Math SL Day 66 Probability Practice [196 marks]
[2 marks]1a.
Events and are independent with and .
Find .
Markschemevalid interpretation (may be seen on a Venn diagram) (M1)
eg
A1 N2
[2 marks]
A B P(A ∩ B) = 0.2 P(A′ ∩ B) = 0.6
P(B)
P(A ∩ B) + P(A′ ∩ B), 0.2 + 0.6
P(B) = 0.8
[4 marks]1b. Find .
Markschemevalid attempt to find (M1)
eg
correct working for (A1)
eg
correct working for (A1)
eg
A1 N3
[4 marks]
P(A ∪ B)
P(A)
P(A ∩ B) = P(A) × P(B), 0.8 × A = 0.2
P(A)
0.25, 0.20.8
P(A ∪ B)
0.25 + 0.8 − 0.2, 0.6 + 0.2 + 0.05
P(A ∪ B) = 0.85
2a. [3 marks]
The following Venn diagram shows the events and , where and
. The values and are probabilities.
(i) Write down the value of .
(ii) Find the value of .
A B
P(A) = 0.4, P(A ∪ B) = 0.8P(A ∩ B) = 0.1 p q
q
p
Markscheme(i)
A1 N1
(ii) appropriate approach (M1)
eg
A1 N2
[3 marks]
q = 0.1
P(A) − q, 0.4 − 0.1
p = 0.3
[3 marks]2b. Find .
Markschemevalid approach (M1)
eg
correct values (A1)
eg
A1 N2
[3 marks]
P(B)
P(A ∪ B) = P(A) + P(B) − P(A ∩ B), P(A ∩ B) + P(B ∩ A′)
0.8 = 0.4 + P(B) − 0.1, 0.1 + 0.4
P(B) = 0.5
3a. [5 marks]
In a class of 21 students, 12 own a laptop, 10 own a tablet, and 3 own neither.
The following Venn diagram shows the events “own a laptop” and “own a tablet”.
The values , , and represent numbers of students.
(i) Write down the value of .
(ii) Find the value of .
(iii) Write down the value of and of .
Markscheme(i)
A1 N1
(ii) valid approach (M1)
eg
A1 N2
(iii) A1A1 N2
p q r s
p
q
r s
p = 3
(12 + 10 + 3) − 21, 22 − 18
q = 4
r = 8, s = 6
3b. [4 marks]
A student is selected at random from the class.
(i) Write down the probability that this student owns a laptop.
(ii) Find the probability that this student owns a laptop or a tablet but not both.
Markscheme(i)
A2 N2
(ii) valid approach (M1)
eg
A1 N2
(= )1221
47
8 + 6, r + s
(= )1421
23
3c. [4 marks]
Two students are randomly selected from the class. Let be the event a “student owns a laptop”.
(i) Copy and complete the following tree diagram. (Do not write on this page.)
(ii) Write down the probability that the second student owns a laptop given that the first owns a laptop.
Markscheme
(i) A1A1A1 N3
(ii) A1 N1
[4 marks]
L
1120
4a. [2 marks]
Let and be independent events, with and , where .
Write down an expression for in terms of.
C
D P(C) = 2k P(D) = 3k2 0 < k < 0.5
P(C ∩ D)k
Markscheme (A1)
A1 N2
[2 marks]
P(C ∩ D) = 2k × 3k2
P(C ∩ D) = 6k3
[3 marks]4b. Find .
MarkschemeMETHOD 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
[3 marks]
Total [7 marks]
P(C′|D)
P(C′ ∩ D)
0.4 × 0.27,0.27 − 0.162,0.108
P(C′|D) = , P(C ′∩D)
0.27
(1−2k)(3k2)
3k2
P(C′|D) = 0.4
P(C′|D) = P(C′)
P(C′) = 1 − P(C)
1 − 2k, 1 − 0.6
P(C′|D) = 0.4
[2 marks]5a.
Ann and Bob play a game where they each have an eight-sided die. Ann’s die has three green faces and five red faces; Bob’s die hasfour green faces and four red faces. They take turns rolling their own die and note what colour faces up. The first player to roll greenwins. Ann rolls first. Part of a tree diagram of the game is shown below.
Find the probability that Ann wins on her first roll.
Markschemerecognizing Ann rolls green (M1)
eg
A1 N2
[2 marks]
P(G)
38
Find the probability that Ann wins the game.
5b. [7 marks]
Markschemerecognize the probability is an infinite sum (M1)
eg Ann wins on her roll or roll or roll…,
recognizing GP (M1)
(seen anywhere) A1
(seen anywhere) A1
correct substitution into infinite sum of GP A1
eg
correct working (A1)
eg
A1 N1
[7 marks]
Total [15 marks]
1st 2nd 3rd S∞
u1 = 38
r = 2064
, ( ) , 38
1− 516
38
1
1−( × )58
48
1
1− 516
, ×38
1116
38
1611
P (Ann wins) = (= )4888
611
[2 marks]6a.
Adam travels to school by car ( ) or by bicycle ( ). On any particular day he is equally likely to travel by car or by bicycle.
The probability of being late ( ) for school is if he travels by car.
The probability of being late for school is if he travels by bicycle.
This information is represented by the following tree diagram.
Find the value of .
Markschemecorrect working (A1)
eg
A1 N2
[2 marks]
C B
L 16
13
p
1 − 16
p = 56
[2 marks]6b. Find the probability that Adam will travel by car and be late for school.
Markschememultiplying along correct branches (A1)
eg
A1 N2
[2 marks]
×12
16
P(C ∩ L) = 112
[4 marks]6c. Find the probability that Adam will be late for school.
Markschememultiplying along the other branch (M1)
eg
adding probabilities of their mutually exclusive paths (M1)
eg
correct working (A1)
eg
A1 N3
[4 marks]
×12
13
2
× + ×12
16
12
13
+112
16
P(L) = (= )312
14
[3 marks]6d. Given that Adam is late for school, find the probability that he travelled by car.
Markschemerecognizing conditional probability (seen anywhere) (M1)
eg
correct substitution of their values into formula (A1)
eg
A1 N2
[3 marks]
P(C|L)
1123
12
P(C|L) = 13
6e. [4 marks]Adam will go to school three times next week.
Find the probability that Adam will be late exactly once.
Markschemevalid approach (M1)
eg , three ways it could happen
correct substitution (A1)
eg
correct working (A1)
eg
A1 N2
[4 marks]
Total [15 marks]
X ∼ B (3, ) , ( ) ( )2, ( 3
1)1
414
34
( 31
) ( )1( )2, × × + × × + × ×1
434
14
34
34
34
14
34
34
34
14
3 ( ) ( ) , + +14
916
964
964
964
2764
[2 marks]7a.
The following cumulative frequency graph shows the monthly income, dollars, of families.
Find the median monthly income.
I 2000
Markschemerecognizing that the median is at half the total frequency (M1)
eg
A1 N2
[2 marks]
20002
m = 2500 (dollars)
7b. [4 marks](i) Write down the number of families who have a monthly income of dollars or less.
(ii) Find the number of families who have a monthly income of more than dollars.
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
Note: If working shown, award M1A1A1 for , using the table.
[4 marks]
2000
4000
500 2000
1850
2000
2000 − 1850
150 4000
128 + 22 = 150
7c. [2 marks]The families live in two different types of housing. The following table gives information about the number of families living ineach type of housing and their monthly income .
Find the value of .
Markschemecorrect calculation (A1)
eg (A1)
A1 N2
[2 marks]
2000I
p
2000 − (436 + 64 + 765 + 28 + 122), 1850 − 500 − 765
p = 585
7d. [2 marks]A family is chosen at random.
(i) Find the probability that this family lives in an apartment.
(ii) Find the probability that this family lives in an apartment, given that its monthly income is greater than dollars.4000
Markscheme(i) correct working (A1)
eg
(exact) A1 N2
(ii) correct working/probability for number of families (A1)
eg
A1 N2
[4 marks]
436 + 765 + 28
0.6145
, 0.615 [0.614, 0.615]12292000
122 + 28, , 0.0751502000
0.186666
(= ) , 0.187 [0.186, 0.187]28150
1475
[2 marks]7e. Estimate the mean monthly income for families living in a villa.
Markschemeevidence of using correct mid-interval values ( ) (A1)
attempt to substitute into (M1)
eg
A1 N2
[3 marks]
Total [15 marks]
1500,3000,4500
∑ fx
∑ f
1500×64+3000×p+4500×12264+585+122
3112.84
3110 [3110, 3120] (dollars)
8. [6 marks]Celeste wishes to hire a taxicab from a company which has a large number of taxicabs.
The taxicabs are randomly assigned by the company.
The probability that a taxicab is yellow is 0.4.
The probability that a taxicab is a Fiat is 0.3.
The probability that a taxicab is yellow or a Fiat is 0.6.
Find the probability that the taxicab hired by Celeste is not a yellow Fiat.
Markschemerecognize 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
A1
recognize need for complement of
(M1)
eg
A1 N3
[6 marks]
P(Y ∩ F), 0.3 × 0.4
P(Y ∩ F)
P(Y ) + P(F) − P(Y ∪ F)
0.4 + 0.3 − 0.6
P(Y ∩ F) = 0.1
Y ∩ F
1 − P(Y ∩ F), 1 − 0.1
P ((Y ∩ F)′) = 0.9
9a. [3 marks]
Bill and Andrea play two games of tennis. The probability that Bill wins the first game is.
If Bill wins the first game, the probability that he wins the second game is.
If Bill loses the first game, the probability that he wins the second game is.
Copy and complete the following tree diagram. (Do not write on this page.)
45
56
23
Markscheme
A1A1A1 N3
Note: Award A1 for each correct bold probability.
[3 marks]
[2 marks]9b. Find the probability that Bill wins the first game and Andrea wins the second game.
Markschememultiplying along the branches (may be seen on diagram) (M1)
eg
A1 N2
[2 marks]
×45
16
( )430
215
[4 marks]9c. Find the probability that Bill wins at least one game.
MarkschemeMETHOD 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 2
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]
× , × , ×45
56
45
16
15
23
× + × + × , + ×45
56
45
16
15
23
45
15
23
+ + , + +2030
430
215
23
215
215
(= )2830
1415
= ×15
13
1 − p, 1 − 115
1415
[5 marks]9d. Given that Bill wins at least one game, find the probability that he wins both games.
MarkschemeP (B wins both)
A1
evidence of recognizing conditional probability (R1)
eg
correct substitution (A2)
eg
A1 N3
[5 marks]
× (= )45
56
23
P(A |B ), P (Bill wins both |Bill wins at least one), tree diagram
×45
56
1415
(= )2028
57
10a. [2 marks]
Let
and
be independent events, where
and
.
Find
.
A
B
P(A) = 0.3
P(B) = 0.6
P(A ∩ B)
Markschemecorrect substitution (A1)
eg
A1 N2
[2 marks]
0.3 × 0.6
P(A ∩ B) = 0.18
10b. [2 marks]Find
.
Markschemecorrect substitution (A1)
eg
A1 N2
[2 marks]
P(A ∪ B)
P(A ∪ B) = 0.3 + 0.6 − 0.18
P(A ∪ B) = 0.72
10c. [1 mark]On the following Venn diagram, shade the region that represents
.
Markscheme
A1 N1
A ∩ B′
10d. [2 marks]Find
.
Markschemeappropriate approach (M1)
eg
(may be seen in Venn diagram) A1 N2
[2 marks]
P(A ∩ B′)
0.3 − 0.18, P(A) × P(B′)
P(A ∩ B′) = 0.12
[4 marks]11a.
Samantha goes to school five days a week. When it rains, the probability that she goes to school by bus is 0.5. When it
does not rain, the probability that she goes to school by bus is 0.3. The probability that it rains on any given day is 0.2.
On a randomly selected school day, find the probability that Samantha goes to school by bus.
Markschemeappropriate approach (M1)
eg
, tree diagram,
one correct multiplication (A1)
eg
correct working (A1)
eg
A1 N3
[4 marks]
P(R ∩ B) + P(R′ ∩ B)
0.2 × 0.5, 0.24
0.2 × 0.5 + 0.8 × 0.3, 0.1 + 0.24
P(bus) = 0.34(exact)
[3 marks]11b. Given that Samantha went to school by bus on Monday, find the probability that it was raining.
Markschemerecognizing conditional probability (R1)
eg
correct working A1
eg
A1 N2
[3 marks]
P(A|B) =P(A∩B)
P(B)
0.2×0.50.34
P(R|B) = , 0.294517
11c. [2 marks]In a randomly chosen school week, find the probability that Samantha goes to school by bus on exactly three
days.
Markschemerecognizing binomial probability (R1)
eg
,
A1 N2
[2 marks]
X ∼ B(n, p)
( 53
)(0.34)3, (0.34)3(1 − 0.34)2
P(X = 3) = 0.171
11d. [5 marks]After
school days, the probability that Samantha goes to school by bus at least once is greater than
. Find the smallest value of
.
n
0.95
n
MarkschemeMETHOD 1
evidence of using complement (seen anywhere) (M1)
eg
valid approach (M1)
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]
1 − P (none), 1 − 0.95
1 − P (none) > 0.95, P (none) < 0.05, 1 − P (none) = 0.95
1 − (0.66)n > 0.95, (0.66)n = 0.05
n > 7.209 (accept n = 7.209)
n = 8
P(X ⩾ 1)
n = 7, P(X ⩾ 1) = 0.9454, n = 8, P(X ⩾ 1) = 0.939
0.9639 > 0.95
n = 8
12a. [2 marks]
Two events
and
are such that
and
.
Given that
and
are mutually exclusive, find
.
Markschemecorrect approach (A1)
eg
A1 N2
[2 marks]
A
B
P(A) = 0.2
P(A ∪ B) = 0.5
A
B
P(B)
0.5 = 0.2 + P(B), P(A ∩ B) = 0
P(B) = 0.3
12b. [4 marks]Given that
and
are independent, find
.
A
B
P(B)
MarkschemeCorrect expression for
(seen anywhere) A1
eg
attempt to substitute into correct formula for
(M1)
eg
correct working (A1)
eg
A1 N3
[4 marks]
P(A ∩ B)
P(A ∩ B) = 0.2P(B), 0.2x
P(A ∪ B)
P(A ∪ B) = 0.2 + P(B) − P(A ∩ B), P(A ∪ B) = 0.2 + x − 0.2x
0.5 = 0.2 + P(B) − 0.2P(B), 0.8x = 0.3
P(B) = (= 0.375, exact)38
13a. [5 marks]
Jar A contains three red marbles and five green marbles. Two marbles are drawn from the jar, one after the other,without replacement.
Find the probability that
(i) none of the marbles are green;
(ii) exactly one marble is green.
Markscheme(i) attempt to find
(M1)
eg ,
,
A1 N2
(ii) attempt to find (M1)
eg ,
,
recognizing two ways to get one red, one green (M1)
eg , ,
A1 N2
[5 marks]
P(red) × P(red)
×38
27
×38
38
×38
28
P(none green) = 656
(= )328
P(red) × P(green)
×58
37
×38
58
1556
2P(R) × P(G)× + ×5
837
38
57
× × 238
58
P(exactly one green) = 3056
(= )1528
[3 marks]13b. Find the expected number of green marbles drawn from the jar.
Markscheme (seen anywhere) (A1)
correct substitution into formula for A1
eg ,
expected number of green marbles is
A1 N2
[3 marks]
P(both green) = 2056
E(X)
0 × + 1 × + 2 ×656
3056
2056
+3064
5064
7056
(= )54
13c. [2 marks]
Jar B contains six red marbles and two green marbles. A fair six-sided die is tossed. If the score is or, a marble is drawn from jar A. Otherwise, a marble is drawn from jar B.
(i) Write down the probability that the marble is drawn from jar B.
(ii) Given that the marble was drawn from jar B, write down the probability that it is red.
Markscheme(i)
A1 N1
(ii)
A1 N1
[2 marks]
12
P(jar B) = 46
(= )23
P(red| jar B) = 68
(= )34
[6 marks]13d. Given that the marble is red, find the probability that it was drawn from jar A.
Markschemerecognizing 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
[6 marks]
P(A|R)P(jar A and red)
P(red)
P(jar A and red) = ×13
38
(= )18
P(jar B and red) = ×23
68
(= )12
P(red) = × + ×13
38
23
68
P(jar A|red) =×1
338
× + ×13
38
23
68
1858
P(jar A|red) = 15
14a. [4 marks]
A running club organizes a race to select girls to represent the club in a competition.
The times taken by the group of girls to complete the race are shown in the table below.
Find the value of and of
.pq
Markschemeattempt to find
(M1)
eg ,
A1 N2
attempt to find (M1)
eg ,
A1 N2
[4 marks]
p
120 − 7050 + 20 + x = 120
p = 50
q
180 − 20200 − 20 − 20
q = 160
14b. [3 marks]A girl is chosen at random.
(i) Find the probability that the time she takes is less than minutes.
(ii) Find the probability that the time she takes is at least minutes.
Markscheme(i)
A1 N1
(ii) valid approach (M1)
eg ,
A1 N2
[3 marks]
14
26
70200
(= )720
20 + 20200 − 16040200
(= )15
14c. [4 marks]A girl is selected for the competition if she takes less than minutes to complete the race.
Given that of the girls are not selected,
(i) find the number of girls who are not selected;
(ii) find .
x
40%
x
Markscheme(i) attempt to find number of girls (M1)
eg ,
are not selected A1 N2
(ii) are selected (A1)
A1 N2
[4 marks]
0.4× 20040
100
80
120
x = 20
14d. [4 marks]Girls who are not selected, but took less than minutes to complete the race, are allowed another chance to be selected. The new times taken by these girls are
shown in the cumulative frequency diagram below.
(i) Write down the number of girls who were allowed another chance.
(ii) Find the percentage of the whole group who were selected.
25
Markscheme(i)
given second chance A1 N1
(ii) took less than minutes (A1)
attempt to find their selected total (may be seen in calculation) (M1)
eg
, their answer from (i)
() A1 N3
[4 marks]
30
2020
%
120 + 20(= 140)120+
70%
[2 marks]15a.
A Ferris wheel with diameter metres rotates clockwise at a constant speed. The wheel completes rotations every hour. The bottom of the wheel is
metres above the ground.
A seat starts at the bottom of the wheel.
Find the maximum height above the ground of the seat.
Markschemevalid approach (M1)
eg ,
maximum height (m) A1 N2
[2 marks]
1222.413
13 + diameter13 + 122
= 135
15b. [2 marks]
After t minutes, the height metres above the ground of the seat is given by
(i) Show that the period of is minutes.
(ii) Write down the exact value of .
Markscheme(i) period
A1
period minutes AG N0
(ii)
A1 N1
[2 marks]
h
h = 74 + a cos bt.
h25
b
= 602.4
= 25
b = 2π25
(= 0.08π)
15c. [3 marks]Find the value of .
MarkschemeMETHOD 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]
a
max − 74|a| = 135−13
274 − 13
|a| = 61a = 61
a = −61
135 = 74 + a cos( )2π×12.525
135 = 74 + a cos(π)13 = 74 + a
a = −61
15d. [4 marks]Sketch the graph of , for
.h0 ≤ t ≤ 50
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]
2
15e. [5 marks]In one rotation of the wheel, find the probability that a randomly selected seat is at least metres above the ground.
Markschemesetting 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]
105
h > 105105 = 74 + a cos bty = 105
t = 8.371 …t = 16.628 …t = 33.371 …t = 41.628 …
16.628−8.37125
t1−t2
252×8.257
502(12.5−8.371)
25
p = 0.330
16a. [2 marks]At a large school, students are required to learn at least one language, Spanish or French. It is known that of the students learn Spanish, and learn French.
Find the percentage of students who learn both Spanish and French.
75%40%
Markschemevalid approach (M1)
e.g. Venn diagram with intersection, union formula,
(accept) A1 N2
[2 marks]
P(S ∩ F) = 0.75 + 0.40 − 1
1515%
16b. [2 marks]At a large school, students are required to learn at least one language, Spanish or French. It is known that of the students learn Spanish, and learn French.
Find the percentage of students who learn Spanish, but not French.
Markschemevalid approach involving subtraction (M1)
e.g. Venn diagram,
60 (accept) A1 N2
[2 marks]
75%40%
75 − 15
60%
16c. [5 marks]At a large school, students are required to learn at least one language, Spanish or French. It is known that of the students learn Spanish, and learn French.
At this school, of the students are girls, and of the girls learn Spanish.
A student is chosen at random. Let G be the event that the student is a girl, and let S be the event that the student learnsSpanish.
(i) Find .
(ii) Show that G and S are not independent.
75%40%
52%85%
P(G ∩ S)
Markscheme(i) valid approach (M1)
e.g. tree diagram, multiplying probabilities,
correct calculation (A1)
e.g.
(exact) A1 N3
(ii) valid reasoning, with words, symbols or numbers (seen anywhere) R1
e.g. ,
, not equal,
one correct value A1
e.g. ,
,
G and S are not independent AG N0
[5 marks]
P(S|G) × P(G)
0.52 × 0.85
P(G ∩ S) = 0.442
P(G) × P(S) ≠ P(G ∩ S)P(S|G) ≠ P(S)
P(G) × P(S) = 0.39P(S|G) = 0.850.39 ≠ 0.442
16d. [6 marks]At a large school, students are required to learn at least one language, Spanish or French. It is known that of the students learn Spanish, and learn French.
At this school, of the students are girls, and of the girls learn Spanish.
A boy is chosen at random. Find the probability that he learns Spanish.
75%40%
52%85%
MarkschemeMETHOD 1
are boys (seen anywhere) A1
e.g.
appropriate approach (M1)
e.g.
correct approach to find P(boy and Spanish) (A1)
e.g. ,
, 0.308
correct substitution (A1)
e.g. ,
correct manipulation (A1)
e.g.
,
A1 N3
[6 marks]
METHOD 2
are boys (seen anywhere) A1
e.g. 0.48 used in tree diagram
appropriate approach (M1)
e.g. tree diagram
correctly labelled branches on tree diagram (A1)
e.g. first branches are boy/girl, second branches are Spanish/not Spanish
correct substitution (A1)
e.g.
correct manipulation (A1)
e.g. ,
,
[6 marks]
48%
P(B) = 0.48
P(girl and Spanish) + P(boy and Spanish) = P(Spanish)
P(B ∩ S)= P(S) − P(G ∩ S)P(B ∩ S)= P(S|B) × P(B)
0.442 + 0.48x = 0.750.48x = 0.308
P(S|B) = 0.3080.48
P(Spanish|boy) = 0.641666 …0.6416̄
P(Spanish|boy) = 0.642[0.641, 0.642]
48%
0.442 + 0.48x = 0.75
0.48x = 0.308P(S|B) = 0.308
0.48
P(Spanish|boy) = 0.641666 …0.6416̄
P(Spanish|boy) = 0.642[0.641, 0.642]
[1 mark]17a.
Events A and B are such that , and
.
The values q , r , s and t represent probabilities.
Write down the value of t .
Markscheme A1 N1
[1 mark]
P(A) = 0.3P(B) = 0.6P(A ∪ B) = 0.7
t = 0.3
17b. [3 marks](i) Show that .
(ii) Write down the value of q and of s .
Markscheme(i) correct values A1
e.g. ,
AG N0
(ii) , A1A1 N2
[3 marks]
r = 0.2
0.3 + 0.6 − 0.70.9 − 0.7
r = 0.2
q = 0.1s = 0.4
17c. [3 marks](i) Write down .
(ii) Find .
Markscheme(i)
A1 N1
(ii) A2 N2
[3 marks]
P(B′)
P(A|B′)
0.4
P(A|B′) = 14
18a. [5 marks]Bag A contains three white balls and four red balls. Two balls are chosen at random without replacement.
(i) Copy and complete the following tree diagram.
(ii) Find the probability that two white balls are chosen.
Markscheme(i)
A1A1A1 N3
(ii) multiplying along the correct branches (may be seen on diagram) (A1)
e.g.
A1 N2
[5 marks]
, and ( , and )46
36
36
23
12
12
×37
26
(= )642
17
18b. [5 marks]Bag A contains three white balls and four red balls. Two balls are chosen at random without replacement.
Bag B contains four white balls and three red balls. When two balls are chosen at random without replacement from bag B,the probability that they are both white is
.
A standard die is rolled. If 1 or 2 is obtained, two balls are chosen without replacement from bag A, otherwise they arechosen from bag B.
Find the probability that the two balls are white.
27
Markscheme ,
(seen anywhere) (A1)(A1)
appropriate approach (M1)
e.g.
correct calculation A1
e.g. ,
A1 N3
[5 marks]
P(bagA) = (= )26
13
P(bagB) = (= )46
23
P(WW ∩ A) + P(WW ∩ B)
× + ×13
17
23
27
+242
842
P(2W) = (= )60252
521
18c. [4 marks]Bag A contains three white balls and four red balls. Two balls are chosen at random without replacement.
Bag B contains four white balls and three red balls. When two balls are chosen at random without replacement from bag B,the probability that they are both white is
.
A standard die is rolled. If 1 or 2 is obtained, two balls are chosen without replacement from bag A, otherwise they arechosen from bag B.
Given that both balls are white, find the probability that they were chosen from bag A.
27
Printed for North Hills Preparatory
© International Baccalaureate Organization 2018
International Baccalaureate® - Baccalauréat International® - Bachillerato Internacional®
Markschemerecognizing conditional probability (M1)
e.g.
,
correct numerator (A1)
e.g.
correct denominator (A1)
e.g.
probability
A1 N3
[4 marks]
P(A∩B)P(B)
P(A|WW) =P(WW∩A)
P(WW)
P(A ∩ WW) = × ,642
26
121
,6252
521
(= )84420
15