IB DP PhysicsIB DP Maths IB DP ChemistryIB DP Biology

Question

When Geraldine travels to work she can travel either by car (C), bus (B) or train (T). She travels by car on one day in five. She uses the bus 50 % of the time. The probabilities of her being late (L) when travelling by car, bus or train are 0.05, 0.12 and 0.08 respectively.

It is not necessary to use graph paper for this question.

Copy the tree diagram below and fill in all the probabilities, where NL represents not late, to represent this information.

[5]

i.a.

Find the probability that Geraldine travels by bus and is late.[1]

i.b.

Find the probability that Geraldine is late.[3]

i.c.

Find the probability that Geraldine travelled by train, given that she is late.[3]

i.d.

Sketch the curve of the function \(f (x) = x^3 − 2x^2 + x − 3\) for values of \(x\) from −2 to 4, giving the intercepts with both axes.[3]

ii.a.

On the same diagram, sketch the line \(y = 7 − 2x\) and find the coordinates of the point of intersection of the line with the curve.[3]

ii.b.

Find the value of the gradient of the curve where \(x = 1.7\) .[2]

ii.c.

Answer/Explanation

Markscheme

Award (A1) for 0.5 at B, (A1) for 0.3 at T, then (A1) for each correct pair. Accept fractions or percentages. (A5)[5 marks]

i.a.

0.06 (accept \(0.5 \times 0.12\) or 6%) (A1)(ft)[1 mark]

i.b.

for a relevant two-factor product, either \(C \times L\) or \(T \times L\) (M1)

for summing three two-factor products (M1)

\((0.2 \times 0.05 + 0.06 + 0.3 \times 0.08)\)

0.094 (A1)(ft)(G2)[3 marks]

i.c.

\(\frac{{0.3 \times 0.08}}{{0.094}}\) (M1)(A1)(ft)

award (M1) for substituted conditional probability formula seen, (A1)(ft) for correct substitution

= 0.255 (A1)(ft)(G2)[3 marks]

i.d.

(G3)[3 marks]

ii.a.

line drawn with –ve gradient and +ve y-intercept (G1)

(2.45, 2.11) (G1)(G1)[3 marks]

ii.b.

\(f ‘ (1.7) = 3(1.7)^2 – 4(1.7) + 1\) (M1)

award (M1) for substituting in their \(f’ (x)\)

2.87 (A1)(G2)[2 marks]

ii.c.

Question

A group of 50 students completed a questionnaire for a Mathematical Studies project. The following data was collected.

\(18\) students own a digital camera (D)
\(15\) students own an iPod (I)
\(26\) students own a cell phone (C)
\(1\) student owns all three items
\(5\) students own a digital camera and an iPod but not a cell phone
\(2\) students own a cell phone and an iPod but not a digital camera
\(3\) students own a cell phone and a digital camera but not an iPod

Claire and Kate both wish to go to the cinema but one of them has to stay at home to baby-sit.

The probability that Kate goes to the cinema is \(0.2\). If Kate does not go Claire goes.

If Kate goes to the cinema the probability that she is late home is \(0.3\).

If Claire goes to the cinema the probability that she is late home is \(0.6\).

Represent this information on a Venn diagram.[4]

i.a.

Calculate the number of students who own none of the items mentioned above.[2]

i.b.

If a student is chosen at random, write down the probability that the student owns a digital camera only.[1]

i.c.

If two students are chosen at random, calculate the probability that they both own a cell phone only.[3]

i.d.

If a student owns an iPod, write down the probability that the student also owns a digital camera.[2]

i.e.

Copy and complete the probability tree diagram below.

[3]

ii.a.

Calculate the probability that

(i) Kate goes to the cinema and is not late;

(ii) the person who goes to the cinema arrives home late.[5]

ii.b.

Answer/Explanation

Markscheme

(A1)(A1)(A1)(A1)(ft)

Note: (A1) for rectangle with 3 intersecting circles, (A1) for \(1\), (A1) for \(5\), \(3\), \(2\), (A1)(ft) for \(9\), \(7\), \(20\) if subtraction is carried out, or \(18\), \(15\), \(26\) seen by the letters D, I and C.[4 marks]

i.a.

\(50 – 47\) (M1)

Note: (M1) for subtracting their value from \(50\).

\( = 3\) (A1)(ft)(G2)[2 marks]

i.b.

\(\frac{9}{{50}}\) (A1)(ft)[1 mark]

i.c.

\(\frac{{20}}{{50}} \times \frac{{19}}{{49}}\) (A1)(ft)(M1)

\( = \frac{{38}}{{245}}{\text{ }}\left( {\frac{{380}}{{2450}}{\text{, }}0.155{\text{, }}15.5\% } \right)\) (A1)(ft)(G2)

Notes: (A1)(ft) for correct fractions from their Venn diagram
(M1) for multiplying their fractions
(A1)(ft) for correct answer.[3 marks]

i.d.

\(\frac{6}{{15}}{\text{ }}\left( {\frac{2}{5}{\text{, }}0.4{\text{, }}40\% } \right)\) (A1)(ft)(A1)(ft)

Note: (A1)(ft) for numerator, (A1)(ft) denominator.[2 marks]

i.e.

(A1)(A1)(A1)

Note: (A1) for \(0.8\), (A1) for \(0.7\), (A1) for \(0.6\) and \(0.4\).[3 marks]

ii.a.

(i) \(0.2 \times 0.7 = 0.14\) (M1)(A1)(ft)(G2)

Note: (M1) for multiplying correct numbers.[2 marks]

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

\( = 0.54\) (A1)(ft)(G3)

Note: (M1) for each correct product (use candidate’s tree), (A1)(ft) for answer.[3 marks]

ii.b.

Question

Sharon and Lisa share a flat. Sharon cooks dinner three nights out of ten. If Sharon does not cook dinner, then Lisa does. If Sharon cooks dinner the probability that they have pasta is 0.75. If Lisa cooks dinner the probability that they have pasta is 0.12.

A survey was carried out in a year 12 class. The pupils were asked which pop groups they like out of the Rockers (R), the Salseros (S), and the Bluers (B). The results are shown in the following diagram.

Copy and complete the tree diagram to represent this information.

[3]

i, a.

Find the probability that Lisa cooks dinner and they do not have pasta.[2]

i, b.

Find the probability that they do not have pasta.[3]

i, c.

Given that they do not have pasta, find the probability that Lisa cooked dinner.[3]

i, d.

Write down \(n(R \cap S \cap B)\).[1]

ii, a.

Find \(n(R’)\).[2]

ii, b.

Describe which groups the pupils in the set \(S \cap B\) like.[2]

ii, c.

Use set notation to describe the group of pupils who like the Rockers and the Bluers but do not like the Salseros.[2]

ii, d.

There are 33 pupils in the class.

Find x.[2]

ii, e, i.

There are 33 pupils in the class.

Find the number of pupils who like the Rockers.[1]

ii, e, ii.

Answer/Explanation

Markscheme

Note: Award (A1) for each correct pair. (A3)[3 marks]

i, a.

\(0.7 \times 0.88 = 0.616{\text{ }}\left( {\frac{{77}}{{125}},{\text{ }}61.6{\text{ }}\% } \right)\) (M1)(A1)(ft)(G2)

Note: Award (M1) for multiplying the correct probabilities.[2 marks]

i, b.

\(0.3 \times 0.25 + 0.7 \times 0.88\) (M1)(M1)

Notes: Award (M1) for a relevant two-factor product, could be \(S \times NP\) OR \(L \times NP\).

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

\({\text{P}} = 0.691{\text{ }}\left( {\frac{{691}}{{1000}},{\text{ }}69.1{\text{ }}\% } \right)\) (A1)(ft)(G2)

Notes: (ft) from their answer to (b).[3 marks]

i, c.

\(\frac{{0.616}}{{0.691}}\) (M1)(A1)

Note: Award (M1) for substituted conditional probability formula, (A1) for correct substitution.

\({\text{P}} = 0.891{\text{ }}\left( {\frac{{616}}{{691}},{\text{ }}89.1{\text{ }}\% } \right)\) (A1)(ft)(G2)[3 marks]

i, d.

3 (A1)[1 mark]

ii, a.

For 5, 4, 7 (0) seen with no extra values (A1)

16 (A1)(G2)[2 marks]

ii, b.

They like (both) the Salseros (S) and they like the Bluers (B) (A1)(A1)

Note: Award (A1) for “and”, (A1) for the correct groups.[2 marks]

ii, c.

\(R \cap B \cap S’\) (A1)(A1)

Note: Award (A1) for \(R \cap B\), (A1) for \( \cap S’\)[2 marks]

ii, d.

\(21+ 3x = 33\) (M1)

\(x = 4\) (A1)(G2)[2 marks]

ii, e, i.

17 (A1)(ft)[1 mark]

ii, e, ii.

Question

Phoebe chooses a biscuit from a blue tin on a shelf. The tin contains one chocolate biscuit and four plain biscuits. She eats the biscuit and chooses another one from the tin. The tree diagram below represents the situation with the four possible outcomes where A stands for chocolate biscuit and B for plain biscuit.

On another shelf there are two tins, one red and one green. The red tin contains three chocolate biscuits and seven plain biscuits and the green tin contains one chocolate biscuit and four plain biscuits. Andrew randomly chooses either the red or the green tin and randomly selects a biscuit.

Write down the value of a.[1]

a.i.

Write down the value of b.[2]

a.ii.

Find the probability that both biscuits are plain.[3]

a.iii.

Copy and complete the tree diagram below.

[3]

Find the probability that he chooses a chocolate biscuit.[3]

c.i.

Find the probability that he chooses a biscuit from the red tin given that it is a chocolate biscuit.[3]

c.ii.

Answer/Explanation

Markscheme

a = 0 \(\left( {\frac{0}{4}} \right)\) (A1)[1 mark]

a.i.

\(b = \frac{3}{4}(0.75,{\text{ }}75\% )\) (A2)(G2)[2 marks]

a.ii.

\(\frac{4}{5} \times \frac{3}{4}\) (M1)(A1)

\(\frac{{12}}{{20}}\left( {\frac{3}{5},{\text{ }}0.6,{\text{ 60% }}} \right)\) (A1)(ft)(G2)

Note: Award (M1) for multiplying two probabilities, (A1) for using their probabilities, (A1) for answer.[3 marks]

a.iii.

(A1)(A1)(A1)

Note: Award (A1) for each pair.[3 marks]

\(\frac{1}{2} \times \frac{3}{{10}} + \frac{1}{2} \times \frac{1}{5}\) (M1)(M1)

\( = \frac{5}{{20}}\left( {\frac{1}{4},{\text{ }}0.25,{\text{ 25% }}} \right)\) (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.[3 marks]

c.i.

\(\frac{{\frac{1}{2} \times \frac{3}{{10}}}}{{\frac{1}{4}}}\) (M1)(A1)

\( = \frac{3}{5}{\text{ }}\left( {0.6,{\text{ 60 }}\% } \right)\) (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).[3 marks]

c.ii.

Question

Part A

100 students are asked what they had for breakfast on a particular morning. There were three choices: cereal (X) , bread (Y) and fruit (Z). It is found that

10 students had all three

17 students had bread and fruit only

15 students had cereal and fruit only

12 students had cereal and bread only

13 students had only bread

8 students had only cereal

9 students had only fruit

Part B

The same 100 students are also asked how many meals on average they have per day. The data collected is organized in the following table.

A \({\chi ^2}\) test is carried out at the 5 % level of significance.

Represent this information on a Venn diagram.[4]

A.a.

Find the number of students who had none of the three choices for breakfast.[2]

A.b.

Write down the percentage of students who had fruit for breakfast.[2]

A.c.

Describe in words what the students in the set \(X \cap Y’\) had for breakfast.[2]

A.d.

Find the probability that a student had at least two of the three choices for breakfast.[2]

A.e.

Two students are chosen at random. Find the probability that both students had all three choices for breakfast.[3]

A.f.

Write down the null hypothesis, H₀, for this test.[1]

B.a.

Write down the number of degrees of freedom for this test.[1]

B.b.

Write down the critical value for this test.[1]

B.c.

Show that the expected number of females that have more than 5 meals per day is 13, correct to the nearest integer.[2]

B.d.

Use your graphic display calculator to find the \(\chi _{calc}^2\) for this data.[2]

B.e.

Decide whether H₀ must be accepted. Justify your answer.[2]

B.f.

Answer/Explanation

Markscheme

(A1) for rectangle and three intersecting circles

(A1) for 10, (A1) for 8, 13 and 9, (A1) for 12, 15 and 17 (A4)[4 marks]

A.a.

100 – (9 +12 +13 +15 +10 +17 + 8) =16 (M1)(A1)(ft)(G2)

Note: Follow through from their diagram.[2 marks]

A.b.

\(\frac{{51}}{{100}}(0.51)\) (A1)(ft)

= 51% (A1)(ft)(G2)

Note: Follow through from their diagram.[2 marks]

A.c.

Note: The following statements are correct. Please note that the connectives are important. It is not the same (had cereal) and (not bread) and (had cereal) or (not bread). The parentheses are not needed but are there to facilitate the understanding of the propositions.

(had cereal) and (did not have bread)

(had cereal only) or (had cereal and fruit only)

(had either cereal or (fruit and cereal)) and (did not have bread) (A1)(A1)

Notes: If the statements are correct but the connectives are wrong then award at most (A1)(A0). For the statement (had only cereal) and (cereal and fruit) award (A1)(A0). For the statement had cereal and fruit award (A0)(A0).[2 marks]

A.d.

\(\frac{{54}}{{100}}(0.54,{\text{ 54 % }})\) (A1)(ft)(A1)(ft)(G2)

Note: Award (A1)(ft) for numerator, follow through from their diagram, (A1)(ft) for denominator. Follow through from total or denominator used in part (c).[2 marks]

A.e.

\(\frac{{10}}{{100}} \times \frac{9}{{99}} = \frac{1}{{110}}(0.00909,{\text{ 0}}{\text{.909 % }})\) (A1)(ft)(M1)(A1)(ft)(G2)

Notes: Award (A1)(ft) for their correct fractions, (M1) for multiplying two fractions, (A1)(ft) for their correct answer. Answer 0.009 with no working receives no marks. Follow through from denominator in parts (c) and (e) and from their diagram.[3 marks]

A.f.

H₀ : The (average) number of meals per day a student has and gender are independent (A1)

Note: For “independent” accept “not associated” but do not accept “not related” or “not correlated”.[1 mark]

B.a.

2 (A1)[1 mark]

B.b.

5.99 (accept 5.991) (A1)(ft)

Note: Follow through from their part (b).[1 mark]

B.c.

\(\frac{{28 \times 45}}{{100}} = 12.6 = 13\) or \(\frac{{28}}{{100}} \times \frac{{25}}{{100}} \times 100 = 12.6 = 13\) (M1)(A1)(AG)

Notes: Award (M1) for correct formula and (A1) for correct substitution. Unrounded answer must be seen for the (A1) to be awarded.[2 marks]

B.d.

0.0321 (G2)

Note: For 0.032 award (G1)(G1)(AP). For 0.03 with no working award (G0).[2 marks]

B.e.

0.0321 < 5.99 or 0.984 > 0.05 (R1)

accept H₀ (A1)(ft)

Note: If reason is incorrect both marks are lost, do not award (R0)(A1).[2 marks]

B.f.

Question

One day the numbers of customers at three cafés, “Alan’s Diner” ( \(A\) ), “Sarah’s Snackbar” ( \(S\) ) and “Pete’s Eats” ( \(P\) ), were recorded and are given below.

     17 were customers of Pete’s Eats only
     27 were customers of Sarah’s Snackbar only
     15 were customers of Alan’s Diner only
     10 were customers of Pete’s Eats and Sarah’s Snackbar but not Alan’s Diner
     8 were customers of Pete’s Eats and Alan’s Diner but not Sarah’s Snackbar

Some of the customers in each café were given survey forms to complete to find out if they were satisfied with the standard of service they received.

Draw a Venn Diagram, using sets labelled \(A\) , \(S\) and \(P\) , that shows this information.[3]

A.a.

There were 48 customers of Pete’s Eats that day. Calculate the number of people who were customers of all three cafés.[2]

A.b.

There were 50 customers of Sarah’s Snackbar that day. Calculate the total number of people who were customers of Alan’s Diner.[3]

A.c.

Write down the number of customers of Alan’s Diner that were also customers of Pete’s Eats.[1]

A.d.

Find \(n[(S \cup P) \cap A’]\).[2]

A.e.

One of the survey forms was chosen at random, find the probability that the form showed “Dissatisfied”;[2]

B.a.

One of the survey forms was chosen at random, find the probability that the form showed “Satisfied” and was completed at Sarah’s Snackbar;[2]

B.b.

One of the survey forms was chosen at random, find the probability that the form showed “Dissatisfied”, given that it was completed at Alan’s Diner.[2]

B.c.

A \({\chi ^2}\) test at the \(5\% \) significance level was carried out to determine whether there was any difference in the level of customer satisfaction in each of the cafés.

Write down the null hypothesis, \({{\text{H}}_0}\) , for the \({\chi ^2}\) test.[1]

B.d.

A \({\chi ^2}\) test at the \(5\% \) significance level was carried out to determine whether there was any difference in the level of customer satisfaction in each of the cafés.

Write down the number of degrees of freedom for the test.[1]

B.e.

A \({\chi ^2}\) test at the \(5\% \) significance level was carried out to determine whether there was any difference in the level of customer satisfaction in each of the cafés.

Using your graphic display calculator, find \({\chi ^2}_{calc}\) .[2]

B.f.

A \({\chi ^2}\) test at the \(5\% \) significance level was carried out to determine whether there was any difference in the level of customer satisfaction in each of the cafés.

State, giving a reason, the conclusion to the test.[2]

B.g.

Answer/Explanation

Markscheme

(A1) for rectangle and three labelled intersecting circles
(A1) for \(15\), \(27\) and \(17\)
(A1) for \(10\) and \(8\) (A3)[3 marks]

A.a.

\(48 – (8 +10 +17)\) or equivalent (M1)

\( = 13\) (A1)(ft)(G2)[2 marks]

A.b.

\(50 – (27 +10 +13)\) (M1)

Note: Award (M1) for working seen.

\( = 0\) (A1)
number of elements in A \(= 36\) (A1)(ft)(G3)

Note: Follow through from (b).[3 marks]

A.c.

\(21\) (A1)(ft)

Note: Follow through from (b) even if no working seen.[1 mark]

A.d.

\(54\) (M1)(A1)(ft)(G2)

Note: Award (M1) for \(17\), \(10\), \(27\) seen. Follow through from (a).[2 marks]

A.e.

\(\frac{{40}}{{120}}{\text{ }}\left( {\frac{1}{3}{\text{, }}0.333{\text{, }}33.3\% } \right)\) (A1)(A1)(G2)

Note: Award (A1) for numerator, (A1) for denominator.[2 marks]

B.a.

\(\frac{{34}}{{120}}{\text{ }}\left( {\frac{{17}}{{60}}{\text{, }}0.283{\text{, }}28.3\% } \right)\) (A1)(A1)(G2)

Note: Award (A1) for numerator, (A1) for denominator.[2 marks]

B.b.

\(\frac{8}{{28}}{\text{ }}\left( {\frac{2}{7}{\text{, }}0.286{\text{, }}28.6\% } \right)\) (A1)(A1)(G2)

Note: Award (A1) for numerator, (A1) for denominator.[2 marks]

B.c.

customer satisfaction is independent of café (A1)

Note: Accept “customer satisfaction is not associated with the café”.[1 mark]

B.d.

\(2\) (A1)[1 mark]

B.e.

\(0.754\) (G2)

Note: Award (G1)(G1)(AP) for \(0.75\) or for correct answer incorrectly rounded to 3 s.f. or more, (G0) for \(0.7\).[2 marks]

B.f.

since \({\chi ^2}_{calc} < {\chi ^2}_{crit}5.991 accept (or Do not reject) H₀ (R1)(A1)(ft)

Note: Follow through from their value in (e).

Accept (or Do not reject) H₀ as \(p\)-value \((0.686) > 0.05\) (R1)(A1)(ft)

Notes: Do not award (A1)(R0). Award the (R1) for comparison of appropriate values.[2 marks]

B.g.

Question

The seniors from Gulf High School are required to participate in exactly one after-school sport. Data were gathered from a sample of 120 students regarding their choice of sport. The following data were recorded.

A \({\chi ^2}\) test was carried out at the 5 % significance level to analyse the relationship between gender and choice of after-school sport.

Write down the null hypothesis, H₀, for this test.[1]

Find the expected value of female footballers.[2]

Write down the number of degrees of freedom.[1]

Write down the critical value of \(\chi ^2\), at the 5 % level of significance.[1]

Use your graphic display calculator to determine the \(\chi _{calc}^2\) value.[2]

Determine whether H₀ should be accepted. Justify your answer.[2]

One student is chosen at random from the 120 students.

Find the probability that this student

(i) is male;

(ii) plays tennis.[2]

Two students are chosen at random from the 120 students.

Find the probability that

(i) both play football;

(ii) neither play basketball.[5]

Answer/Explanation

Markscheme

H₀ : Gender and choice of afterschool sport are independent. (A1)

Note: Accept “not associated”, do not accept “not related”, “not correlated”, or “not linked”. Accept “the relation between gender and sport is independent”.[1 mark]

\(\frac{{85}}{{120}} \times \frac{{48}}{{120}} \times 120\left( {\frac{{85 \times 48}}{{120}}} \right)\) (M1)

Note: Award (M1) for correct expression.

= 34 (A1)(G2)[2 marks]

2 (A1)[1 mark]

5.99 (5.991) (A1)(ft)

Note: Follow through from part (c).[1 mark]

2.42 (2.42094…) (G2)[2 marks]

Since 2.42 < 5.99 therefore accept (do not reject) H₀ (R1)(A1)(ft)

Note: The numerical values need not be seen, but must be consistent with their parts (d) and (e).

p-value 0.298 > 0.05 therefore accept (do not reject) H₀ (R1)(A1)

Note: p-value comparison may not be used as part of a follow through solution. Do not award (A1)(R0). Follow through from parts (c), (d) and (e).[2 marks]

(i) \(\frac{{35}}{{120}}\left( {\frac{7}{{24}},{\text{ }}0.292,{\text{ }}29.2\% } \right)\) (0.291666…) (A1)

(ii) \(\frac{{25}}{{120}}\left( {\frac{5}{{24}},{\text{ }}0.208,{\text{ }}20.8\% } \right)\) (0.208333…) (A1) [2 marks]

(i) \(\frac{{48}}{{120}} \times \frac{{47}}{{119}}\) (A1)(M1)

Note: Award (A1) for two correct fractions, (M1) for multiplying their two fractions.

\( = \frac{{94}}{{595}}(0.158,{\text{ }}15.8\% )\) (0.157983…) (A1)(G2)

(ii) \(\frac{{73}}{{120}} \times \frac{{72}}{{119}}\) (M1)

Note: Award (M1) for multiplying correct fractions. If sampling with replacement has been used in both parts (h)(i) and (h)(ii) do not penalise in part (h)(ii). Award a maximum of (M1)(A1)(ft).

\( = \frac{{219}}{{595}}(0.368,{\text{ }}36.8\% )\) (0.368067…) (A1)(G2)[5 marks]

Question

Leanne goes fishing at her favourite pond. The pond contains four different types of fish: bream, flathead, whiting and salmon. The fish are either undersized or normal. This information is shown in the table below.

Write down the total number of fish in the pond.[1]

Leanne catches a fish.

Find the probability that she

(i) catches an undersized bream;

(ii) catches either a flathead or an undersized fish or both;

(iii) does not catch an undersized whiting;

(iv) catches a whiting given that the fish was normal.[7]

Leanne notices that on windy days, the probability she catches a fish is 0.1 while on non-windy days the probability she catches a fish is 0.65. The probability that it will be windy on a particular day is 0.3.

Copy and complete the probability tree diagram below.

[3]

Calculate the probability that it is windy and Leanne catches a fish on a particular day.[2]

Calculate the probability that Leanne catches a fish on a particular day.[3]

Use your answer to part (e) to calculate the probability that Leanne catches a fish on two consecutive days.[2]

Given that Leanne catches a fish on a particular day, calculate the probability that the day was windy.[3]

Answer/Explanation

Markscheme

90 (A1)[1 mark]

(i) \(\frac{3}{{90}}(0.0\bar 3,{\text{ }}0.0333,{\text{ }}0.0333…,{\text{ }}3.\bar 3\% ,{\text{ }}3.33\% )\) (A1)(ft)

Note: For the denominator follow through from their answer in part (a).

(ii) \(\frac{{53}}{{90}}(0.5\bar 8,{\text{ }}0.588…,{\text{ }}0.589,{\text{ }}58.\bar 8\% ,{\text{ }}58.9\% )\) (A1)(A1)(ft)(G2)

Notes: Award (A1) for the numerator. (A1)(ft) for denominator. For the denominator follow through from their answer in part (a).

(iii) \(\frac{{72}}{{90}}{\text{(0.8, 80}}\%)\) (A1)(ft)(A1)(ft)(G2)

Notes: Award (A1)(ft) for the numerator, (their part (a) –18) (A1)(ft) for denominator. For the denominator follow through from their answer in part (a).

(iv) \(\frac{{24}}{{48}}(0.5,{\text{ 50}}\% )\) (A1)(A1)(G2)

Note: Award (A1) for numerator, (A1) for denominator.[7 marks]

(A1)(A1)(A1)

Notes: Award (A1) for each correct entry. Tree diagram must be seen for marks to be awarded.[3 marks]

\(0.3 \times 0.1 = 0.03\left( {\frac{3}{{100}}} \right)\) (M1)(A1)(G2)

Note: Award (M1) for correct product seen.[2 marks]

\(0.3 \times 0.1+ 0.7\times0.65\) (M1)(M1)

Notes: Award (M1) for \(0.7\times0.65\) (or 0.455) seen, (M1) for adding their 0.03. Follow through from their answers to parts (c) and (d).

\( = 0.485\left( {\frac{{485}}{{1000}},\frac{{97}}{{200}}} \right)\) (A1)(ft)(G2)

Note: Follow through from their tree diagram and their answer to part (d).[3 marks]

\(0.485 \times 0.485\) (M1)

\(0.235\left( {\frac{{9409}}{{40000}}{\text{, }}0.235225} \right)\) (A1)(ft)(G2)

Note: Follow through from their answer to part (e).[2 marks]

\(\frac{{0.03}}{{0.485}}\) (M1)(A1)(ft)

Notes: Award (M1) for substituted conditional probability formula, (A1)(ft) for their (d) as numerator and their (e) as denominator.

\(0.0619\left( {\frac{{6}}{{97}}}\text{, 0.0618556…} \right) \) (A1)(ft)(G2)

Note: Follow through from their parts (d) and (e).[3 marks]

Question

100 students at IB College were asked whether they study Music (M), Chemistry (C), or Economics (E) with the following results.

10 study all three

15 study Music and Chemistry

17 study Music and Economics

12 study Chemistry and Economics

11 study Music only

6 study Chemistry only

Draw a Venn diagram to represent the information above.[4]

Write down the number of students who study Music but not Economics.[1]

There are 22 Economics students in total.

(i) Calculate the number of students who study Economics only.

(ii) Find the number of students who study none of these three subjects.[4]

A student is chosen at random from the 100 that were asked above.

Find the probability that this student

(i) studies Economics;

(ii) studies Music and Chemistry but not Economics;

(iii) does not study either Music or Economics;

(iv) does not study Music given that the student does not study Economics.[7]

Answer/Explanation

Markscheme

(A1) for rectangle and three labelled circles (U need not be seen)

(A1) for 10 in the correct region

(A1) for 2, 7 and 5 in the correct regions

(A1) for 6 and 11 in the correct regions (A4)

16 (A1)(ft)

Note: Follow through from their Venn diagram.

(i) \(10 + 7 + 2\) (M1)

Note: Award (M1) for summing their 10, 7 and 2.

\(22 – 19\)

\(= 3\) (A1)(ft)(G2)

Note: Follow through from their diagram. Award (M1)(A1)(ft) for answers consistent with their diagram irrespective of whether working seen. Award a maximum of (M1)(A0) for a negative answer.

(ii) \(22 + 11+ 5 + 6\) (M1)

Note: Award (M1) for summing 22, and their 11, 5 and 6.

\(100 – 44\)

\(= 56\) (A1)(ft)(G2)

Note: Follow through from their diagram. Award (M1)(A1)(ft) for answers consistent with their diagram and the use of 22 irrespective of whether working seen. If negative values are used or implied award (M0)(A0).

(i) \(\frac{{22}}{{100}}\left( {\frac{{11}}{{50}},0.22,22\% } \right)\) (A1)(G1)

(ii) \(\frac{{5}}{{100}}\left( {\frac{{1}}{{20}},0.05,5\% } \right)\) (A1)(ft)(A1)(G2)

Note: Award (A1)(ft) for their 5 in numerator, (A1) for denominator.

Follow through from their diagram.

(iii) \(\frac{{62}}{{100}}\left( {\frac{{31}}{{50}},0.62,62\% } \right)\) (A1)(ft)(A1)(G2)

Note: Award (A1)(ft) for \(100 – (22 + 11 + {\text{their }}5)\), (A1) for denominator.

Follow through from their diagram.

(iv) \(\frac{{62}}{{78}}\left( {\frac{{31}}{{39}},0.795,79.5\% } \right)\) (0.794871…) (A1)(ft)(A1)(G2)

Note: Award (A1)(ft) for numerator, (A1) for denominator. Follow

through from part (d)(iii) for numerator.

Question

Forty families were surveyed about the places they went to on the weekend. The places were the circus (C), the museum (M) and the park (P).

16 families went to the circus

22 families went to the museum

14 families went to the park

4 families went to all three places

7 families went to both the circus and the museum, but not the park

3 families went to both the circus and the park, but not the museum

1 family went to the park only

Draw a Venn diagram to represent the given information using sets labelled C, M and P. Complete the diagram to include the number of families represented in each region.[4]

Find the number of families who

(i) went to the circus only;

(ii) went to the museum and the park but not the circus;

(iii) did not go to any of the three places on the weekend.[4]

A family is chosen at random from the group of 40 families. Find the probability that the family went to

(i) the circus;

(ii) two or more places;

(iii) the park or the circus, but not the museum;

(iv) the museum, given that they also went to the circus.[8]

Two families are chosen at random from the group of 40 families.

Find the probability that both families went to the circus.[3]

Answer/Explanation

Markscheme

(A1)(A1)(A1)(A1)

Award (A1) for 3 intersecting circles and rectangle, (A1) for 1, 3, 4 and 7, (A1) for 2, (A1) for 6 and 5.

(i) 2 (A1)(ft)

(ii) 6 (A1)(ft)

(iii) 40 − (1 + 6 + 2 + 3 + 4 + 7 + 5) (M1)

Note: Award (M1) for subtracting all their values from 40.

= 12 (A1)(ft)(G2)

Note: Follow through from their Venn diagram for parts (i), (ii) and (iii).

(i) \(\frac{{16}}{{40}}\left( {\frac{2}{5},0.4,40\% } \right)\) (A1)(A1)(G2)

Note: Award (A1) for numerator, (A1) for denominator. Answer must be less than 1 otherwise award (A0)(A0). Award (A0)(A0) if answer is given as incorrect reduced fraction without working.

(ii) \(\frac{{20}}{{40}}\left( {\frac{1}{2},0.5,50\% } \right)\) (A1)(ft) (A1) (G2)

Note: Award (A1)(ft) for numerator, (A1) for denominator. Follow through from their Venn diagram. Answer must be less than 1 otherwise award (A0)(A0). Award (A0)(A0) if answer is given as incorrect reduced fraction without working.

(iii) \(\frac{6}{{40}}\left( {\frac{3}{{20}},0.15,15\% } \right)\) (A1)(ft)(A1)(G2)

(iv) \(\frac{{11}}{{16}}\left( {0.6875,68.75\% } \right)\) (A1)(ft)(A1)(G2)

\(\frac{{16}}{{40}} \times \frac{{15}}{{39}}\) (A1)(A1)(ft)

Note: Award (A1) for multiplication of their probabilities, (A1)(ft) for their correct probabilities.

\(\frac{{240}}{{1560}}\left( {\frac{2}{{13}},0.153846…,15.4\% } \right)\) (A1)(ft)(G2)

Note: Follow through from their answer to part (c)(i). Answer must be less than 1 otherwise award at most (A1)(A1)(A0)(ft).

Question

A group of tourists went on safari to a game reserve. The game warden wanted to know how many of the tourists saw Leopard (\(L\)), Cheetah (\(C\)) or Rhino (\(R\)). The results are given as follows.

5 of the tourists saw all three

7 saw Leopard and Rhino

1 saw Cheetah and Leopard but not Rhino

4 saw Leopard only

3 saw Cheetah only

9 saw Rhino only

Draw a Venn diagram to show this information.[4]

There were 25 tourists in the group and every tourist saw at least one of the three types of animal.

Find the number of tourists that saw Cheetah and Rhino but not Leopard.[2]

There were 25 tourists in the group and every tourist saw at least one of the three types of animal.

Calculate the probability that a tourist chosen at random from the group

(i) saw Leopard;

(ii) saw only one of the three types of animal;

(iii) saw only Leopard, given that he saw only one of the three types of animal.[6]

There were 25 tourists in the group and every tourist saw at least one of the three types of animal.

If a tourist chosen at random from the group saw Leopard, find the probability that he also saw Cheetah.[2]

Answer/Explanation

Markscheme

(A1)(A1)(A1)(A1)

Note: Award (A1) for rectangle and three labelled intersecting circles (the rectangle need not be labelled), (A1) for 5, (A1) for 2 and 1, (A1) for 4, 3 and 9.[4 marks]

\(25 – (5 + 2 + 1 + 4 + 3 + 9)\) (M1)

Notes: Award (M1) for their \(5 + 2 + 1 + 4 + 3 + 9\) seen even if total is greater than \(25\).

Do not award (A1)(ft) if their total is greater than \(25\).

\( = 1\) (A1)(ft)(G2)[2 marks]

(i) \(\frac{{12}}{{25}}{\text{ }}(0.48,{\text{ }}48\% )\) (A1)(ft)(A1)(G2)

Notes: Award (A1)(ft) for numerator, (A1) for denominator.

Follow through from Venn diagram.

(ii) \(\frac{{16}}{{25}}{\text{ }}(0.64,{\text{ }}64\% )\) (A1)(A1)(G2)

Notes: Award (A1) for numerator, (A1) for denominator.

There is no follow through; all information is given.

(iii) \(\frac{4}{{16}}{\text{ }}(0.25,{\text{ }}25\% )\)) (A1)(A1)(ft)(G2)

Notes: Award (A1) for numerator, (A1)(ft) for denominator.

Follow through from part (c)(ii) only.[6 marks]

\(\frac{6}{{12}}{\text{ }}(0.5,{\text{ }}50\% )\) (A1)(A1)(ft)(G2)

Notes: Award (A1) for numerator, (A1)(ft) for denominator.

Follow through from Venn diagram.[2 marks]

Question

Tomek is attending a conference in Singapore. He has both trousers and shorts to wear. He also has the choice of wearing a tie or not.

The probability Tomek wears trousers is \(0.3\). If he wears trousers, the probability that he wears a tie is \(0.8\).

If Tomek wears shorts, the probability that he wears a tie is \(0.15\).

The following tree diagram shows the probabilities for Tomek’s clothing options at the conference.

Find the value of

(i) \({\text{A}}\);

(ii) \({\text{B}}\);

(iii) \({\text{C}}\).[3]

Calculate the probability that Tomek wears

(i) shorts and no tie;

(ii) no tie;

(iii) shorts given that he is not wearing a tie.[8]

The conference lasts for two days.

Calculate the probability that Tomek wears trousers on both days.[2]

The conference lasts for two days.

Calculate the probability that Tomek wears trousers on one of the days, and shorts on the other day.[3]

Answer/Explanation

Markscheme

(i) \(0.7 \left( {\frac{{70}}{{100}},{\text{ }}\frac{7}{{10}},{\text{ 70% }}} \right)\) (A1)

(ii) \(0.2 \left( {\frac{{20}}{{100}},{\text{ }}\frac{2}{{10}},{\text{ }}\frac{1}{5},{\text{ 20% }}} \right)\) (A1)

(iii) \(0.85 \left( {\frac{{85}}{{100}},{\text{ }}\frac{{17}}{{20}},{\text{ 85% }}} \right)\) (A1)[3 marks]

(i) \(0.7 \times 0.85\) (M1)

Note: Award (M1) for multiplying their values from parts (a)(i) and (a)(iii).

\( = 0.595{\text{ }}\left( {\frac{{119}}{{200}},{\text{ 59.5% }}} \right)\) (A1)(ft)(G1)

Note: Follow through from part (a).

(ii) \(0.3 \times 0.2 + 0.7 \times 0.85\) (M1)(M1)

Note: Award (M1) for their two products, (M1) for adding their two products.

\( = 0.655{\text{ }}\left( {\frac{{131}}{{200}},{\text{ 65.5% }}} \right)\) (A1)(ft)(G2)

Note: Follow through from part (a).

(iii) \(\frac{{0.595}}{{0.655}}\) (A1)(ft)(A1)(ft)

Notes: Award (A1)(ft) for correct numerator, (A1)(ft) for correct denominator. Follow through from parts (b)(i) and (ii).

\( = 0.908{\text{ }}\left( {{\text{0.90839}} \ldots ,{\text{ }}\frac{{119}}{{131}},{\text{ 90,8% }}} \right)\) (A1)(ft)(G2)[8 marks]

\(0.3 \times 0.3\) (M1)

\( = 0.09 \left( {\frac{9}{{100}}, 9\%} \right)\) (A1)(G2)[2 marks]

\(0.3 \times 0.7\) (M1)

\(0.3 \times 0.7 \times 2\) OR \((0.3 \times 0.7) + (0.7 \times 0.3)\) (M1)

Note: Award (M1) for their correct product seen, (M1) for multiplying their product by 2 or for adding their products twice.

\( = 0.42 \left( {\frac{{42}}{{100}}, \frac{{21}}{{50}}, 42\%} \right)\) (A1)(ft)(G2)

Note: Follow through from part (a)(i).[3 marks]

Question

Mike, the laboratory mouse, is placed at the starting point, S, of a maze. Some paths in the maze lead to Trap A, some to Trap B, and others to escape doors. Some paths have one and some have two sections. If his path forks, Mike randomly chooses a path forward.

The following tree diagram represents the maze, showing all possible paths, and the probability that Mike chooses a certain section of a path through the maze.

Write down the value of

(i) \(p\) ;

(ii) \(q\) ;

(iii) \(r\).[3]

(i) Find the probability that Mike reaches Trap B.

(ii) Find the probability that Mike reaches Trap A.

(iii) Find the probability that Mike escapes from the maze.[7]

Sonya, a lab assistant, counts the number of paths that lead to traps or escape doors. She believes that the probability that Mike will be trapped is greater than the probability that he will escape.

State whether Sonya is correct. Give a mathematical justification for your conclusion.[2]

During the first trial Mike escapes.

Given that Mike escaped, find the probability that he went directly from S to Escape Door 3.[3]

Answer/Explanation

Markscheme

(i) \(\frac{1}{3}\;\;\;(0.333333…,{\text{ }}33.3333…\% )\) (A1)

(ii) \(\frac{1}{2}\;\;\;(0.5,{\text{ }}50\% )\) (A1)

(iii) \(\frac{1}{4}\;\;\;(0.25,{\text{ }}25\% )\) (A1)

(i) \(\frac{1}{3} \times \frac{1}{4}\) (M1)

\( = \frac{1}{{12}}\;\;\;({\text{0.0833333…, 8.33333…% )}}\) (A1)(G2)

(ii) \(\frac{1}{3} \times \frac{1}{2} + \frac{1}{3} \times \frac{1}{4} + \frac{1}{3} \times \frac{1}{4}\) (A1)(ft)(M1)

Note: Award (A1)(ft) for their three correct products seen, (M1) for addition of their products.

\( = \frac{1}{3}\;\;\;(0.333333…,{\text{ }}33.3333…\% )\) (A1)(ft)(G2)

Note: Follow through from their parts (a)(i) and (a)(iii).

(iii) \(1 – \frac{1}{{12}} – \frac{1}{3}\) (M1)

Note: Follow through from parts (b)(i) and (b)(ii).

\(\frac{1}{3} \times \frac{1}{2} + \frac{1}{3} \times \frac{1}{4} + \frac{1}{3}\) (M1)

Note: Follow through from parts (a)(i) and (a)(ii).

\( = \frac{7}{{12}}\;\;\;(0.583333…,{\text{ }}58.3333…\% )\) (A1)(ft)(G2)

Sonya is not correct. (A1)(ft)

The probability that Mike escapes is \(\frac{7}{{12}}\), which is

greater than \(\frac{5}{{12}}{\text{ }}\left( {{\text{or greater than }}\frac{1}{2}} \right)\). (R1)(ft)

Notes: Do not award (A1)(R0).

Follow through from their answers to part (b).

\(\frac{{\frac{1}{3}}}{{\frac{7}{{12}}}}\) (A1)(A1)(ft)

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

\( = \frac{4}{7}\;\;\;\left( {\frac{{12}}{{21}},{\text{ 0.571428…, 57.1428…% }}} \right)\) (A1)(ft)(G2)

Note: Follow through from their answer to part (b)(iii).

Question

A group of 100 customers in a restaurant are asked which fruits they like from a choice of mangoes, bananas and kiwi fruits. The results are as follows.

15 like all three fruits

22 like mangoes and bananas

33 like mangoes and kiwi fruits

27 like bananas and kiwi fruits

8 like none of these three fruits

\(x\) like only mangoes

Copy the following Venn diagram and correctly insert all values from the above information.

[3]