Save My Exams! – The Home of Revision
For more awesome GCSE and A level resources, visit us at www.savemyexams.co.uk/
Probability distribution table
Mark Scheme 2
Level
International A Level
Subject
Maths
Exam Board
CIE
Topic
Descrete random variables
Sub Topic
Probability distribution table
Booklet
Mark Scheme 2
Time Allowed:
59 minutes
Score:
/ 49
Percentage:
/100
Grade Boundaries:
A*
>85%
A
777.5%
B
C
D
E
U
70%
62.5%
57.5%
45%
<45%
Save My Exams! – The Home of Revision
For more awesome GCSE and A level resources, visit us at www.savemyexams.co.uk/
1
P(0) = 7/10 × 6/9 × 5/8 = 210/720
P(1) = 3/10 × 7/9 × 6/8 × 3C1 = 378/720
P(2) = 3/10 × 2/9 × 7/8 × 3C2 = 126/720
P(3) = 3/10 × 2/9 × 1/8 = 6/720 (1/120)
2
−3p + 2r + 4 × 0.4 = 2.3
(−3)2p + 22r + 42 × 0.4 – 2.32 = 3.01
p + q + r + 0.4 = 1
−3p + 2r = 0.7
9p + 4r = 1.9
so − 9p + 6r = 2.1 or − 6p + 4r = 1.4
4r + 6r = 1.9 + 2.1 or 9p + 6p = 1.9−1.4
1
2
r=
(0.4), p =
(0.0333)
30
5
1
(0.167)
q = 0.6 – 0.4 – 0.0333 =
6
3
(i
B1
B1
B1
B1
Finding P(0, 1, 2, 3)
1 or 2 correct
3 correct
[4] All correct
B1
B1
B1
Correct unsimplified equation, oe
Correct unsimplified equation, oe
Correct equation, oe
M1
Obtain an equation in 1 unknown
A1
One correct answer
A1
6
Remaining two answers correct
M1
Considering values of X of 1, 2, 3, 4
M1
Attempting to find the probability of at
least 2 values of X
P(X = 3) = P(GGGB) × 4C3 = 3/7
A1
One correct probability
P(X = 4) = P(GGGG) × 4C4 = 1/14
A1
All correct
P(1) = 5C1 / 8C4 = 1/14
M1
Considering values of X of 1, 2, 3, 4
P(2) = 3C2 × 5C2 / 8C4 = 3/7
M1
Dividing by 8C4
P(3) = 3C1 × 5C3 / 8C4 = 3/7
A1
One correct probability
P(4) = 5C4 / 8C4 = 1/14
A1
P(X = 1) = P(GBBB) 4 × C1
= 5/8 × 3/7 × 2/6 × 1/5 × 4 = 1/14
P(X = 2) = P(GGBB) × 4C2 = 3/7
OR
(ii) Var(X) = 1/14 + 12/7 + 27/7 + 16/14 – (5/2)2
= 15/28 (0.536)
M1
A1
[4] All correct
Using a variance formula correctly
with mean2 subtracted numerically, no
extra division
[2] Correct final answer
Save My Exams! – The Home of Revision
For more awesome GCSE and A level resources, visit us at www.savemyexams.co.uk/
4
4p +2 5p2 + 1.5p + 2.5p + 1.5p = 1
10p + 19p – 2 = 0
M1
Summing 5 probs to = 1 can be implied
p = 0.1 or –2
A1
For 0.1 seen with or without –2
p = 0.1
A1
Choosing 0.1 must be by rejecting –2
[3]
5
(i)
x
1
Prob k
2
2k
3
3k
5k
15k = 1
k = 1/15 (0.0667)
(i
A1
correct answer
[3]
(ii) E(X)
= k + 4k + 9k + 16k + 25k
= 55k = 11/3 (3.67)
6
M1
1, 2, 3, 4, 5 seen, together with some
probabilities involving k but not x
summing probs involving k to 1
M1
4k
M1
using Σpx no dividing
correct answer, ft on 55k, 0 < k < 1
A1ft
[2]
–0.16 – p + 0.16 + 2q + 0.66 = 1.05
M1
Attempt at Σpx = 1.05 no dividing
– p + 2q = 0.39
p + q = 0.42
q = 0.27
p = 0.15
A1
B1
Correct simplified equation
Accept p = 0.42 – q oe
(ii) Var (X) = 4 × 0.08 + p + 0.16 + 4q
+ 1.98 – (1.05 2
= 2.59
A1
[4] Both answers correct
M1
Subst in Σpx2 – mean2 formula, mean2 subt
numerically, p +ve and < 1
[2] Correct answer
A1
Save My Exams! – The Home of Revision
For more awesome GCSE and A level resources, visit us at www.savemyexams.co.uk/
7
(i) 40 = 120 / 3 so r = 3
P(40) = 3/45 = 1/15 AG
(ii)
x
120
120
40
P(X = x) 1/45 2
24
4
5/45
5/45
2
6
17.14
15
M1
A1
30
(i
B1ft
M1
[1] ft their table
Adding 5 probabilities o.e.
A1
[2] Correct answer
P(odd) = 2/3 or 0.667
P(7) = 8C7 × (2 / 3) 7 (1/ 3)
= 0.156
B1
P(8) = (2/3)8 = 0.0390
M1
P(7 or 8) = 0.195 (1280/6561)
A1
(ii)
x
P(X=x)
2
11/36
/36
x
P(X=x)
9
4/36
4/36
(iii) E(X) =
M1
4
2
4/36
1
4
11
8/36
[3]
8 or 9 values for x, correct to nearest integer
One correct probability apart from 1/15
Correct table
13.3
(iii) 40/3 oe (13.3)
(iv) P(18 < X < 100) = (2 + 3 + 4 + 5
+ 6)/4
= 20/45 (4/9) (0.444)
8
B1
B1
B1
[2]
r = 3 seen or obtained from table
Given answer legit obtained
12
4/36
B1
B2
∑ p i xi
Can be implied if normal approx used with
µ = 5.333(= 8 × 2/3)
Binomial expression with C in and 2/3 and 1/3 in
powers summing to 8
Summing P(7) + P(8) binomial expressions
[4] Correct answer
Values of x all correct in table of probabilities
[3] All probs correct and not duplicated, –1 ee
∑ pi xi , all p < 1 and no further
= 2 × 1/36 + 4 × 2/36 + .....
M1
attempt to find
= 312/36 (26/3) (8.67)
A1
division of any sort
[2] correct answer
(iv) P(X > E(X)) = P(X = 9, 10, 11, 12)
M1
= 20/36 (5/9) (0.556)
A1
attempt to add their relevant probs
[2] correct answer