposite event inside the correct group; 11/20 is 44/80, adding the 30 girls and the 14 boys who walk and dividing by the whole group, which throws away the information that the pupil picked is a girl.
- (b) 0.355 — Method: use the law of total probability across the two Monday branches: P(rain Tue) = P(rain Mon) × P(rain Tue | rain Mon) + P(no rain Mon) × P(rain Tue | no rain Mon). Working: P(no rain Mon) = 1 − 0.3 = 0.7. P(rain Tue) = (0.3 × 0.6) + (0.7 × 0.25) = 0.18 + 0.175 = 0.355. Answer: 0.355. Watch out: using only the rain-Monday branch (0.3 × 0.6) or only the no-rain-Monday branch (0.7 × 0.25) accounts for just one of the two ways Tuesday can turn out rainy — both branches must be added. And swapping which weekday-probability multiplies which branch (0.7 with the rain branch, 0.3 with the no-rain branch) uses the right numbers on the wrong branches.
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