AGC # 1. Markscheme
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Transcript of AGC # 1. Markscheme
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7/28/2019 AGC # 1. Markscheme
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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
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7/28/2019 AGC # 1. Markscheme
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(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]
IB Questionbank Mathematical Studies 3rd edition 2
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7/28/2019 AGC # 1. Markscheme
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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]
IB Questionbank Mathematical Studies 3rd edition 3
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7/28/2019 AGC # 1. Markscheme
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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
IB Questionbank Mathematical Studies 3rd edition 4
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7/28/2019 AGC # 1. Markscheme
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(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]
IB Questionbank Mathematical Studies 3rd edition 5
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7/28/2019 AGC # 1. Markscheme
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6. (a)
(A1)
(A1)
(A1)
(A1)
Note: Award (A1) for correct tree structure, (A1) for each
complementary pair.
IB Questionbank Mathematical Studies 3rd edition 6
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7/28/2019 AGC # 1. Markscheme
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(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)
IB Questionbank Mathematical Studies 3rd edition 7
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7/28/2019 AGC # 1. Markscheme
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(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]
IB Questionbank Mathematical Studies 3rd edition 8