The General Addition Law - Worksheets, Questions and Revision

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H26 The General Addition Law

EDEXCEL 1ST0 · Calculator allowed · about 90 minutes

Total Marks

Name: _______________________________ Date: ____ / ____ / ______

Answer ALL questions. Show all your working.

For each statement about the general addition law, write down whether it is True or False.

(i)The general addition law can be written as P(A or B) = P(A) + P(B) - P(A and B).(1)

(ii)If two events A and B are mutually exclusive, then P(A and B) = 1.(1)

(iii)For any two events A and B, P(A or B) = P(A) + P(B) always gives the correct probability.(1)

(Total for Question 1 is 3 marks)

Use the general addition law, P(A or B) = P(A) + P(B) - P(A and B), to answer each part.

(a)Events A and B are such that P(A) = 0.6, P(B) = 0.35 and P(A and B) = 0.2. Find P(A or B).(2)

(b)Events C and D are mutually exclusive, with P(C) = 0.28 and P(D) = 0.55. Find P(C or D).(2)

(Total for Question 2 is 4 marks)

As part of a customer survey, a bakery asked 120 customers whether they bought bread and whether they bought milk that day. The two-way table shows the results.

	Milk	No milk	Total
Bread	42	18	60
No bread	24	36	60
Total	66	54	120

A customer is chosen at random from the 120 surveyed. Let event A be "the customer bought bread" and event B be "the customer bought milk".

(a)Write down P(A and B).(1)

(b)Write down P(A) and P(B).(2)

(c)Use the general addition law to find P(A or B), then confirm your answer by working directly from the table.(2)

(Total for Question 3 is 5 marks)

The Venn diagram shows the number of people, out of 30 surveyed, in events A and B.

(a)Write down P(A and B).(1)

(b)Use the general addition law to find P(A or B).(2)

(c)Verify your answer to part (b) by counting directly from the Venn diagram.(1)

(Total for Question 4 is 4 marks)

A fair 10-sided spinner is numbered 1 to 10 and spun once. Event A: the spinner lands on an even number. Event B: the spinner lands on a multiple of 5. Event C: the spinner lands on a multiple of 3.

(a)List the outcomes in "A and B", and hence write down P(A and B).(1)

(b)Use the general addition law to find P(A or B).(2)

(c)Explain, with reference to the outcomes, whether events A and C are mutually exclusive.(1)

(Total for Question 5 is 4 marks)

A counter is drawn at random from a bag. Event R: the counter is red, with P(R) = 0.35. Event Y: the counter is yellow, with P(Y) = 0.4. A counter cannot be both red and yellow.

(a)Write down P(R and Y).(1)

(b)Find P(R or Y).(1)

(c)Explain why, for mutually exclusive events, the general addition law simplifies to P(A or B) = P(A) + P(B).(1)

(Total for Question 6 is 3 marks)

For two events A and B, P(A) = 0.52, P(B) = 0.3 and P(A or B) = 0.66.

(a)Rearrange the general addition law to make P(A and B) the subject.(1)

(b)Hence find P(A and B).(2)

(Total for Question 7 is 3 marks)

A school surveyed 200 students on whether they study French and whether they study Spanish. The two-way table shows the results, with one value missing. Let event A be "the student studies French" and event B be "the student studies Spanish".

	Spanish	No Spanish	Total
French	34	46	80
No French	?	94	120
Total	60	140	200

(a)Work out the missing value in the table, marked ?.(2)

(b)A student is chosen at random from the 200 students. Write down P(A and B).(1)

(c)Use the general addition law to find P(A or B), the probability that the student studies French or Spanish (or both).(2)

(Total for Question 8 is 5 marks)

In a class of 40 students, every student studies Art, Drama, both, or neither. The Venn diagram shows the number of students in each region, in terms of x. Let event A be "studies Art" and event B be "studies Drama".

(a)Find the value of x.(2)

(b)Hence find P(A) and P(B).(2)

(c)Use the general addition law to find P(A or B).(2)

(Total for Question 9 is 6 marks)

A bag contains 5 red counters and 3 blue counters. Sofia takes a counter at random, replaces it, then takes a second counter at random. The tree diagram shows the probabilities for the two draws. Event A: the first counter is red. Event B: the second counter is red.

(a)Use the tree diagram to find P(A and B), the probability that both counters are red.(2)

(b)Given that P(A) = P(B) = 5/8, use the general addition law to find P(A or B).(2)

(c)Hence find P(neither counter is red).(1)

(Total for Question 10 is 5 marks)

For two events A and B, P(A) = 0.4, P(B | A) = 0.25 and P(B) = 0.3.

(a)Find P(A and B).(2)

(b)Use the general addition law to find P(A or B).(2)

(Total for Question 11 is 4 marks)

A student is investigating whether there is a link, for Year 11 pupils, between walking to school and attending the breakfast club. She first considers surveying only pupils who arrive at school before 8am, but decides against this and instead surveys a random sample of 90 pupils from the whole year group. The two-way table shows the results of her survey. Let event A be "the pupil walks to school" and event B be "the pupil attends the breakfast club".

	Breakfast club	No breakfast club	Total
Walks	21	24	45
Does not walk	9	36	45
Total	30	60	90

(a)Explain why surveying only pupils who arrive before 8am, as she first considered, might not give a fair test of her question.(1)

(b)Find P(A), P(B) and P(A and B).(2)

(c)Use the general addition law to find the probability that a pupil chosen at random walks to school or attends the breakfast club (or both).(2)

(d)Interpret your answer to part (c) in the context of the survey.(1)

(Total for Question 12 is 6 marks)

For two events A and B, P(A or B) = 0.72, P(A) = 0.5 and P(A and B) = 0.18.

(a)Rearrange the general addition law to make P(B) the subject.(1)

(b)Hence find P(B).(2)

(Total for Question 13 is 3 marks)

In a fitness club with 60 members, every member uses the swimming pool, the running track, both, or neither. The Venn diagram shows the number of members in each region, in terms of x. Let event A be "uses the swimming pool" and event B be "uses the running track".

(a)Find the value of x.(2)

(b)Find P(A and B).(1)

(c)Use the general addition law to find P(A or B). Give your answer as a fraction in its simplest form.(2)

(d)One member is chosen at random from those outside both A and B. State the probability that this member is in event A.(1)

(Total for Question 14 is 6 marks)

Use the general addition law to decide whether each pair of events is mutually exclusive, showing your working.

(a)For events A and B, P(A) = 0.45, P(B) = 0.3 and P(A or B) = 0.75. Determine whether A and B are mutually exclusive.(3)

(b)For events C and D, P(C) = 0.4, P(D) = 0.35 and P(C or D) = 0.68. Determine whether C and D are mutually exclusive.(2)

(Total for Question 15 is 5 marks)

Two events A and B are such that P(A or B) = P(A) + P(B). Show that A and B must be mutually exclusive.

(Total for Question 16 is 3 marks)

For two events A and B, P(A) = x, P(B) = x + 0.1, P(A and B) = 0.05 and P(A or B) = 0.65.

(a)Form an equation in x using the general addition law.(1)

(b)Solve your equation to find the value of x.(2)

(c)Hence write down P(A) and P(B).(2)

(Total for Question 17 is 5 marks)

A box contains 10 tickets: 6 marked WIN and 4 marked LOSE. Two tickets are drawn at random, one after another, without replacement. The tree diagram shows the probabilities for the two draws. Event A: the first ticket is WIN. Event B: the second ticket is WIN.

(a)Write down P(A).(1)

(b)Use the tree diagram to find P(A and B), the probability that both tickets are WIN.(2)

(c)Given that P(B) = 3/5, use the general addition law to find P(A or B).(2)

(d)Hence find P(neither ticket is WIN).(1)

(Total for Question 18 is 6 marks)

Mark scheme · H26 The General Addition Law

Question 1

(i) B1 True cao
(i) Answer: True
(ii) B1 False cao
(ii) Answer: False
(iii) B1 False cao
(iii) Answer: False

Question 2

(a) M1 0.6 + 0.35 - 0.2, oe
(a) A1 0.75 cao
(a) Answer: 0.75
(b) M1 0.28 + 0.55, oe (P(C and D) = 0 since mutually exclusive)
(b) A1 0.83 cao
(b) Answer: 0.83

Question 3

(a) B1 42/120 oe cao
(a) Answer: 7/20
(b) B1 P(A) = 60/120 oe cao
(b) B1 P(B) = 66/120 oe cao
(b) Answer: P(A) = 1/2, P(B) = 11/20
(c) M1 1/2 + 11/20 - 7/20, oe
(c) A1 7/10 oe cao, confirmed by (120 - 36)/120 = 84/120 = 7/10 from the table
(c) Answer: 7/10

Question 4

(a) B1 6/30 oe cao
(a) Answer: 1/5
(b) M1 15/30 + 16/30 - 6/30, oe (P(A) = (9+6)/30, P(B) = (6+10)/30)
(b) A1 5/6 oe cao
(b) Answer: 5/6
(c) B1 9 + 6 + 10 = 25, 25/30 = 5/6, ft from (b)
(c) Answer: 25/30 = 5/6

Question 5

(a) B1 {10} identified, P(A and B) = 1/10 cao
(a) Answer: 1/10
(b) M1 1/2 + 1/5 - 1/10, oe
(b) A1 3/5 oe cao
(b) Answer: 3/5
(c) B1 not mutually exclusive because 6 is both even and a multiple of 3 (A and C share the outcome 6)
(c) Answer: Not mutually exclusive; 6 is in both A and C.

Question 6

(a) B1 0 cao
(a) Answer: 0
(b) B1 0.75 cao (0.35 + 0.4 - 0)
(b) Answer: 0.75
(c) B1 mutually exclusive events cannot both occur, so P(A and B) = 0, and subtracting 0 from P(A) + P(B) leaves P(A) + P(B)
(c) Answer: Because P(A and B) = 0 for mutually exclusive events, so the term being subtracted disappears.

Question 7

(a) B1 P(A and B) = P(A) + P(B) - P(A or B), oe
(a) Answer: P(A and B) = P(A) + P(B) - P(A or B)
(b) M1 0.52 + 0.3 - 0.66, oe
(b) A1 0.16 cao
(b) Answer: 0.16

Question 8

(a) M1 60 - 34, or 120 - 94, oe
(a) A1 26 cao
(a) Answer: 26
(b) B1 34/200 oe cao
(b) Answer: 17/100
(c) M1 80/200 + 60/200 - 34/200, oe
(c) A1 53/100 oe cao
(c) Answer: 53/100

Question 9

(a) M1 x + 2x + (x + 5) + 7 = 40, oe
(a) A1 x = 7 cao
(a) Answer: x = 7
(b) B1 P(A) = 21/40 cao
(b) B1 P(B) = 26/40 oe cao
(b) Answer: P(A) = 21/40, P(B) = 13/20
(c) M1 21/40 + 26/40 - 14/40, oe
(c) A1 33/40 cao
(c) Answer: 33/40

Question 10

(a) M1 5/8 x 5/8, oe
(a) A1 25/64 cao
(a) Answer: 25/64
(b) M1 5/8 + 5/8 - 25/64, oe (40/64 + 40/64 - 25/64)
(b) A1 55/64 cao
(b) Answer: 55/64
(c) B1 9/64, ft as 1 - their (b), oe
(c) Answer: 9/64

Question 11

(a) M1 0.4 x 0.25, oe
(a) A1 0.1 cao
(a) Answer: 0.1
(b) M1 0.4 + 0.3 - 0.1, oe
(b) A1 0.6 cao
(b) Answer: 0.6

Question 12

(a) B1 valid reason, e.g. pupils arriving before 8am are more likely to walk and/or attend breakfast club, so the sample would not represent all Year 11 pupils (biased/unrepresentative sample)
(a) Answer: The sample would be biased, as pupils who arrive early are more likely to walk or attend the breakfast club than the whole year group.
(b) B1 P(A) = 45/90 and P(B) = 30/90, both oe
(b) B1 P(A and B) = 21/90 oe
(b) Answer: P(A) = 1/2, P(B) = 1/3, P(A and B) = 7/30
(c) M1 1/2 + 1/3 - 7/30, oe (15/30 + 10/30 - 7/30)
(c) A1 3/5 oe cao
(c) Answer: 3/5
(d) B1 correct interpretation in context, e.g. 3/5 (or 60%) of the 90 pupils surveyed either walk to school, attend the breakfast club, or do both
(d) Answer: 3/5 (60%) of the pupils surveyed walk to school, attend the breakfast club, or both.

Question 13

(a) B1 P(B) = P(A or B) - P(A) + P(A and B), oe
(a) Answer: P(B) = P(A or B) - P(A) + P(A and B)
(b) M1 0.72 - 0.5 + 0.18, oe
(b) A1 0.4 cao
(b) Answer: 0.4

Question 14

(a) M1 2x + (x + 4) + (3x - 2) + 10 = 60, oe
(a) A1 x = 8 cao
(a) Answer: x = 8
(b) B1 12/60 oe cao
(b) Answer: 1/5
(c) M1 28/60 + 34/60 - 12/60, oe
(c) A1 5/6 cao
(c) Answer: 5/6
(d) B1 0, with reason: outside both A and B means not in A
(d) Answer: 0

Question 15

(a) M1 0.45 + 0.3 - 0.75, oe
(a) A1 P(A and B) = 0
(a) B1 correct conclusion: A and B are mutually exclusive, since P(A and B) = 0, ft
(a) Answer: P(A and B) = 0, so A and B are mutually exclusive.
(b) M1 0.4 + 0.35 - 0.68, oe
(b) A1 P(C and D) = 0.07, so not mutually exclusive (since P(C and D) is not 0)
(b) Answer: P(C and D) = 0.07, so C and D are not mutually exclusive.

Question 16

M1 equates the general addition law to the given statement: P(A) + P(B) - P(A and B) = P(A) + P(B), oe
A1 rearranges to P(A and B) = 0
A1 correct conclusion: since P(A and B) = 0, events A and B cannot both occur, so they are mutually exclusive
Answer: P(A and B) = 0, so A and B are mutually exclusive.

Question 17

(a) B1 x + (x + 0.1) - 0.05 = 0.65, oe
(a) Answer: x + (x + 0.1) - 0.05 = 0.65
(b) M1 2x + 0.05 = 0.65, oe
(b) A1 x = 0.3 cao
(b) Answer: x = 0.3
(c) B1 P(A) = 0.3 cao
(c) B1 P(B) = 0.4 cao
(c) Answer: P(A) = 0.3, P(B) = 0.4

Question 18

(a) B1 3/5 oe cao
(a) Answer: 3/5
(b) M1 3/5 x 5/9, oe
(b) A1 1/3 cao
(b) Answer: 1/3
(c) M1 3/5 + 3/5 - 1/3, oe (9/15 + 9/15 - 5/15)
(c) A1 13/15 cao
(c) Answer: 13/15
(d) B1 2/15, ft as 1 - their (c), oe
(d) Answer: 2/15

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