A LEVEL MATHEMATICS ANSWERS AND MARKSCHEMES

PROBABILITY

1. a) i) 9 possible numbers 11, 12, 13 21, 22, 23 31, 32, 33 B1

P(prime) = B1[2]

ii) B1[1]

b) P(prime | >13) = M1

A1[4]

2. a) 12 – 6 – 4 = 2 B1[1]

b) M1

M1(V. dgm or other)

P(B) = a + b = B1

P(C) = b +

Using : A1

b + = 3b b = M1 (solving) A1So 3 had both vegetables B1

[6]

Page 1

126

122

a b

B C

A LEVEL MATHEMATICS ANSWERS AND MARKSCHEMES

PROBABILITY

3. a) The probability that A or B or both happens is equal to the probability that A occurs B1 (for plus the probability that B occurs subtract the probability that both A and B occur B1 (for

[2]

b)

P( B) = P(A B) = 0 B1

P(A|B) = M1

= 0 A1[3]

c) Mutually exclusive B1[1]

d) M1 A1The probability that neither A nor B happens B1

[3]

Page 2

21

31

0

A B

A LEVEL MATHEMATICS ANSWERS AND MARKSCHEMES

PROBABILITY

4. a)

P(positive) = 0.002 1 + 0.998 0.1 M1 A1= 0.1018 A1

[3]

b) P(disease | positive) = M1

=

A1 cao

[4]

c) Someone with a positive result has a small probability of actually having the disease, so not a very good test. B1

[1]

Page 3

0.002

0.998

1

0

0.1

0.9

Disease

No Disease

positive

positive

negative

negative

A LEVEL MATHEMATICS ANSWERS AND MARKSCHEMES

PROBABILITY

5. a)

P(large) = 0.2 0.2 + 0.3 0.3 + 0.5 0.5 M1 A1= 0.38 A1

[3]

b) P(lamb | small) = M1

=

= (=0.403) A1[5]

c) 3! 0.2 0.3 0.5 M1 A1 (3!)= 0.18 A1

[3]

Page 4

0.2

L

L

S

L

S

T

0.5

0.2

0.5

0.8

0.5

L

S

0.3

0.7

0.3C

A LEVEL MATHEMATICS ANSWERS AND MARKSCHEMES

PROBABILITY

6. a) P(HN) = 0.4 M1 (conditionals)P(NH) = 0.3 A1

M1 (V dgm or other)

a + b + c = 1, since all play at least one sport B1

P(HN) = 0.4

M1

b = 0.4b + 0.4c0.6b = 0.4c3b = 2c A1

P(NH) = 0.3

b = 0.3a + 0.3b0.7b = 0.3a7b = 3a A1

Substitution into M1

b = A1[9]

b) a = = M1 A1[2]

Page 5

b

H N

a c

A LEVEL MATHEMATICS ANSWERS AND MARKSCHEMES

PROBABILITY

7. a) P(RRR) = M1 A1

P(GGG) = A1

P(all the same) = B1[4]

b) P(RYG) = B1 (one combination)3! = 6 arrangements M1so is the probability A1

[3]

iii) M1

P(2 red) + P(3 red) M1P(2 red) = M1 A1

A1

So A1

[6]

Page 6

A LEVEL MATHEMATICS ANSWERS AND MARKSCHEMES

PROBABILITY

8.

a) 0.5 0.6 0.6 + 0.5 0.4 0.3 + 0.5 0.3 0.6 + 0.5 0.7 0.3 M1 A1 A1= 0.435 A1

[4]

b) P( 2)= P(2) + P(3) M1P(3) = 0.5 x 0.7 x 0.7 = 0.245 B1P(2) = 0.5 x 0.4 x 0.7 + 0.5 x 0.3 x 0.4 + 0.5 x 0.7 x 0.3 M1

= 0.305 A1P( 2) = 0.55 A1

[5]

c) P( 2 1) = M1

P( 1)= 1 – P(0) M1= 1 – 0.5 x 0.6 x 0.6 = 0.82 A1

P( 2 1) = A1 ft (0.55)

= (=0.671) A1[5]

Page 7

A

D

A

D

A

D

0.5

0.5

0.6

0.3

0.4

0.7

A

D0.6

0.4

A

D0.6

0.4

A

D0.3

0.7

A

D0.3

0.7

A LEVEL MATHEMATICS ANSWERS AND MARKSCHEMES

PROBABILITY

9. a) P(H) = 0.55P(WH) = B1 (conditional)

P( ) =

P(WH) = M1

P (W and H) = A1[3]

b) P( ) = M1

A1 (0.45)

A1[3]

c) P(W and ) = P( ) – P( and ) M1= 0.45 –0.25= 0.2 A1 ft

[2]

d) Options are:

1st couple 2nd coupleH and W and

and W H and B1H and and W (different cases)

and H and W

P( H) = B1

So probability = M1 A1

= A1[5]

Page 8

A LEVEL MATHEMATICS ANSWERS AND MARKSCHEMES

PROBABILITY

10.a) P( B) = M1

so

P( and B) = A1[2]

b) Need P(A and ) M1

P(A ) =

B1

P(A and ) = B1

So P(just one) = M1 A1[5]

c) P(AB) =

P(A and B) + P( and B) = P(B) M1P(A and B) = A1

P(AB) = B1

[3]

Page 9

A LEVEL MATHEMATICS ANSWERS AND MARKSCHEMES

PROBABILITY

11.a) M1 A1[2]

b) P(both the same) = M1 A1

= A1

P(both toffeesboth the same) = M1

= A1 ft

= A1[6]

c) “Given” first 2 toffees means can assume 1st 2 sweets were toffees M1 (or conditional) So left with 4 toffees, 5 caramels, 3 chocolatesSo P(4th a toffee) = M1 A1

= A1[4]

12.a) P(A B) = 0 B1P(A B) = P(A) + P(B) B1

= 0.7 [2]

b) P(A B) = P(A)P(B) M1= 0.4 0.3= 0.12 A1

P(A B) = P(A) + P(B) – P(A B) M1= 0.4 + 0.3 – 0.12= 0.58 A1

[4]

Page 10

A LEVEL MATHEMATICS ANSWERS AND MARKSCHEMES

PROBABILITY

13.a) P(A) = B1

B 1 2 3 4 5 61 2 3 4 5 6 72 3 4 5 6 7 83 4 5 6 7 8 9 M1 A14 5 6 7 8 9 105 6 7 8 9 10 116 7 8 9 10 11 12

P(B) = B1[4]

b) P(BA) = M1

A1 cao[4]

c) Independent B1Because P(BA) = P(B) B1

[2]

Page 11

A LEVEL MATHEMATICS ANSWERS AND MARKSCHEMES

PROBABILITY

14.a) 100 – 2 – 20 = 78% B1[1]

b)

x = 0.2 B1So 40% read Private Eye B1

[2]

c) y = 0.58 B1

P(just Private Eyejust one) M1

A1

(=0.256) A1[4]

15.a) xy = 0.12 B1 (Use of indept)x + y – xy = 0.58 M1 A1

[3]

b) y = 0.58 + 0.12 - x= 0.7 – x M1 (solving)

So: x(0.7 – x) = 0.120.7x – x2 = 0.12x2 – 0.7x + 0.12 = 0(x – 0.3)(x – 0.4) = 0 M1 (solving)x = 0.3, 0.4 A2So P(A) = 0.4; P(B) = 0.3 B1

[5]

c) P(AB) = P(A) = 0.4 B1[1]

Page 12

0.2

P N

x y

0.02

A LEVEL MATHEMATICS ANSWERS AND MARKSCHEMES

PROBABILITY

16.a) P(W1 W2) = M1 A1

= A1[3]

b) P(R1 W2) = M1

= A1

P(W2)= M1

= A1[4]

c) P(W1R2) = M1

=

= A1[5]

Page 13

A LEVEL MATHEMATICS ANSWERS AND MARKSCHEMES

PROBABILITY

17.a)

P(pass M1) = 0.5p + 0.3(1-p) M1 A1= 0.2p + 0.3 A1

[3]

b) P(fails P1passes M1) = M1

= A1 A1 ft

= A1

[4]

c)

36 – 36p = 14p + 21 M1 A115 = 50p0.3 = p A1

[3]

d) P(passes at least one) = 1 – P(fails both) M1= 1 - (0.7 x 0.7)= 0.51 A1 f.t.

P(passes P1passes at least 1) = M1

=

= (=0.588) A1[6]

Page 14

p

1 - p8

0.5

0.5

0.3

0.7

pass P1

fail P1

pass M1

pass M1

fail M1

fail M1

A LEVEL MATHEMATICS ANSWERS AND MARKSCHEMES

PROBABILITY

18.a) 0.3 0.1 = 0.03 M1 A1[2]

b) 0.3 0.1 + 0.5 0.7 + 0.1 0.2 M1 A1= 0.03 + 0.35 + 0.02= 0.4 A1

[3]

c) P(for Tonifor someone who’s at home) = M1

=

= A1 cao[4]

d) 3! 0.1 0.7 0.2 B1 (3!) M1= 0.084 A1

[3]

19.a) P(R ) is the probability R happens and Q does not B1P( ) is the probability neither R nor Q happen B1

[2]

b) P(R ) + P( ) = P( ) M1P( ) = 1 – q B1P( ) = 1 – q – x A1

[3]

c) P(R) = P(R Q) + P (R ) M1= 0.1 + x A1

Q and R independent P(R Q) = P(R)P(Q) M10.1 = (0.1 + x)q

q = A1

[4]

d) q = 0.25 B1P( Q) = P(Q) – P(R Q) M1

= 0.25 – 0.1 = 0.15 A1[3]

Page 15

A LEVEL MATHEMATICS ANSWERS AND MARKSCHEMES

PROBABILITY

20.a) 0.4 0.75 + 0.5 0.8 + 0.1 0.7 M1 A1= 0.77 A1

[3]

b) P(Bnot prompt) = M1

=

= (= 0.0435) A1[5]

c) 1 – P(never used) M1= 1 – 0.94 A1= 0.3439 A1

[3]

d) P(used 3 timesused at least once) = M1

P(3 times) = 4 0.13 0.9 M1 A1

So P(3 timesat least once) =

= (= 0.0105) A1

[4]

21.a) 0.3 x 0.76 M1 A1= 0.03529 A1

[3]

b) 7 x 0.36 x 0.7 M1 B1 (7)= 0.003572 A1

[3]

c) Probability of it raining on one day is independent of whether it rains on others. B1[1]

Page 16

A LEVEL MATHEMATICS ANSWERS AND MARKSCHEMES

PROBABILITY

22.

a) P(2nd is red) = M1 A1 A1

M1

4 = R A1[5]

b) P(1st red2nd red) = M1

= A1

= A1[3]

Page 17

red

green8

red

green

red

green

10R

53

10R10

52

53

52

A LEVEL MATHEMATICS ANSWERS AND MARKSCHEMES

PROBABILITY

23.a) P( ) = 0.3 P(B) = 0.3 P(BA) = 0.2 B1 (conditional)

M1 (V. dgm or other)

b + c = 0.3

= 0.2 M1 A1

a + b + c = 0.7

and a = 0.4 B1So, using

= 0.2 M1

b = 0.2b + 0.080.08b = 0.08b = 0.1 A1

[7]

b) c = 0.2 B1P(just one) = 0.2 + 0.4 = 0.6 M1 A1

[3]

c) P( B) = M1

= A1

= A1[3]

d) Cases are:

A B B A

A B A B

M1 A1 (cases)

So probability = 2 x 0.1 x 0.3 + 2 x 0.2 x 0.4 M1 A1= 0.22 A1

[5]

Page 18

b

A B

a c

0.3