AGC # 1. Markscheme

7/28/2019 AGC # 1. Markscheme

1/8

1. (a)

(A1)(A1)(A1) (C3)

Note: Award (A1) for 0.25, (A1) for 0.1 and 0.9, (A1) for 0.05

and 0.95.

(b) P(late) = 0.25 0.1 + 0.75 0.05 (A1)(ft)(M1)

Note: Award (A1)(ft) for two correct products from their

diagram and award (M1) for addition of their two products.

= 0.0625

%25.6,16

1

(A1)(ft) (C3)[6]

2. (a) (i) a= 0

4

0

(A1)

(ii) b= 4

3

(0.75, 75%) (A2)(G2)

(iii) 4

3

5

4

(M1)(A1)

%60,6.0,5

3

20

12

(A1)(ft)(G2)

Note: Award (M1) for multiplying two probabilities, (A1) for

using their probabilities, (A1) for answer.

IB Questionbank Mathematical Studies 3rd edition 1


2/8

(b)

(A1)(A1)(A1)

Note: Award (A1) for each pair.

(c) (i) 5

1

2

1

10

3

2

1+

(M1)(M1)

=

%25,25.0,4

1

20

5

(A1)(ft)(G2)

Note: Award (M1) for two products seen with numbers from the

problem, (M1) for adding two products. Follow through from

their tree diagram.

(ii) 4

1

10

3

2

1

(M1)(A1)

= 5

3

(0.6, 60%) (A1)(ft)(G2)

Note: Award (M1) for substituted conditional probability

formula, (A1) for correct substitution. Follow through from

their part (b) and part (c) (i).[15]



3/8

3. (a)

(A3)

Note: Award (A1) for each correct pair.

(b) 0.7 0.88 = 0.616

%6.61,125

77

(M1)(A1)(ft)(G2)

Note: Award (M1) for multiplying the correct probabilities.

(c) 0.3 0.25 + 0.7 0.88 (M1)(M1)

Notes: Award (M1) for a relevant two-factor product, could beS NPORL NP.

Award (M1) for summing 2 two-factor products.

P = 0.691

%1.69,1000

691

(A1)(ft)(G2)

Notes: (ft) from their answer to (b).

(d) 691.0

616.0

(M1)(A1)

Note: Award (M1) for substituted conditional probability

formula, (A1) for correct substitution.

P = 0.891

%1.89,691

616

(A1)(ft)(G2)[11]



4/8

4. (a)

(A1)(A1)(A1) (C3)

Note: Award (A1) for each correct pair.

(b) 0.7 0.1

= 0.07 (100

7

, 7%) (A1)(ft) (C1)

(c) 0.3 0.8 + 0.07 (M1)

= 0.31 (100

31

, 31%) (A1)(ft) (C2)

Note: In (b) and (c) follow through from sensible answers only

i.e. not a probability greater than one.

[6]

5. (a) (i) P (a dog is grey and has the yellow bowl)

)111.0(9

1

3

1

3

1===

(M1)(A1)(G2)

Note: The (M1) is for multiplying two values along any branch

of the tree.

(ii) P (dog does not get yellow bowl) =)6.0orsf)3(667.0(

3

2

=(A1) 3



5/8

(b) (i) The tree diagram should show the values 2

1,

2

1

for the brown branch(A1)

and 4

3,

4

1

in the correct positions for the grey branch. (A1)(ft)

Note: Follow through if the branches are interchanged.

(ii) P (the dog is grey or is playing with a stick, but not both)

2

1

3

2

4

3

3

1+=

(M1)

)583.0(12

7==

(A1)(ft)(G1)

Notes: The (M1) is for showing two correct products (whether

added or not).

Follow through from b(i).

Award (M1) for 4

1

3

1+

(must be a sum).

(iii) P (dog is grey given that it is playing with stick)

12

5

12

1or

4

1

3

1

2

1

3

2

4

1

3

1

)(

)(

+

=

SP

SGp

(M1)(A1)(ft)

Note: (M1) for substituted conditional probability formula,

(A1) for correct substitutions.

)2.0(5

1==

(A1)(ft)(G2)

(iv) P (grey and fed from yellow bowl and not playing with stick)

= 12

1

4

3

3

1

3

1 =

sf).30833.036

3( ==

(M1)

(A1)(ft)(G1) 9

Note: (M1) is for product of 3 reasonable probability values.[12]



6/8

6. (a)

(A1)

(A1)

(A1)

(A1)

Note: Award (A1) for correct tree structure, (A1) for each

complementary pair.



7/8

(b) (i) 10

1

2

1

(M1)

%)5,05.0(20

1

(A1)(ft)(G2)

Note:Award (M1) for correct product.

(ii) 5

1

2

1

10

1

2

1 +(M1)(M1)

Note: Award (M1) for finding two products, (M1) for adding

two products.

)15%0.15,(20

3=

(A1)(ft)(G2)

(c) 20

3

5

1

2

1

(M1)

Note: Award (M1) for using the conditional probability

formula.

)667.0(,3

2=

(A1)(A1)(ft)

or (G3)

Note: Award (A1) for correct numerator, (A1) for correct

denominator.

[12]

7. (a)

1

0 . 6

0 . 40 . 8

0 . 1

0 . 2

0 . 6 5

0 . 3 5

0 . 9

(A1)(A1) (C2)



8/8

(b) 0.65 0.1 (= 0.065) (A1)

0.35 0.8 0.4 (= 0.112) (A1)

0.35 0.2 1 the 1 can be implied(= 0.07) (A1)(ft)

0.247 (A1)(ft) (C4)

Note: No (ft) for any probabilities greater than 1.[6]

8. (a)

(A1)(A1)(A1)

Note: (A1) for 0.8, (A1) for 0.7, (A1) for 0.6 and 0.4.

(b) (i) 0.2 0.7 = 0.14 (M1)(A1)(ft)

Note: (M1) for multiplying correct numbers. (G2)

(ii) 0.2 0.3 + 0.8 0.6 (M1)(M1)

= 0.54 (A1)(ft)(G2)

Note: (M1) for each correct product (use candidates

tree), (A1)(ft) for answer.[8]


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AGC # 1. Markscheme