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5.5 SL

1a. [2 marks]

Markscheme

valid approach (M1)

eg 0.30 − 0.1, + 0.1 = 0.3

= 0.2 A1 N2

[2 marks]

Examiners report

[N/A]

1b. [2 marks]

Markscheme

valid approach (M1)

eg 1 − (0.3 + 0.4), 1 − 0.4 − 0.1 −

= 0.3 A1 N2

[2 marks]

Examiners report

[N/A]

1c. [2 marks]

Markscheme

valid approach (M1)

eg , , ,

1

A1 N2

[2 marks]

Examiners report

[N/A]

2a. [2 marks]

Markscheme

valid approach to find P(one red) (M1)

eg , , ,

listing all possible cases for exactly one red (may be indicated on tree diagram)

P(1 red) = 0.444 [0.444, 0.445] A1 N2

[3 marks] [5 maximum for parts (a.i) and (a.ii)]

Examiners report

[N/A]

2b. [3 marks]

Markscheme

valid approach (M1)

eg P( ) + P( ), 1 − P( ≤ 1), binomcdf

2

correct working (A1)

eg , 0.222 + 0.037 ,

0.259259

P(at least two red) = 0.259 A1 N3

[3 marks] [5 maximum for parts (a.i) and (a.ii)]

Examiners report

[N/A]

2c. [5 marks]

Markscheme

recognition that winning $10 means rolling exactly one green (M1)

recognition that winning $10 also means rolling at most 1 red (M1)

eg “cannot have 2 or more reds”

correct approach A1

eg P(1G ∩ 0R) + P(1G ∩ 1R), P(1G) − P(1G ∩ 2R),

“one green and two yellows or one of each colour”

Note: Because this is a “show that” question, do not award this A1 for purely numerical expressions.

one correct probability for their approach (A1)

eg , , , ,

correct working leading to A1

eg , ,

probability = AG N0

3

[5 marks]

Examiners report

[N/A]

2d. [1 mark]

Markscheme

, 0.259 (check FT from (a)(ii)) A1 N1

[1 mark]

Examiners report

[N/A]

2e. [2 marks]

Markscheme

evidence of summing probabilities to 1 (M1)

eg , ,

0.148147 (0.148407 if working with their value to 3 sf)

(exact), 0.148 A1 N2

[2 marks]

Examiners report

[N/A]

2f. [3 marks]

Markscheme

correct substitution into the formula for expected value (A1)

eg

correct critical value (accept inequality) A14

eg = 34.2857 , > 34.2857

$40 A1 N2

[3 marks]

Examiners report

[N/A]

3a. [1 mark]

Markscheme

A1 N1

[1 mark]

Examiners report

[N/A]

3b. [3 marks]

Markscheme

correct probability for one of the draws A1

eg P(not blue first) = , blue second =

valid approach (M1)

eg recognizing loss on first in order to win on second, P(B' then B), P(B') × P(B | B'), tree diagram

correct expression in terms of A1 N3

eg , ,

[3 marks]

5

Examiners report

[N/A]

3c. [2 marks]

Markscheme


eg

A1 N2

[2 marks]

Examiners report

[N/A]

3d. [2 marks]

Markscheme


eg

A1 N2

[2 marks]

Examiners report

[N/A]

3e. [7 marks]

Markscheme

6

correct probabilities (seen anywhere) (A1)(A1)

eg , (may be seen on tree diagram)

valid approach to find E (M) or expected winnings using their probabilities (M1)

eg ,

correct working to find E (M) or expected winnings (A1)

eg ,

correct equation for fair game A1

eg ,

correct working to combine terms in (A1)

eg , ,

= 5 A1 N0

Note: Do not award the final A1 if the candidate’s FT probabilities do not sum to 1.

[7 marks]

Examiners report

[N/A]

4a. [2 marks]

Markscheme

valid approach (M1)

, , 11 + 6

7

number of students taking art class = 17 A1 N2

[2 marks]

Examiners report

[N/A]

4b. [2 marks]

Markscheme

valid approach (M1)

13 + 5, 35 − 17, 18, 1 − P(A)

0.514285

P(A') = (exact), 0.514 A1 N2

[2 marks]

Examiners report

[N/A]

4c. [2 marks]

Markscheme

valid approach (M1)

11 + 13, 35 − 6 − 5, 24

0.685714

P(A or M but not both) = (exact), 0.686 A1 N2

[2 marks]

8

Examiners report

[N/A]

5a. [3 marks]

Markscheme

A1A1A1 N3

Note: Award A1 for each bold fraction.

[3 marks]

Examiners report

[N/A]

5b. [2 marks]

Markscheme

multiplying along correct branches (A1)

eg

P(leaves before 07:00 ∩ late) = A1 N2

[2 marks]

Examiners report

[N/A]

9

5c. [3 marks]

Markscheme

multiplying along other “late” branch (M1)

eg

adding probabilities of two mutually exclusive late paths (A1)

eg

A1 N2

[3 marks]

Examiners report

[N/A]

5d. [3 marks]

Markscheme

recognizing conditional probability (seen anywhere) (M1)

eg

correct substitution of their values into formula (A1)

eg

A1 N2

[3 marks]

Examiners report

[N/A]

5e. [3 marks]

10

Markscheme

valid approach (M1)

eg 1 − P(not late twice), P(late once) + P(late twice)


eg

A1 N2

[3 marks]

Examiners report

[N/A]

6a. [2 marks]

Markscheme

valid approach

eg Venn diagram, P(A) − P (A ∩ B), 0.62 − 0.18 (M1)

P(A ∩ B' ) = 0.44 A1 N2

[2 marks]

Examiners report

[N/A]

6b. [4 marks]

Markscheme

valid approach to find either P(B′ ) or P(B) (M1)

eg (seen anywhere), 1 − P(A ∩ B′ ) − P((A ∪ B)′ )

correct calculation for P(B′ ) or P(B) (A1)11

eg 0.44 + 0.19, 0.81 − 0.62 + 0.18

correct substitution into (A1)

eg

0.698412

P(A | B′ ) = (exact), 0.698 A1 N3

[4 marks]

Examiners report

[N/A]

7a. [3 marks]

Markscheme

correct probabilities

A1A1A1 N3

Note: Award A1 for each correct bold answer.

[3 marks]

Examiners report

12

[N/A]

7b. [3 marks]

Markscheme

multiplying along branches (M1)

eg

adding probabilities of correct mutually exclusive paths (A1)

eg

A1 N2

[3 marks]

Examiners report

[N/A]

8a. [2 marks]

Markscheme

valid approach (M1)

eg

A1 N2

[2 marks]

Examiners report

[N/A]

8b. [2 marks]

Markscheme

valid approach (M1)

13

eg

A1 N2

[2 marks]

Examiners report

[N/A]

8c. [2 marks]

Markscheme

valid approach (M1)

eg

A1 N2

[2 marks]

Examiners report

[N/A]

9a. [2 marks]

Markscheme

correct substitution into formula (A1)

eg

, 0.0333 A1 N2

[2 marks]

Examiners report

[N/A]

9b. [2 marks]

14

Markscheme

evidence of summing probabilities to 1 (M1)

eg

A1 N2

[2 marks]

Examiners report

[N/A]

9c. [1 mark]

Markscheme

A1 N1

[1 mark]

Examiners report

[N/A]

9d. [1 mark]

Markscheme

valid reasoning R1

eg

AG N0

[1 mark]

Examiners report

[N/A]

9e. [3 marks]

Markscheme15

valid method (M1)

eg

correct equation A1

eg

A1 N2

[3 marks]

Examiners report

[N/A]

9f. [4 marks]

Markscheme

recognizing one prize in first seven attempts (M1)

eg


eg

correct approach (A1)

eg

0.065119

0.0651 A1 N2

[4 marks]

Examiners report

[N/A]

16

10. [2 marks]

Markscheme

valid approach (M1)

eg

0.279081

0.279 A1 N2

[2 marks]

Examiners report

[N/A]

11a. [2 marks]

Markscheme

valid interpretation (may be seen on a Venn diagram) (M1)

eg

A1 N2

[2 marks]

Examiners report

[N/A]

11b. [4 marks]

Markscheme

valid attempt to find (M1)

eg

correct working for (A1)

17

eg

correct working for (A1)

eg

A1 N3

[4 marks]

Examiners report

[N/A]

12a. [3 marks]

Markscheme

(i) A1 N1

(ii) appropriate approach (M1)

eg

A1 N2

[3 marks]

Examiners report

This question was well done by most candidates. In part (b), the intersection was

sometimes overlooked, incorrectly using instead.

12b. [3 marks]

Markscheme

valid approach (M1)

eg

correct values (A1)

eg

18

A1 N2

[3 marks]

Examiners report

This question was well done by most candidates. In part (b), the intersection was

sometimes overlooked, incorrectly using instead.

13a. [5 marks]

Markscheme

(i) A1 N1

(ii) valid approach (M1)

eg

A1 N2

(iii) A1A1 N2

Examiners report

On the whole, candidates were very successful on this question, with the majority of candidates earning

most of the available marks.

13b. [4 marks]

Markscheme

(i) A2 N2


eg

A1 N2

Examiners report

19


most of the available marks.

13c. [4 marks]

Markscheme

(i) A1A1A1 N3

(ii) A1 N1

[4 marks]

Examiners report


most of the available marks. The most common error was seen in part (c)(ii), where many candidates

did not earn the mark. It is also interesting to note that many of the candidates who answered this part

correctly did so by using the formula for conditional probability, rather than recognizing that the

required probability is given to them in the second branch of the tree diagram.

14a. [2 marks]

Markscheme

(A1)

A1 N2

[2 marks]

20

Examiners report

[N/A]

14b. [2 marks]

Markscheme

their correct equation (A1)

eg

A1 N2

Examiners report

[N/A]

14c. [3 marks]

Markscheme

METHOD 1

finding their (seen anywhere) (A1)

eg

correct substitution into conditional probability formula (A1)

eg

A1 N2

METHOD 2

recognizing A1

finding their (only if first line seen) (A1)

eg

A1 N2

21

[3 marks]

Total [7 marks]

Examiners report

[N/A]

15a. [2 marks]

Markscheme

recognizing Ann rolls green (M1)

eg

A1 N2

[2 marks]

Examiners report

Some teachers’ comments suggested that the word ‘loses’ in the diagram was misleading. But candidate

scripts did not indicate any adverse effect.

a) Very well answered.

b) i) Probabilities and were typically found correctly. ii) Fewer candidates identified the common

ratio and number of rolls correctly.

Few candidates recognized that this was an infinite geometric sum although some did see that a

geometric progression was involved.

15b. [7 marks]

Markscheme

recognize the probability is an infinite sum (M1)

eg Ann wins on her roll or roll or roll…,

recognizing GP (M1)

(seen anywhere) A1

22

(seen anywhere) A1

correct substitution into infinite sum of GP A1

eg


eg

A1 N1

[7 marks]

Total [15 marks]

Examiners report

Some teachers’ comments suggested that the word ‘loses’ in the diagram was misleading, But candidate

scripts did not indicate any adverse effect.

a) Very well answered.

b) i) Probabilities and were typically found correctly. ii) Fewer candidates identified the common

ratio and number of rolls correctly.

Few candidates recognized that this was an infinite geometric sum although some did see that a

geometric progression was involved.

16a. [2 marks]

Markscheme


eg

A1 N2

[2 marks]23

Examiners report

Parts (a) and (b) of this question were answered correctly by nearly all candidates, and the majority

earned full marks on part (c), as well. Unfortunately, there were a number of candidates who made

arithmetic errors when multiplying or adding fractions.

16b. [2 marks]

Markscheme

multiplying along correct branches (A1)

eg

A1 N2

[2 marks]

Examiners report



arithmetic errors when multiplying or adding fractions. Candidates were not as successful in parts (d)

and (e) of this question. Although many knew that conditional probability was necessary in part (d),

many did not know to use their values from parts (b) and (c), and started from scratch with brand new,

and often incorrect, calculations for the numerator and denominator. A majority of candidates did not

recognize that binomial probability was needed in part (e), not realizing that there were three ways for

Adam to be "late exactly once". A very common incorrect solution to part (e) was .

16c. [4 marks]

Markscheme

multiplying along the other branch (M1)

eg

adding probabilities of their mutually exclusive paths (M1)

eg

24


eg

A1 N3

[4 marks]

Examiners report









16d. [3 marks]

Markscheme

recognizing conditional probability (seen anywhere) (M1)

eg

correct substitution of their values into formula (A1)

eg

A1 N2

[3 marks]

Examiners report



25







16e. [4 marks]

Markscheme

valid approach (M1)

eg , three ways it could happen

correct substitution (A1)

eg


eg

A1 N2

[4 marks]

Total [15 marks]

Examiners report







26



17a. [2 marks]

Markscheme

recognizing that the median is at half the total frequency (M1)

eg

A1 N2

[2 marks]

Examiners report

This question was well handled by most candidates. Except for miscalculations and incorrect readings

from the cumulative frequency graph, the processes and concepts seemed to be well understood by the

majority.

A number of students did not gain full marks in parts (bii) and (e), for not showing their process. In part

(c), some candidates wrote things like “using GDC”, without showing relevant work, and so lost marks.

Those who chose a formulaic approach to the conditional probability question in (dii) were often not as

successful as those who could interpret the question in terms of the table values.

A large number of candidates could not find the mean value in (e). Some used the incorrect mid-interval

values and others did not consider their use.

17b. [4 marks]

Markscheme

(i) families have a monthly income less than A1 N1

(ii) correct cumulative frequency, (A1)

subtracting their cumulative frequency from (M1)

eg

families have a monthly income of more than dollars A1 N2

27

Note: If working shown, award M1A1A1 for , using the table.

[4 marks]

Examiners report



majority.







17c. [2 marks]

Markscheme

correct calculation (A1)

eg (A1)

A1 N2

[2 marks]

Examiners report



majority.







28

17d. [2 marks]

Markscheme

(i) correct working (A1)

eg

(exact) A1 N2

(ii) correct working/probability for number of families (A1)

eg

A1 N2

[4 marks]

Examiners report



majority.







17e. [2 marks]

Markscheme

evidence of using correct mid-interval values ( ) (A1)

29

attempt to substitute into (M1)

eg

A1 N2

[3 marks]

Total [15 marks]

Examiners report



majority.







18. [6 marks]

Markscheme

recognize need for intersection of Y and F (R1)

eg

valid approach to find (M1)

eg , Venn diagram

correct working (may be seen in Venn diagram) (A1)

eg

30

A1

recognize need for complement of (M1)

eg

A1 N3

[6 marks]

Examiners report

[N/A]

19a. [3 marks]

Markscheme

A1A1A1 N3

Note: Award A1 for each correct bold probability.

31

[3 marks]

Examiners report

[N/A]

19b. [2 marks]

Markscheme

multiplying along the branches (may be seen on diagram) (M1)

eg

A1 N2

[2 marks]

Examiners report

[N/A]

19c. [4 marks]

Markscheme

METHOD 1

multiplying along the branches (may be seen on diagram) (M1)

eg

adding their probabilities of three mutually exclusive paths (M1)

eg

correct simplification (A1)

eg

A1 N3

METHOD 232

recognizing “Bill wins at least one” is complement of “Andrea wins 2” (R1)

eg finding P (Andrea wins 2)

P (Andrea wins both) (A1)

evidence of complement (M1)

eg

A1 N3

[4 marks]

Examiners report

[N/A]

19d. [5 marks]

Markscheme

P (B wins both) A1

evidence of recognizing conditional probability (R1)

eg


eg

A1 N3

[5 marks]

Examiners report

[N/A]

20a. [2 marks]

33

Markscheme


eg

A1 N2

[2 marks]

Examiners report

[N/A]

20b. [2 marks]

Markscheme


eg

A1 N2

[2 marks]

Examiners report

[N/A]

20c. [1 mark]

Markscheme

A1 N1

Examiners report

34

[N/A]

20d. [2 marks]

Markscheme

appropriate approach (M1)

eg

(may be seen in Venn diagram) A1 N2

[2 marks]

Examiners report

[N/A]

21a. [4 marks]

Markscheme


eg , tree diagram,

one correct multiplication (A1)

eg


eg

A1 N3

[4 marks]

Examiners report

[N/A]

21b. [3 marks]

Markscheme

35

recognizing conditional probability (R1)

eg

correct working A1

eg

A1 N2

[3 marks]

Examiners report

[N/A]

21c. [2 marks]

Markscheme

recognizing binomial probability (R1)

eg ,

A1 N2

[2 marks]

Examiners report

[N/A]

21d. [5 marks]

Markscheme

METHOD 1

evidence of using complement (seen anywhere) (M1)

eg

valid approach (M1)

36

eg

correct inequality (accept equation) A1

eg

(A1)

A1 N3

METHOD 2

valid approach using guess and check/trial and error (M1)

eg finding for various values of n

seeing the “cross over” values for the probabilities A1A1

recognising (R1)

A1 N3

[5 marks]

Examiners report

[N/A]

22a. [5 marks]

Markscheme

(i) attempt to find (M1)

eg , ,

A1 N2

(ii) attempt to find (M1)

37

eg , ,

recognizing two ways to get one red, one green (M1)

eg , ,

A1 N2

[5 marks]

Examiners report

Many candidates correctly found the probability of selecting no green marbles in two draws, although

some candidates treated the second draw as if replacing the first. When finding the probability for

exactly one green marble, candidates often failed to recognize two pathways for selecting one of each

color.

22b. [3 marks]

Markscheme

(seen anywhere) (A1)

correct substitution into formula for A1

eg ,

expected number of green marbles is A1 N2

[3 marks]

Examiners report

Few candidates understood the concept of expected value in this context, often leaving this blank or

treating as if a binomial experiment. Successful candidates often made a distribution table before

making the final calculation.

22c. [2 marks]

38

Markscheme

(i) A1 N1

(ii) A1 N1

[2 marks]

Examiners report

Most candidates answered part (c) correctly. However, many overcomplicated (c)(ii) by using the

conditional probability formula. Those with a clear understanding of the concept easily followed the

“write down” instruction.

22d. [6 marks]

Markscheme

recognizing conditional probability (M1)

eg , , tree diagram

attempt to multiply along either branch (may be seen on diagram) (M1)

eg

attempt to multiply along other branch (M1)

eg

adding the probabilities of two mutually exclusive paths (A1)

eg

correct substitution

eg , A1

A1 N3

39

[6 marks]

Examiners report

Only a handful of candidates correctly applied conditional probability to find in part (d). While

some wrote down the formula, or drew a tree diagram, few correctly calculated . A

common error was to combine the marbles in the two jars to give .

23a. [4 marks]

Markscheme

attempt to find (M1)

eg ,

A1 N2

attempt to find (M1)

eg ,

A1 N2

[4 marks]

Examiners report

Overall, candidates were very successful in parts (a), (b) and (c) of this question. Most of the errors in

these parts had to do with candidates not understanding terms such as "at least" or "less than".

23b. [3 marks]

Markscheme

(i) A1 N1


eg ,

40

A1 N2

[3 marks]

Examiners report



23c. [4 marks]

Markscheme

(i) attempt to find number of girls (M1)

eg ,

are not selected A1 N2

(ii) are selected (A1)

A1 N2

[4 marks]

Examiners report



23d. [4 marks]

Markscheme

(i) given second chance A1 N1

(ii) took less than minutes (A1) 41

attempt to find their selected total (may be seen in calculation) (M1)

eg , their answer from (i)

( ) A1 N3

[4 marks]

Examiners report

Part (d) was quite challenging for candidates, who may not have read the question carefully and

studied the values in the diagram. Many seemed confused by the idea that not all the girls who were

given a second chance were selected. In part (d)(ii), many did not find the percentage of the whole

group, but rather the percentage of the girls who were given a second chance.

24a. [2 marks]

Markscheme

valid approach (M1)

eg ,

maximum height (m) A1 N2

[2 marks]

Examiners report

Most candidates were successful with part (a).

24b. [2 marks]

Markscheme

(i) period A1

period minutes AG N0

(ii) A1 N1

42

[2 marks]

Examiners report

A surprising number had difficulty producing enough work to show that the period was ; writing

down the exact value of also overwhelmed a number of candidates. In part (c), candidates did not

recognize that the seat on the Ferris wheel is a minimum at thereby making the value of a

negative. Incorrect values of were often seen with correct follow through obtained when sketching

the graph in part (d). Graphs again frequently failed to show key features in approximately correct

locations and candidates lost marks for incorrect domains and ranges.

24c. [3 marks]

Markscheme

METHOD 1

valid approach (M1)

eg , ,

(accept ) (A1)

A1 N2

METHOD 2

attempt to substitute valid point into equation for h (M1)

eg

correct equation (A1)

eg ,

A1 N2

[3 marks]

Examiners report

43







24d. [4 marks]

Markscheme

A1A1A1A1 N4

Note: Award A1 for approximately correct domain, A1 for approximately correct range,

A1 for approximately correct sinusoidal shape with cycles.

Only if this last A1 awarded, award A1 for max/min in approximately correct positions.

[4 marks]

Examiners report





44



24e. [5 marks]

Markscheme

setting up inequality (accept equation) (M1)

eg , , sketch of graph with line

any two correct values for t (seen anywhere) A1A1

eg , , ,

valid approach M1

eg , , ,

A1 N2

[5 marks]

Examiners report

Part (e) was very poorly done for those who attempted the question and most did not make the

connection between height, time and probability. The idea of linking probability with a real-life

scenario proved beyond most candidates. That said, there were a few novel approaches from the

strongest of candidates using circles and angles to solve this part of question 10.

25a. [8 marks]

Markscheme

METHOD 1

(i) appropriate approach (M1)

eg , ,

A1 N2

45

(ii) multiplying one pair of gold and silver probabilities (M1)

eg , , 0.24

adding the product of both pairs of gold and silver probabilities (M1)

eg ,

A1 N3

(iii)

(seen anywhere) (A1)

correct substitution into formula for (A1)

eg ,

A1 N3

METHOD 2

(i) evidence of recognizing binomial (may be seen in part (ii)) (M1)

eg ,

correct probability for use in binomial (A1)

eg , ,

A1 N3

(ii) correct set up (A1)

46

eg

A1 N2

(iii)

attempt to substitute into (M1)

eg

correct substitution into (A1)

eg

A1 N3

[8 marks]

Examiners report

Parts (a)(i) and (ii) were generally well done, with candidates either using a tree diagram or a binomial

approach. Part (a)(iii) proved difficult, with many either having trouble finding or using

.

25b. [2 marks]

Markscheme

Let be the number of gold balls drawn from the bag.

evidence of recognizing binomial (seen anywhere) (M1)

eg ,

47

A1 N2

[2 marks]

Examiners report

A great majority were confident solving part (b) with the GDC, although some did write the binomial

term.

25c. [2 marks]

Markscheme

recognize need to find (M1)

A1 N2

[2 marks]

Examiners report

Those candidates who did not use the binomial function on the GDC had more difficulty in part (c),

although a pleasing number were still able to identify that they were seeking .

25d. [3 marks]

Markscheme

Let be the number of gold balls drawn from the bag.


eg , , ,

(A1)

(to dp) A1 N2

48

[3 marks]

Examiners report

While most candidate knew to use conditional probability in part (d), fewer were able to do so

successfully, and even fewer still correctly rounded their answer to two decimal places. The most

common error was to multiply probabilities to find the intersection needed for the conditional

probability formula. Overall, candidates seemed better prepared for probability.

26a. [1 mark]

Markscheme

A1 N1

[1 mark]

Examiners report

Parts (a), (b), and (c)(i) of this Venn diagram probability question were answered quite well with

candidates consistently earning full marks.

26b. [3 marks]

Markscheme

(i) correct values A1

e.g. ,

AG N0

(ii) , A1A1 N2

[3 marks]

Examiners report

Parts (a), (b), and (c)(i) of this Venn diagram probability question were answered quite well with

candidates consistently earning full marks. Only a few candidates worked backwards from the

given in the "show that" portion of part (b).

26c. [3 marks]

Markscheme49

(i) A1 N1

(ii) A2 N2

[3 marks]

Examiners report

Many candidates struggled on part (c)(ii), either not recognizing conditional probability or multiplying

probabilities to find the numerator as if the events were independent. A number of candidates who

successfully found the probability in part (c)(ii) left their incomplete answer of .

27a. [5 marks]

Markscheme

(i)

A1A1A1 N3

(ii) multiplying along the correct branches (may be seen on diagram) (A1)

e.g.

A1 N2

50

[5 marks]

Examiners report

Part (a) of this question was answered correctly by the large majority of candidates. There were some

who did not follow the instruction to copy and complete the tree diagram on their separate paper, and

simply filled in the blanks on the exam paper.

27b. [5 marks]

Markscheme

, (seen anywhere) (A1)(A1)


e.g.

correct calculation A1

e.g. ,

A1 N3

[5 marks]

Examiners report

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In part (b), many candidates struggled with finding the compound probability, and did not use the

provided information in the appropriate manner. Quite a few candidates seemed to be confused about

when they should add the probabilities or when they should multiply.

27c. [4 marks]

Markscheme


e.g. ,

correct numerator (A1)

e.g.

correct denominator (A1)

e.g.

probability A1 N3

[4 marks]

Examiners report

In part (c), many recognized that the question dealt with conditional probability, and many tried to use

the formula from the information booklet, but failed to realize that they had already found the required

values for the numerator and denominator in their working for part (b).

Throughout this question, it was discouraging to see the large number of candidates making arithmetic

errors. There were a surprising number of candidates who multiplied fractions incorrectly, or found an

incorrect value for simple multiplication such as or .

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