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.A box holds 5 blue pens and 7 black pens. Two pens are taken at random, one at a time, and are not put back. The first pen taken is black. Work out the probability that the second pen taken is blue.
- 2.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.
- 3.A box contains 9 red balls and 11 green balls. Two balls are taken out at random, one after the other, without being replaced. Given that both balls taken out are the same colour, work out the probability that both balls are red.
- 4.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.
- 5.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.
- 6.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.
- 7.A ferry company finds that on 20% of days the sea is rough. If the sea is rough, the probability that a crossing is delayed is 0.75. If the sea is calm, the probability that a crossing is delayed is 0.1. Given that a crossing was delayed, work out the probability that the sea was rough that day.
- 8.At a garden centre, 3/5 of the plants for sale are perennials. 1/4 of the perennials are in flower. Work out the probability that a plant picked at random from the garden centre is a perennial and is in flower.
- 9.In a class of 30 pupils, 18 study French, 14 study German and 7 study both French and German. A pupil who studies French is picked at random. Work out the probability that this pupil also studies German.
- 10.In a year group of 60 pupils, 25 play football, 18 play tennis and 8 play both football and tennis. A pupil who plays football or tennis or both is picked at random. Work out the probability that this pupil plays both sports.
- 11.In a survey, 120 adults were asked whether they have a driving licence. 70 of the adults are women and 50 are men. 45 of the women and 35 of the men have a driving licence. One of the adults who has a driving licence is picked at random. Work out the probability that this adult is a man.
- 12.Two ordinary fair dice are rolled and the two scores are added together. Given that the total is an even number, work out the probability that the total is 8.
- 13.A drawer contains 9 black socks and 4 white socks. Three socks are taken out at random, one after another, without being replaced. Given that at least two of the three socks taken out are black, work out the probability that all three are black.
- 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.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.
Answer key
- (d) 5/11 — Method: the pen already taken was black, so update the contents of the box before working out the second probability. Working: the box held 12 pens and one black pen has gone, so 11 pens remain. None of the blue pens has been taken, so all 5 are still there, and the probability is 5/11, which will not cancel. Answer: the probability is 5/11. The distractors: 5/12 uses the box as it was at the start, which is only correct if the first pen is put back; 4/11 takes one off the blue count as well as the total, as though the pen removed had been blue; 6/11 gives the probability that the second pen is black, carrying on with the colour of the first pen instead of the colour asked for.
- (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) 36/91 — Method: P(both red | same colour) = P(both red) ÷ P(same colour), where P(same colour) = P(both red) + P(both green). Working: P(both red) = 9/20 × 8/19 = 72/380 = 18/95. P(both green) = 11/20 × 10/19 = 110/380 = 11/38. P(same colour) = 18/95 + 11/38 = 36/190 + 55/190 = 91/190. P(both red | same colour) = (36/190) ÷ (91/190) = 36/91. Answer: 36/91. Watch out: stopping at 18/95 gives P(both red) itself, without dividing by the probability that the colours matched at all. Working out 55/91 finds the same-colour probability for green instead of red — check which colour's count you are putting on top. And 9/20 is just the chance the first ball drawn is red, which ignores the second draw and the without-replacement condition completely.
- (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.
- (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.
- (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) 15/23 — Method: find P(rough and delayed) and the overall P(delayed) using the tree, then divide. Working: P(rough and delayed) = 0.2 × 0.75 = 0.15. P(calm and delayed) = 0.8 × 0.1 = 0.08. P(delayed) = 0.15 + 0.08 = 0.23. P(rough | delayed) = 0.15 ÷ 0.23 = 15/23. Answer: 15/23. Watch out: leaving the answer as 0.15 (3/20) gives P(rough and delayed) itself, without dividing by the overall probability that a crossing is delayed. Giving 0.75 (3/4) is the probability you were told to start with — that a crossing is delayed GIVEN the sea is rough — which is the reverse of what's being asked. And 0.2 (1/5) is just the original probability that the sea is rough, before you take the fact that the crossing was delayed into account.
- (c) 3/20 — Method: the second fraction is quoted for the perennials only, so it is a conditional probability and the two fractions multiply. Working: the probability that a plant is a perennial is 3/5, and given that it is a perennial the probability that it is in flower is 1/4. Multiplying gives 3 × 1 over 5 × 4, which is 3/20. Answer: the probability is 3/20. The distractors: 17/20 comes from adding the fractions, 12/20 plus 5/20, instead of multiplying, which would be right only for two outcomes that cannot both happen; 4/9 comes from adding the numerators and the denominators separately, the classic 3 + 1 over 5 + 4; 1/4 quotes the flowering fraction on its own, as though every plant in the garden centre were a perennial, so the 3/5 is never used.
- (d) 7/18 — Method: the pupil picked is known to study French, so the sample space shrinks to the 18 French students; divide the number who study both languages by 18. Working: 7 of the pupils study both French and German, and all 7 of them are among the 18 French students, so the probability is 7/18, which will not cancel. Answer: the probability is 7/18. The distractors: 7/30 divides by the whole class, keeping the restricted numerator but the full denominator; 1/2 is 7/14, which conditions on the German students instead, answering the probability that a German student also studies French; 7/25 uses 18 + 14 minus 7 = 25, the number who study at least one language, which is a larger group than the one the question restricts you to.
- (b) 8/35 — Method: the pupil picked is known to play at least one of the two sports, so first count how many pupils that is, then divide the number who play both by it. Working: 25 play football and 18 play tennis, but the 8 who play both have been counted in each figure, so the number who play at least one sport is 25 + 18 minus 8, which is 35. The pupils who play both give 8/35, which will not cancel. Answer: the probability is 8/35. The distractors: 2/15 is 8/60, dividing by the whole year group instead of by the 35 pupils who play at least one sport; 8/43 uses 25 + 18 as the denominator, forgetting that the 8 pupils who play both have been counted twice; 8/25 conditions on the footballers alone, answering the probability that a footballer also plays tennis rather than using every pupil who plays a sport.
- (a) 7/16 — Method: the adult picked is known to have a driving licence, so the sample space is everyone with a licence; divide the number of men with a licence by that total. Working: 45 women and 35 men have a licence, so 80 adults have one. The men with a licence give 35/80, and dividing the numerator and the denominator by 5 gives 7/16. Answer: the probability is 7/16. The distractors: 7/10 is 35/50, the probability that an adult has a licence given that he is a man, which is the condition and the event the wrong way round; 7/24 is 35/120, dividing by all 120 adults surveyed instead of by the 80 who have a licence; 5/12 is 50/120, the probability that an adult picked from the whole survey is a man, which uses none of the licence information the question supplies.
- (c) 5/18 — Method: knowing the total is even cuts the 36 equally likely outcomes down to the even ones, so count those first and then count how many of them give 8. Working: the even totals occur as 2 once, 4 three times, 6 five times, 8 five times, 10 three times and 12 once, which is 18 outcomes. The total is 8 for 2 and 6, 3 and 5, 4 and 4, 5 and 3, and 6 and 2, which is 5 outcomes. The probability is 5/18, which will not cancel. Answer: the probability is 5/18. The distractors: 5/36 keeps the right count of ways to make 8 but divides by all 36 outcomes, ignoring the fact that the odd totals have already been ruled out; 1/6 treats the six even totals 2, 4, 6, 8, 10 and 12 as equally likely and picks one of them, which they are not; 1/11 treats the eleven possible totals from 2 to 12 as equally likely and uses neither the counting nor the condition.
- (c) 7/19 — Method: 'at least two black' covers two cases — all three black, and exactly two black. Work out the probability of each along a tree, add them, then use P(all three black | at least two black) = P(all three black) ÷ P(at least two black). Working: P(all three black) = 9/13 × 8/12 × 7/11 = 504/1716 = 42/143. For exactly two black, one order is black, black, white = 9/13 × 8/12 × 4/11 = 288/1716; the white sock could be drawn first, second or third, so there are 3 such orders, giving 3 × 288/1716 = 864/1716 = 72/143. P(at least two black) = 42/143 + 72/143 = 114/143. P(all three black | at least two black) = (42/143) ÷ (114/143) = 42/114 = 7/19. Answer: 7/19. Watch out: stopping at 42/143 gives the unconditioned probability that all three are black — it ignores that you already know at least two of them are. Dividing by the 'exactly two black' probability on its own gives 7/12, and forgets that the all-black outcomes are themselves part of the 'at least two black' group, so they must be inside the denominator, not left out of it. And 7/11 answers a different, easier question — the probability the THIRD sock is black given the FIRST TWO specifically are black — not 'at least two of the three, in any order, are black'.
- (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.
- (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.
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