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.
Higher only
Conditional probability worksheet — GCSE Higher
MathsUKwww.geekhero.co.uk
- 1.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.
- 2.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.
- 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.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.
- 5.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.
- 6.For two events A and B, P(A) = 0.6 and P(A and B) = 0.15. Work out P(B | A).
- 7.250 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.
- 8.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.
- 9.Of the 70 students on the Year 10 geography trip, 42 are girls and the rest are boys. 24 of the girls and 10 of the boys brought a packed lunch. A student who did not bring a packed lunch is chosen at random. Work out the probability that this student is a girl.
- 10.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.
- 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.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.
- 13.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.
- 14.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.
- 15.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.
Answer key
- (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.
- (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.
- (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) 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) 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.
- (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.
- (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.
- (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.
- (c) 1/2 — Method: restrict to the students who did NOT bring a packed lunch, then find what fraction of that group are girls. Girls without lunch = 42 − 24 = 18. Boys = 70 − 42 = 28, so boys without lunch = 28 − 10 = 18. Total without lunch = 18 + 18 = 36. Working: P(girl | no lunch) = 18 ÷ 36 = 1/2. Answer: 1/2. Watch out: dividing 18 by 42 (the total number of girls) instead of by 36 finds P(no lunch | girl), the reverse conditional. Dividing by 70 (the whole trip) ignores that you already know the student did not bring a lunch. And using the 'brought a lunch' numbers (24 out of 34) answers the question for the wrong group entirely — you were asked about the students who did NOT bring one.
- (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.
- (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.
- (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.
- (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.
- (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'.
- (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.
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