# Probability ```Probability
[166 marks]
1. Let A and B be events such that
P (A ∪ B) = 0.6.
Find
P (A) = 0.5, P (B) = 0.4 and
[5 marks]
P ( A | B).
Markscheme
attempt to substitute into
P (A ∪ B) = P (A) + P (B) − P (A ∩ B)
(M1)
Note: Accept use of Venn diagram or other valid method.
0.6 = 0.5 + 0.4 − P (A ∩ B)
(A1)
P (A ∩ B) = 0.3 (seen anywhere)
attempt to substitute into
=
A1
P ( A | B) =
P(A∩B )
P(B )
(M1)
0.3
0.4
P ( A | B) = 0.75 (= 34 ) A1
[5 marks]
At a school, 70% of the students play a sport and 20% of the students are
involved in theatre. 18% of the students do neither activity.
A student is selected at random.
2a. Find the probability that the student plays a sport and is involved in
theatre.
[2 marks]
Markscheme
EITHER
P(S)+P(T )+P(S' ∩ T ')−P(S ∩ T )= 1 OR P(S ∪ T )= P((S' ∩ T ')') (M1)
0. 7 + 0. 2 + 0. 18 − P(S ∩ T )= 1 OR P(S ∪ T )= 1 − 0. 18
OR
a clearly labelled Venn diagram (M1)
THEN
P (S ∩ T )= 0. 08 (accept 8%) A1
Note: To obtain the M1 for the Venn diagram all labels must be correct and in
the correct sections. For example, do not accept 0. 7 in the area corresponding
to S ∩ T ' .
[2 marks]
2b. Find the probability that the student is involved in theatre, but does not [2 marks]
play a sport.
Markscheme
EITHER
P(T ∩ S')= P(T )−P(T ∩ S)(= 0. 2 − 0. 08) OR
P(T ∩ S')= P(T ∪ S)−P(S)(= 0. 82 − 0. 7) (M1)
OR
a clearly labelled Venn diagram including
P(S), P(T ) and P(S ∩ T ) (M1)
THEN
= 0. 12 (accept 12%) A1
[2 marks]
48%
25%
At the school 48% of the students are girls, and 25% of the girls are involved in
theatre.
A student is selected at random. Let G be the event “the student is a girl” and let
T be the event “the student is involved in theatre”.
2c. Find
P(G ∩ T ).
[2 marks]
Markscheme
P(G ∩ T )= P(T /G)P(G)(0. 25 &times; 0. 48) (M1)
= 0. 12 A1
[2 marks]
2d. Determine if the events G and
Markscheme
METHOD 1
P(G)&times;P(T )(= 0. 48 &times; 0. 2)= 0. 096 A1
P(G)&times;P(T )≠P(G ∩ T )⇒ G and T are not independent R1
METHOD 2
P(T G )= 0. 25 A1
P(T G )≠ P(T )⇒ G and T are not independent R1
Note: Do not award A0R1.
[2 marks]
30
19
3
6
In a class of 30 students,
do not play either sport.
19 play tennis, 3 play both tennis and volleyball, and 6
The following Venn diagram shows the events “plays tennis” and “plays
volleyball”. The values t and v represent numbers of students.
[2 marks]
3a. Find the value of t.
Markscheme
valid approach to find t
(M1)
eg t + 3 = 19, 19 − 3
t = 16 (may be seen on Venn diagram)
A1 N2
[2 marks]
[2 marks]
3b. Find the value of v.
Markscheme
valid approach to find v
(M1)
eg t + 3 + v + 6 = 30, 30 − 19 − 6
v = 5 (may be seen on Venn diagram)
A1 N2
[2 marks]
3c. Find the probability that a randomly selected student from the class
plays tennis or volleyball, but not both.
[2 marks]
Markscheme
valid approach
(M1)
3+6
eg 16 + 5, 21 students, 1 − 30 ,
21
30
(=
7
)
10
A1 N2
[2 marks]
Andre will play in the semi-final of a tennis tournament.
If Andre wins the semi-final he will progress to the final. If Andre loses the semifinal, he will not progress to the final.
If Andre wins the final, he will be the champion.
The probability that Andre will win the semi-final is p. If Andre wins the semi-final,
then the probability he will be the champion is 0. 6.
4a. Complete the values in the tree diagram.
[1 mark]
Markscheme
* This question is from an exam for a previous syllabus, and may contain
minor differences in marking or structure. It appeared in a paper that
permitted the use of a calculator, and so might not be suitable for all forms of
practice.
(A1)
(C1)
Note: Award (A1) for the correct pair of probabilities.
[1 mark]
The probability that Andre will not be the champion is 0. 58.
4b. Find the value of p.
[2 marks]
Markscheme
p &times; 0. 4 +(1 − p)= 0. 58
(M1)
Note: Award (M1) for multiplying and adding correct probabilities for losing
equated to 0. 58.
OR
p &times; 0. 6 = 1 − 0. 58
(M1)
Note: Award (M1) for multiplying correct probabilities for winning equated to
1 − 0. 58 or 0. 42.
(p =) 0. 7
(A1)(ft)
(C2)
Note: Follow through from their part (a). Award the final (A1)(ft) only if their
p is within the range 0 &lt; p &lt; 1.
[2 marks]
4c. Given that Andre did not become the champion, find the probability that [3 marks]
he lost in the semi-final.
Markscheme
0.3
0.58
( 1−0.7
)
0.58
(A1)(ft)(A1)
Note: Award (A1)(ft) for their correct numerator. Follow through from part
(b). Award (A1) for the correct denominator.
OR
0.3
0.3+0.7&times;0.4
(A1)(ft)(A1)(ft)
Note: Award (A1)(ft) for their correct numerator. Follow through from part
(b). Award (A1)(ft) for their correct calculation of Andre losing the semi-final
or winning the semi-final and then losing in the final. Follow through from their
parts (a) and (b).
15
29
(0. 517, 0. 517241 … , 51. 7%)
(A1)(ft)
(C3)
Note: Follow through from parts (a) and (b).
[3 marks]
On a school excursion, 100 students visited an amusement park. The amusement
park’s main attractions are rollercoasters (R), water slides (W), and virtual reality
rides (V).
The students were asked which main attractions they visited. The results are
shown in the Venn diagram.
A total of
74 students visited the rollercoasters or the water slides.
5a. Find the value of a .
[2 marks]
Markscheme
74 −(32 + 12 + 10 + 9 + 5) OR 74 − 68
(M1)
Note: Award (M1) for setting up a correct expression.
(a =) 6
(A1)(G2)
[2 marks]
[2 marks]
5b. Find the value of b.
Markscheme
100 −(74 + 18)
(M1)
OR
100 − 92
(M1)
OR
100 −(32 + 9 + 5 + 12 + 10 + 18 + 6)
(M1)
Note: Award (M1) for setting up a correct expression. Follow through from
part (a)(i) but only for a ≥ 0.
(b =) 8
(A1)(ft)(G2)
Note: Follow through from part(a)(i). The value of
to zero for the (A1)(ft) to be awarded.
b must be greater or equal
[2 marks]
5c. Find the number of students who visited at least two types of main
attraction.
[2 marks]
Markscheme
9 + 5 + 12 + 10
(M1)
36
9, 5, 12 and 10.
(A1)(G2)
[2 marks]
5d. Write down the value of n(
R ∩ W) .
[1 mark]
Markscheme
14
(A1)
[1 mark]
5e. Find the probability that a randomly selected student visited the
rollercoasters.
[2 marks]
Markscheme
58
100
( 29
, 0. 58, 58%)
50
(A1)(A1)(G2)
Note: Award (A1) for correct numerator. Award (A1) for the correct
denominator. Award (A0) for 58 only.
[2 marks]
5f. Find the probability that a randomly selected student visited the virtual
reality rides.
[1 mark]
Markscheme
45
100
9
( 20
, 0. 45, 45%)
(A1)(ft)
Note: Follow through from their denominator from part (d)(i).
[1 mark]
5g. Hence determine whether the events in parts (d)(i) and (d)(ii) are
[2 marks]
Markscheme
they are not independent
58
100
&times;
45
100
≠
17
OR
100
(A1)(ft)
0. 261 ≠ 0. 17
(R1)
Note: Comparison of numerical values must be seen for (R1) to be awarded.
Do not award (A1)(R0). Follow through from parts (d)(i) and (d)(ii).
[2 marks]
In a class of 30 students, 18 are fluent in Spanish, 10 are fluent in French, and 5
are not fluent in either of these languages. The following Venn diagram shows the
events “fluent in Spanish” and “fluent in French”.
The values
m, n, p and q represent numbers of students.
6a. Write down the value of q.
[1 mark]
Markscheme
q=5
A1 N1
[1 mark]
[2 marks]
6b. Find the value of n.
Markscheme
valid approach
(M1)
eg (18 + 10 + 5) − 30, 28 − 25, 18 + 10 − n = 25
n=3
A1 N2
[2 marks]
[3 marks]
6c. Write down the value of m and of p.
Markscheme
valid approach for finding m or p (may be seen in part (b))
(M1)
18 − 3, 3 + p = 10
m = 15, p = 7
A1A1 N3
eg
[3 marks]
P ( ) = 0.3
The following Venn diagram shows the events
values shown are probabilities.
A and B, where P (A) = 0.3. The
[2 marks]
7a. Find the value of p.
Markscheme
* This question is from an exam for a previous syllabus, and may contain
minor differences in marking or structure.
(M1)
valid approach
eg
0.30 − 0.1,
p = 0.2
p + 0.1 = 0.3
A1 N2
[2 marks]
[2 marks]
7b. Find the value of q.
Markscheme
valid approach
eg
(M1)
1 − (0.3 + 0.4), 1 − 0.4 − 0.1 −
q = 0.3
p
A1 N2
[2 marks]
7c. Find
P (A′ ∪ B).
[2 marks]
Markscheme
valid approach
eg
P (A′
(M1)
0.7 + 0.5 − 0.3, p + q + 0.4, 1 − 0.1,
∪ B) = P (A′ ) + P (B) − P (A′ ∩ B)
P (A′ ∪ B) = 0.9
[2 marks]
A1 N2
The diagram shows a circular horizontal board divided into six equal sectors. The
sectors are labelled white (W), yellow (Y) and blue (B).
A pointer is pinned to the centre of the board. The pointer is to be spun and when
it stops the colour of the sector on which the pointer stops is recorded. The
pointer is equally likely to stop on any of the six sectors.
Eva will spin the pointer twice. The following tree diagram shows all the possible
outcomes.
8a. Find the probability that both spins are yellow.
[2 marks]
Markscheme
* This question is from an exam for a previous syllabus, and may contain
minor differences in marking or structure.
1
3
&times;
1
OR
3
2
( 13 ) (M1)
Note: Award (M1) for multiplying correct probabilities.
1
(0.111, 0.111111…, 11.1%)
9
(A1)
(C2)
[2 marks]
[3 marks]
8b. Find the probability that at least one of the spins is yellow.
Markscheme
( 12 &times; 13 ) + ( 16 &times; 13 ) +
1
3
Note: Award (M1) for ( 12
(M1)(M1)
&times; 13 ) and ( 16 &times; 13 ) or equivalent, and (M1) for
and adding only the three correct probabilities.
1
3
OR
2
1 − ( 23 )
(M1)(M1)
2
Note: Award (M1) for 23 seen and (M1) for subtracting ( 23 ) from 1. This may
be shown in a tree diagram with “yellow” and “not yellow” branches.
5
(0.556, 0.555555…, 55.6%)
9
(A1)(ft) (C3)
Note: Follow through marks may be awarded if their answer to part (a) is
used in a correct calculation.
[3 marks]
8c. Write down the probability that the second spin is yellow, given that the
first spin is blue.
[1 mark]
Markscheme
1
(0.333, 0.333333…, 33.3%)
3
(A1) (C1)
[1 mark]
9a. Place the numbers
Venn diagram.
3
2π, − 5, 3−1 and 2 2 in the correct position on the
[4 marks]
Markscheme
* This question is from an exam for a previous syllabus, and may contain
minor differences in marking or structure.
(A1)(A1)(A1)(A1)
(C4)
Note: Award (A1) for each number in the correct position.
[4 marks]
9b. In the table indicate which two of the given statements are true by
placing a tick (✔) in the right hand column.
[2 marks]
Markscheme
(A1)(A1)
(C2)
Note: Award (A1) for each correctly placed tick.
[2 marks]
A school caf&eacute; sells three flavours of smoothies: mango (M ), kiwi fruit (K) and
banana (B).
85 students were surveyed about which of these three flavours they like.
35 students liked mango, 37 liked banana, and 26 liked kiwi fruit
2 liked all three flavours
20 liked both mango and banana
14 liked mango and kiwi fruit
3 liked banana and kiwi fruit
10a. Using the given information, complete the following Venn diagram.
[2 marks]
Markscheme
* This question is from an exam for a previous syllabus, and may contain
minor differences in marking or structure.
(A1)(A1) (C2)
Note: Award (A1) for 18, 12 and 1 in correct place on Venn diagram, (A1) for
3, 16 and 11 in correct place on Venn diagram.
[2 marks]
10b. Find the number of surveyed students who did not like any of the three [2 marks]
flavours.
Markscheme
85 − (3 + 16 + 11 + 18 + 12 + 1 + 2)
(M1)
Note: Award (M1) for subtracting the sum of their values from 85.
22
(A1)(ft) (C2)
Note: Follow through from their Venn diagram in part (a).
If any numbers that are being subtracted are negative award (M1)(A0).
[2 marks]
10c. A student is chosen at random from the surveyed students.
[2 marks]
Find the probability that this student likes kiwi fruit smoothies given that they like
mango smoothies.
Markscheme
14
35
( 25 , 0.4, 40% )
(A1) (ft)(A1)(ft) (C2)
Note: Award (A1) for correct numerator; (A1) for correct denominator. Follow
through from their Venn diagram.
[2 marks]
At Penna Airport the probability, P(A), that all passengers arrive on time for a
flight is 0.70. The probability, P(D), that a flight departs on time is 0.85. The
probability that all passengers arrive on time for a flight and it departs on time is
0.65.
11a. Show that event A and event D are not independent.
[2 marks]
Markscheme
* This question is from an exam for a previous syllabus, and may contain
minor differences in marking or structure.
METHOD 1
multiplication of P(A) and P(D)
(A1)
eg 0.70 &times; 0.85, 0.595
correct reasoning for their probabilities
R1
eg 0.595 ≠ 0.65 , 0.70 &times; 0.85 ≠ P (A ∩ D)
A and D are not independent
AG N0
METHOD 2
calculation of P (D |A )
eg
(A1)
13
, 0.928
14
correct reasoning for their probabilities
eg 0.928 ≠ 0.85 ,
R1
0.65
P (D)
0.7
A and D are not independent
AG N0
[2 marks]
11b. Find
P (A ∩ D′ ).
[2 marks]
Markscheme
(A1)
correct working
eg P (A) − P (A ∩ D) , 0.7 − 0.65 , correct shading and/or value on Venn
diagram
P (A ∩ D′ ) = 0.05
A1 N2
[2 marks]
11c. Given that all passengers for a flight arrive on time, find the probability [3 marks]
that the flight does not depart on time.
Markscheme
recognizing conditional probability (seen anywhere)
eg
P(D′∩ A )
,
P(A )
P (A | B )
correct working
eg
(M1)
(A1)
0.05
0.7
0.071428
P (D′ | A ) =
1
14 , 0.0714
A1 N2
[3 marks]
The number of hours that pilots fly per week is normally distributed with a mean
of 25 hours and a standard deviation σ. 90 % of pilots fly less than 28 hours in a
week.
11d. Find the value of σ.
[3 marks]
Markscheme
finding standardized value for 28 hours (seen anywhere)
(A1)
eg z = 1.28155
(A1)
correct working to find σ
eg
1.28155 =
28−25
28−25
,
σ
1.28155
2.34091
σ = 2.34
A1 N2
[3 marks]
11e. All flights have two pilots. Find the percentage of flights where both
pilots flew more than 30 hours last week.
Markscheme
P (X &gt; 30) = 0.0163429
(A1)
valid approach (seen anywhere)
2
(M1)
eg [P (X &gt; 30)] , (0.01634)2 , B(2, 0.0163429) , 2.67E-4 , 2.66E-4
0.0267090
0.0267 %
[4 marks]
A2 N3
[4 marks]
A bag contains 5 red and 3 blue discs, all identical except for the colour. First,
Priyanka takes a disc at random from the bag and then Jorg&eacute; takes a disc at
random from the bag.
[3 marks]
12a. Complete the tree diagram.
Markscheme
* This question is from an exam for a previous syllabus, and may contain
minor differences in marking or structure.
(A1)(A1)(A1) (C3)
Note: Award (A1) for each correct pair of branches.
[3 marks]
12b. Find the probability that Jorg&eacute; chooses a red disc.
[3 marks]
Markscheme
5
8
&times;
4
7
+
3
8
&times;
5
7
(A1)(ft)(M1)
Note: Award (A1)(ft) for their two correct products from their tree diagram.
Follow through from part (a), award (M1) for adding their two products. Award
= 58
( 35
, 0.625, 62.5 % )
56
(A1)(ft) (C3)
Note: Follow through from their tree diagram, only if probabilities are [0,1].
[3 marks]
In a group of 35 students, some take art class (A) and some take music class (M).
5 of these students do not take either class. This information is shown in the
following Venn diagram.
13a. Write down the number of students in the group who take art class.
[2 marks]
Markscheme
* This question is from an exam for a previous syllabus, and may contain
minor differences in marking or structure.
valid approach (M1)
(A ∩ M ′ ) + (A ∩ M),
17
, 11 + 6
35
number of students taking art class = 17 A1 N2
[2 marks]
One student from the group is chosen at random. Find the probability that
13b. the student does not take art class.
[2 marks]
Markscheme
valid approach (M1)
13 + 5, 35 − 17, 18, 1 − P(A)
0.514285
P(A') = 18
(exact), 0.514 A1 N2
35
[2 marks]
13c. the student takes either art class or music class, but not both.
Markscheme
valid approach (M1)
11 + 13, 35 − 6 − 5, 24
0.685714
P(A or M but not both) = 24
(exact), 0.686 A1 N2
35
[2 marks]
[2 marks]
160 students attend a dual language school in which the students are taught only
in Spanish or taught only in English.
A survey was conducted in order to analyse the number of students studying
Biology or Mathematics. The results are shown in the Venn diagram.
Set S represents those students who are taught in Spanish.
Set B represents those students who study Biology.
Set M represents those students who study Mathematics.
14a. Find the number of students in the school that are taught in Spanish.
[2 marks]
Markscheme
10 + 40 + 28 + 17 (M1)
= 95 (A1)(G2)
Note: Award (M1) for each correct sum (for example: 10 + 40 + 28 + 17)
seen.
[2 marks]
14b. Find the number of students in the school that study Mathematics in
English.
[2 marks]
Markscheme
20 + 12 (M1)
= 32 (A1)(G2)
Note: Award (M1) for each correct sum (for example: 10 + 40 + 28 + 17)
seen.
[2 marks]
14c. Find the number of students in the school that study both Biology and
Mathematics.
[2 marks]
Markscheme
12 + 40 (M1)
= 52 (A1)(G2)
Note: Award (M1) for each correct sum (for example: 10 + 40 + 28 + 17)
seen.
[2 marks]
14d. Write down
n (S ∩ (M ∪ B)).
[1 mark]
Markscheme
78 (A1)
[1 mark]
14e. Write down
n (B ∩ M ∩ S ′ ).
Markscheme
12 (A1)
[1 mark]
[1 mark]
A student from the school is chosen at random.
14f. Find the probability that this student studies Mathematics.
[2 marks]
Markscheme
100
160
( 58 , 0.625, 62.5 %) (A1)(A1) (G2)
Note: Throughout part (c), award (A1) for correct numerator, (A1) for correct
denominator. All answers must be probabilities to award (A1).
[2 marks]
14g. Find the probability that this student studies neither Biology nor
Mathematics.
[2 marks]
Markscheme
42
160
( 21
, 0.263 (0.2625) , 26.3 % (26.25 %)) (A1)(A1) (G2)
80
Note: Throughout part (c), award (A1) for correct numerator, (A1) for correct
denominator. All answers must be probabilities to award (A1).
[2 marks]
14h. Find the probability that this student is taught in Spanish, given that
the student studies Biology.
[2 marks]
Markscheme
50
70
( 57 , 0.714 (0.714285 …) , 71.4 % (71.4285 … %)) (A1)(A1) (G2)
Note: Throughout part (c), award (A1) for correct numerator, (A1) for correct
denominator. All answers must be probabilities to award (A1).
[2 marks]
A group of 60 sports enthusiasts visited the PyeongChang 2018 Winter Olympic
games to watch a variety of sporting events.
The most popular sports were snowboarding (S), figure skating (F) and ice hockey
(H).
For this group of 60 people:
4 did not watch any of the most popular sports,
x watched all three of the most popular sports,
9 watched snowboarding only,
11 watched figure skating only,
15 watched ice hockey only,
7 watched snowboarding and figure skating,
13 watched figure skating and ice hockey,
11 watched snowboarding and ice hockey.
15. Find the value of x.
[2 marks]
Markscheme
4 + 9 + 11 + 15 + x + (7 − x) + (11 − x) + (13 − x) = 60
(M1)
Note: Award (M1) for equating the sum of at least seven of the entries in
their Venn diagram to 60.
(x =) 5
(A1)(ft) (C2)
Note: Follow through from part (a), but only if answer is positive.
[2 marks]
Pablo drives to work. The probability that he leaves home before 07:00 is 34 .
If he leaves home before 07:00 the probability he will be late for work is 18 .
If he leaves home at 07:00 or later the probability he will be late for work is 58 .
16a. Copy and complete the following tree diagram.
[3 marks]
Markscheme
* This question is from an exam for a previous syllabus, and may contain
minor differences in marking or structure.
A1A1A1 N3
Note: Award A1 for each bold fraction.
[3 marks]
16b. Find the probability that Pablo leaves home before 07:00 and is late for [2 marks]
work.
Markscheme
multiplying along correct branches
eg 34 &times; 18
(A1)
3
P(leaves before 07:00 ∩ late) = 32
A1 N2
[2 marks]
16c. Find the probability that Pablo is late for work.
[3 marks]
Markscheme
(M1)
multiplying along other “late” branch
eg 14 &times; 58
adding probabilities of two mutually exclusive late paths
3
5
eg ( 34 &times; 18 ) + ( 14 &times; 58 ) , 32
+ 32
P (L) =
8
32
(A1)
(= 14 ) A1 N2
[3 marks]
16d. Given that Pablo is late for work, find the probability that he left home
before 07:00.
[3 marks]
Markscheme
recognizing conditional probability (seen anywhere)
eg P (A|B) , P (before 7|late)
correct substitution of their values into formula
eg
3
32
1
4
3
8
P (left before 07:00|late) =
(M1)
(A1)
A1 N2
[3 marks]
16e. Two days next week Pablo will drive to work. Find the probability that
he will be late at least once.
Markscheme
valid approach
(M1)
eg 1 − P(not late twice), P(late once) + P(late twice)
correct working
(A1)
3
3
eg 1 − ( 4 &times; 4 ) , 2 &times; 14
7
16
A1 N2
[3 marks]
&times;
3
4
+
1
4
&times;
1
4
[3 marks]
In an international competition, participants can answer questions in only one of
the three following languages: Portuguese, Mandarin or Hindi. 80 participants took
part in the competition. The number of participants answering in Portuguese,
Mandarin or Hindi is shown in the table.
[1 mark]
17a. State the number of boys who answered questions in Portuguese.
Markscheme
* This question is from an exam for a previous syllabus, and may contain
minor differences in marking or structure.
20
(A1) (C1)
[1 mark]
A boy is chosen at random.
17b. Find the probability that the boy answered questions in Hindi.
[2 marks]
Markscheme
5
43
(0.11627 … , 11.6279 … % )
(A1)(A1) (C2)
Note: Award (A1) for correct numerator, (A1) for correct denominator.
[2 marks]
17c. Two girls are selected at random.
[3 marks]
Calculate the probability that one girl answered questions in Mandarin and the
Markscheme
7
37
&times;
12
36
+
12
37
&times;
7
36
(A1)(M1)
Note: Award (A1) for first or second correct product seen, (M1) for adding
their two products or for multiplying their product by two.
=
14
111
( 0.12612 … , 12.6126 % )
(A1) (C3)
[3 marks]
Two events A and B are such that P(A) = 0.62 and P(A ∩ B) = 0.18.
18a. Find P(A ∩ B′ ).
[2 marks]
Markscheme
* This question is from an exam for a previous syllabus, and may contain
minor differences in marking or structure.
valid approach
eg Venn diagram, P(A) − P (A ∩ B), 0.62 − 0.18
P(A ∩ B' ) = 0.44
(M1)
A1 N2
[2 marks]
18b. Given that P((A ∪ B)′ ) = 0.19, find P(A | B′ ).
[4 marks]
Markscheme
valid approach to find either P(B′ ) or P(B)
eg
(seen anywhere), 1 − P(A ∩ B′ ) − P((A ∪ B)′ )
correct calculation for P(B′ ) or P(B)
(A1)
eg 0.44 + 0.19, 0.81 − 0.62 + 0.18
correct substitution into
P(A∩B ′)
P(B ′)
(A1)
0.44
0.44
eg 0.19+0.44
, 1−0.37
0.698412
P(A | B′ ) = 44
(exact), 0.698
63
[4 marks]
(M1)
A1 N3
Contestants in a TV gameshow try to get through three walls by passing through
doors without falling into a trap. Contestants choose doors at random.
If they avoid a trap they progress to the next wall.
If a contestant falls into a trap they exit the game before the next contestant
plays.
Contestants are not allowed to watch each other attempt the game.
The first wall has four doors with a trap behind one door.
Ayako is a contestant.
19a. Write down the probability that Ayako avoids the trap in this wall.
[1 mark]
Markscheme
* This question is from an exam for a previous syllabus, and may contain
minor differences in marking or structure.
3
(0.75, 75%)
4
(A1)
[1 mark]
Natsuko is the second contestant.
19b. Find the probability that only one of Ayako and Natsuko falls into a trap [3 marks]
while attempting to pass through a door in the first wall.
Markscheme
3
4
&times;
1
4
+
1
4
&times;
3
OR
4
2&times;
3
4
&times;
1
4
(M1)(M1)
Note: Award (M1) for their product 14 &times; 34 seen, and (M1) for adding their
two products or multiplying their product by 2.
=
3
8
6
( 16
, 0.375, 37.5% )
(A1)(ft) (G3)
Note: Follow through from part (a), but only if the sum of their two fractions is
1.
[3 marks]
The second wall has five doors with a trap behind two of the doors.
The third wall has six doors with a trap behind three of the doors.
The following diagram shows the branches of a probability tree diagram for a
contestant in the game.
19c. Copy the probability tree diagram and write down the relevant
probabilities along the branches.
[3 marks]
Markscheme
(A1)(ft)(A1)(A1)
Note: Award (A1) for each correct pair of branches. Follow through from part
(a).
[3 marks]
19d. A contestant is chosen at random. Find the probability that this
contestant fell into a trap while attempting to pass through a door in
the second wall.
[2 marks]
Markscheme
3
4
&times;
2
5
(M1)
Note: Award (M1) for correct probabilities multiplied together.
=
3
10
(0.3, 30% )
(A1)(ft) (G2)
Note: Follow through from their tree diagram or part (a).
[2 marks]
19e. A contestant is chosen at random. Find the probability that this
contestant fell into a trap.
[3 marks]
Markscheme
1−
3
4
2
5
&times;
&times;
3
OR
6
1
4
+
3
4
&times;
2
5
+
3
4
&times;
3
5
&times;
3
6
(M1)(M1)
Note: Award (M1) for 34 &times; 35 &times; 36 and (M1) for subtracting their correct
probability from 1, or adding to their 14 + 34 &times; 25 .
=
93
120
( 31
, 0.775, 77.5% )
40
(A1)(ft) (G2)
Note: Follow through from their tree diagram.
[3 marks]
[3 marks]
19f. 120 contestants attempted this game.
Find the expected number of contestants who fell into a trap while attempting to
pass through a door in the third wall.
Markscheme
3
4
&times;
3
5
&times;
3
6
&times; 120
(M1)(M1)
Note: Award (M1) for 34
for multiplying by 120.
= 27
&times;
3
5
&times;
3
6
( 34 &times;
3
5
&times;
3
6
OR
27
120
OR
9
) and (M1)
40
(A1)(ft) (G3)
Note: Follow through from their tree diagram or their 34
calculation in part (d)(ii).
[3 marks]
&times;
3
5
&times;
3
from their
6
In a company it is found that 25 % of the employees encountered traffic on their
way to work. From those who encountered traffic the probability of being late for
work is 80 %.
From those who did not encounter traffic, the probability of being late for work is
15 %.
The tree diagram illustrates the information.
20a. Write down the value of a.
[1 mark]
Markscheme
a = 0.2
(A1)
[1 mark]
20b. Write down the value of b.
[1 mark]
Markscheme
b = 0.85 (A1)
[1 mark]
20c. Use the tree diagram to find the probability that an employee
encountered traffic and was late for work.
[2 marks]
Markscheme
0.25 &times; 0.8 (M1)
Note: Award (M1) for a correct product.
= 0.2 ( 15 , 20%) (A1)(G2)
[2 marks]
20d. Use the tree diagram to find the probability that an employee was late [3 marks]
for work.
Markscheme
0.25 &times; 0.8 + 0.75 &times; 0.15 (A1)(ft)(M1)
Note: Award (A1)(ft) for their (0.25 &times; 0.8) and (0.75 &times; 0.15), (M1) for adding
two products.
= 0.313 (0.3125,
5
,
16
31.3%) (A1)(ft)(G3)
Note: Award the final (A1)(ft) only if answer does not exceed 1. Follow
through from part (b)(i).
[3 marks]
20e. Use the tree diagram to find the probability that an employee
encountered traffic given that they were late for work.
[3 marks]
Markscheme
0.25&times;0.8
(A1)(ft)(A1)(ft)
0.25&times;0.8+0.75&times;0.15
Note: Award (A1)(ft) for a correct numerator (their part (b)(i)), (A1)(ft) for a
correct denominator (their part (b)(ii)). Follow through from parts (b)(i) and (b)
(ii).
= 0.64 ( 16
, 64%) (A1)(ft)(G3)
25
Note: Award final (A1)(ft) only if answer does not exceed 1.
[3 marks]
The company investigates the different means of transport used by their
employees in the past year to travel to work. It was found that the three most
common means of transport used to travel to work were public transportation (P ),
car (C ) and bicycle (B ).
The company finds that 20 employees travelled by car, 28 travelled by bicycle
and 19 travelled by public transportation in the last year.
Some of the information is shown in the Venn diagram.
20f. Find the value of x.
[1 mark]
Markscheme
(x =) 3 (A1)
[1 Mark]
20g. Find the value of y.
[1 mark]
Markscheme
(y =) 10 (A1)(ft)
Note: Following through from part (c)(i) but only if their x is less than or equal
to 13.
[1 Mark]
There are 54 employees in the company.
20h. Find the number of employees who, in the last year, did not travel to
work by car, bicycle or public transportation.
[2 marks]
Markscheme
54 − (10 + 3 + 4 + 2 + 6 + 8 + 13) (M1)
Note: Award (M1) for subtracting their correct sum from 54. Follow through
from their part (c).
= 8 (A1)(ft)(G2)
Note: Award (A1)(ft) only if their sum does not exceed 54. Follow through
from their part (c).
[2 marks]
20i. Find
n ((C ∪ B) ∩ P ′ ).
[2 marks]
Markscheme
6 + 8 + 13 (M1)
Note: Award (M1) for summing 6, 8 and 13.
27 (A1)(G2)
[2 marks]
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