practice: probability 2 - mr kenneth...

practice: probability_2 [281 marks]

1a. [4 marks]

In a college 450 students were surveyed with the following results

150 have a television

205 have a computer

220 have an iPhone

75 have an iPhone and a computer

60 have a television and a computer

70 have a television and an iPhone

40 have all three.

Draw a Venn diagram to show this information. Use T to represent the set of students who have a television, C the set ofstudents who have a computer and I the set of students who have an iPhone.

Markscheme

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

Notes: Award (A1) for labelled sets T, C, and I included inside an enclosed universal set. (Label U is not essential.) Award (A1) forcentral entry 40. (A1) for 20, 30 and 35 in the other intersecting regions. (A1) for 60, 110 and 115 or T(150), C(205), I(220).

[4 marks]

1b. [2 marks]Write down the number of students that

(i) have a computer only;

(ii) have an iPhone and a computer but no television.

MarkschemeIn parts (b), (c) and (d) follow through from their diagram.

(i) 110 (A1)(ft)

(ii) 35 (A1)(ft)

[2 marks]

[1 mark]1c. Write down .n[T ∩ (C ∪ I ])′


60 (A1)(ft)

[2 marks]

[2 marks]1d. Calculate the number of students who have none of the three.


450 − (60 + 20 + 40 + 30 + 115 + 35 + 110) (M1)

Note: Award (M1) for subtracting all their values from 450.

= 40 (A1)(ft)(G2)

[2 marks]

1e. [6 marks]Two students are chosen at random from the 450 students. Calculate the probability that

(i) neither student has an iPhone;

(ii) only one of the students has an iPhone.

Markscheme(i) (A1)(M1)

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

(A1)(G2)

Note: Follow through from their Venn diagram in part (a).

(ii) (A1)(A1)

Note: Award (A1) for addition of their products, (A1) for two correct products.

OR

(A1)(A1)

Notes: Award (A1) for their product of two fractions multiplied by 2, (A1) for correct product of two fractions multiplied by 2.Award (A0)(A0) if correct product is seen not multiplied by 2.

(A1)(G2)

Note: Follow through from their Venn diagram in part (a) and/or their 230 used in part (e)(i).

Note: For consistent use of replacement in parts (i) and (ii) award at most (A0)(M1)(A0) in part (i) and (A1)(ft)(A1)(A1)(ft) in part(ii).

[6 marks]

×230450

229449

( , 0.261, 26.1% )(0.26067...)52670202050

526720205

× + ×220450

230449

230450

220449

× × 2230450

220449

(0.501, 50.1% )(0.50086. . .)20244041

1f. [3 marks]The students are asked to collect money for charity. In the first month, the students collect x dollars and the students collect ydollars in each subsequent month. In the first 6 months, they collect 7650 dollars. This can be represented by the equation x + 5y = 7650.

In the first 10 months they collect 13 050 dollars.

(i) Write down a second equation in x and y to represent this information.

(ii) Write down the value of x and of y .

Markscheme(i) x + 9y = 13050 (A1)

(ii) x = 900 (A1)(ft)

y = 1350 (A1)(ft)

Notes: Follow through from their equation in (f)(i). Do not award (A1)(ft) if answer is negative. Award (M1)(A0) for an attempt atsolving simultaneous equations algebraically but incorrect answer obtained.

[3 marks]

1g. [3 marks]The students are asked to collect money for charity. In the first month, the students collect x dollars and the students collect ydollars in each subsequent month. In the first 6 months, they collect 7650 dollars. This can be represented by the equation x + 5y = 7650.

In the first 10 months they collect 13 050 dollars.

Calculate the number of months that it will take the students to collect 49 500 dollars.

Markscheme49500 = 900 + 1350n (A1)(ft)

Notes: Award (A1)(ft) for setting up correct equation. Follow through from candidate’s part (f).

n = 36 (A1)(ft)

The total number of months is 37. (A1)(ft)(G2)

Note: Award (G1) for 36 seen as final answer with no working. The value of n must be a positive integer for the last two (A1)(ft) tobe awarded.

OR

49500 = 900 + 1350(n − 1) (A2)(ft)

Notes: Award (A2)(ft) for setting up correct equation. Follow through from candidate’s part (f).

n = 37 (A1)(ft)(G2)

Note: The value of n must be a positive integer for the last (A1)(ft) to be awarded.

[3 marks]

2a. [6 marks]

A store recorded their sales of televisions during the 2010 football World Cup. They looked at the numbers of televisions bought bygender and the size of the television screens.

This information is shown in the table below; S represents the size of the television screen in inches.

The store wants to use this information to predict the probability of selling these sizes of televisions for the 2014 football World Cup.

Use the table to find the probability that

(i) a television will be bought by a female;

(ii) a television with a screen size of 32 < S ≤ 46 will be bought;

(iii) a television with a screen size of 32 < S ≤ 46 will be bought by a female;

(iv) a television with a screen size greater than 46 inches will be bought, given that it is bought by a male.

Markscheme(i) (A1)(G1)

(ii) (A1)(G1)

(iii) (A1)(A1)(G2)

(iv) (A1)(A1)(G2)

Note: Award (A1) for numerator, (A1) for denominator. Award (A0)(A0) if answers are given as incorrect reduced fractions withoutworking.

[6 marks]

( , 0.44, 44%)220500

1125

( , 0.36, 36%)180500

925

( , 0.08, 8%)40500

225

( , 0.196, 19.6%)55500

1156

2b. [1 mark]The manager of the store wants to determine whether the screen size is independent of gender. A Chi-squared test is performedat the 1 % significance level.

Write down the null hypothesis.

Markscheme“The size of the television screen is independent of gender.” (A1)

Note: Accept “not associated”, do not accept “not correlated”.

[1 mark]

2c. [2 marks]The manager of the store wants to determine whether the screen size is independent of gender. A Chi-squared test is performedat the 1 % significance level.

Show that the expected frequency for females who bought a screen size of 32 < S ≤ 46, is 79, correct to the nearest integer.

Markscheme OR (M1)

= 79.2 (A1)

= 79 (AG)

Note: Both the unrounded and the given answer must be seen for the final (A1) to be awarded.

[2 marks]

× × 500180500

220500

180×220500

2d. [1 mark]The manager of the store wants to determine whether the screen size is independent of gender. A Chi-squared test is performedat the 1 % significance level.

Write down the number of degrees of freedom.

Markscheme3 (A1)

[1 mark]

2e. [2 marks]The manager of the store wants to determine whether the screen size is independent of gender. A Chi-squared test is performedat the 1 % significance level.

Write down the calculated value.

Markscheme = 104(103.957...) (G2)

Note: Award (M1) if an attempt at using the formula is seen but incorrect answer obtained.

[2 marks]

χ2

χ2calc

2f. [1 mark]The manager of the store wants to determine whether the screen size is independent of gender. A Chi-squared test is performedat the 1 % significance level.

Write down the critical value for this test.

Markscheme11.345 (A1)(ft)

Notes: Follow through from their degrees of freedom.

[1 mark]

2g. [2 marks]The manager of the store wants to determine whether the screen size is independent of gender. A Chi-squared test is performedat the 1 % significance level.

Determine if the null hypothesis should be accepted. Give a reason for your answer.

Markscheme > OR p < 0.01 (R1)

Do not accept H . (A1)(ft)

Note: Do not award (R0)(A1)(ft). Follow through from their parts (d), (e) and (f).

[2 marks]

χ2calc

χ2crit

0

3a. [1 mark]

Merryn plans to travel to a concert tomorrow. Due to bad weather, there is a 60 % chance that all flights will be cancelled tomorrow.If the flights are cancelled Merryn will travel by car.

If she travels by plane the probability that she will be late for the concert is 10 %.

If she travels by car, the probability that she will not be late for the concert is 25 %.

Complete the tree diagram below.

Markscheme

(A1) (C1)

Note: Award (A1) for 0.9 and 0.75.

[1 mark]

[3 marks]3b. Find the probability that Merryn will not be late for the concert.

Markscheme0.4 × 0.9 + 0.6 × 0.25 (M1)(M1)

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

(A1)(ft) (C3)

Note: Follow through from their answers to part (a).

[3 marks]

0.51( , 51%)51100

3c. [2 marks]Merryn was not late for the concert the next day.

Given that, find the probability that she travelled to the concert by car.

Markscheme (M1)

Note: Award (M1) for correctly substituted conditional probability formula.

(A1)(ft) (C2)

Note: Follow through from their tree diagram and their part (b).

[2 marks]

0.6×0.250.51

0.294( , 0.294117...)517

[1 mark]4a.

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

Write down the total number of fish in the pond.

Markscheme90 (A1)

[1 mark]

4b. [7 marks]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.

Markscheme(i) (A1)(ft)

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

(ii) (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) (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 theiranswer in part (a).

(iv) (A1)(A1)(G2)

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

[7 marks]

(0.0 , 0.0333, 0.0333..., 3. %, 3.33%)390

3̄ 3̄

(0.5 , 0.588..., 0.589, 58. %, 58.9%)5390

8̄ 8̄

(0.8, 80%)7290

(0.5, 50%)2448

4c. [3 marks]Leanne notices that on windy days, the probability she catches a fish is 0.1 while on non-windy days the probability shecatches 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.

Markscheme

(A1)(A1)(A1)

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

[3 marks]

4d. [2 marks]Leanne notices that on windy days, the probability she catches a fish is 0.1 while on non-windy days the probability shecatches a fish is 0.65. The probability that it will be windy on a particular day is 0.3.

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

Markscheme (M1)(A1)(G2)

Note: Award (M1) for correct product seen.

[2 marks]

0.3 × 0.1 = 0.03( )3100

4e. [3 marks]Leanne notices that on windy days, the probability she catches a fish is 0.1 while on non-windy days the probability shecatches a fish is 0.65. The probability that it will be windy on a particular day is 0.3.

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

Markscheme (M1)(M1)

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

(A1)(ft)(G2)

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

[3 marks]

0.3 × 0.1 + 0.7 × 0.65

0.7 × 0.65

= 0.485( , )4851000

97200

[2 marks]4f. Use your answer to part (e) to calculate the probability that Leanne catches a fish on two consecutive days.

Markscheme (M1)

(A1)(ft)(G2)

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

[2 marks]

0.485 × 0.485

0.235( , 0.235225)940940000

4g. [3 marks]Leanne notices that on windy days, the probability she catches a fish is 0.1 while on non-windy days the probability shecatches a fish is 0.65. The probability that it will be windy on a particular day is 0.3.

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

Markscheme (M1)(A1)(ft)

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

(A1)(ft)(G2)

Note: Follow through from their parts (d) and (e).

[3 marks]

0.030.485

0.0619( , 0.0618556...)697

3

5a. [3 marks]

The probability that Tanay eats lunch in the school cafeteria is .

If he eats lunch in the school cafeteria, the probability that he has a sandwich is .

If he does not eat lunch in the school cafeteria the probability that he has a sandwich is .


Markscheme

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

Note: Award (A1) for each correct pair of branches.

35

310

910

[3 marks]5b. Find the probability that Tanay has a sandwich for his lunch.

Markscheme (A1)(ft)(M1)

Notes: Award (A1)(ft) for their two correct products, (M1) for addition of their products. Follow through from their tree diagram.

(A1)(ft) (C3)

× + ×35

310

25

910

= (0.54,54%)2750

[4 marks]6a.

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.

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)

[1 mark]6b. Write down the number of students who study Music but not Economics.

Markscheme16 (A1)(ft)

Note: Follow through from their Venn diagram.

6c. [4 marks]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.

Markscheme(i) (M1)

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

(A1)(ft)(G2)

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

(ii) (M1)

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

(A1)(ft)(G2)

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

10 + 7 + 2

22 − 19

= 3

22 + 11 + 5 + 6

100 − 44

= 56

6d. [7 marks]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.

Markscheme(i) (A1)(G1)

(ii) (A1)(ft)(A1)(G2)

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

Follow through from their diagram.

(iii) (A1)(ft)(A1)(G2)

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

Follow through from their diagram.

(iv) (0.794871...) (A1)(ft)(A1)(G2)

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

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

( ,0.22,22%)22100

1150

( ,0.05,5%)5100

120

( ,0.62,62%)62100

3150

100 − (22 + 11 + their 5)

( ,0.795,79.5%)6278

3139

7a. [4 marks]

Forty families were surveyed about the places they went to on the weekend. The places were the circus (C), the museum (M) and thepark (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 thenumber of families represented in each region.

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.

7b. [4 marks]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.

Markscheme(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).

7c. [8 marks]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.

Markscheme(i) (A1)(A1)(G2)

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

(ii) (A1)(ft) (A1) (G2)

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

(iii) (A1)(ft)(A1)(G2)


(iv) (A1)(ft)(A1)(G2)


( ,0.4,40%)1640

25

( ,0.5,50%)2040

12

( ,0.15,15%)640

320

(0.6875,68.75%)1116

7d. [3 marks]Two families are chosen at random from the group of 40 families.

Find the probability that both families went to the circus.

Markscheme (A1)(A1)(ft)

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

(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).

×1640

1539

( ,0.153846...,15.4%)2401560

213

[1 mark]8a.

Alan’s laundry basket contains two green, three red and seven black socks. He selects one sock from the laundry basket at random.

Write down the probability that the sock is red.

Markscheme (A1) (C1)( ,0.25,25%)3

1214

8b. [2 marks]Alan returns the sock to the laundry basket and selects two socks at random.

Find the probability that the first sock he selects is green and the second sock is black.

Markscheme (M1)

Note: Award (M1) for correct product.

(A1) (C2)

( ) × ( )212

711

= ( ,0.10606...,10.6%)14132

766

8c. [3 marks]Alan returns the socks to the laundry basket and again selects two socks at random.

Find the probability that he selects two socks of the same colour.

Markscheme (M1)(M1)

Note: Award (M1) for addition of their 3 products, (M1) for 3 correct products.

(A1) (C3)

( × ) + ( × ) + ( × )212

111

312

211

712

611

= ( ,0.37878...,37.9%)50132

2566

9a. [3 marks]

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 . If he wears trousers, the probability that he wears a tie is .

If Tomek wears shorts, the probability that he wears a tie is .

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

Find the value of

(i) ;

(ii) ;

(iii) .

Markscheme(i) (A1)

(ii) (A1)

(iii) (A1)

[3 marks]

0.3 0.8

0.15

A

B

C

0.7( , , 70% )70100

710

0.2( , , , 20% )20100

210

15

0.85( , , 85% )85100

1720

9b. [8 marks]Calculate the probability that Tomek wears

(i) shorts and no tie;

(ii) no tie;

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

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

(A1)(ft)(G1)

Note: Follow through from part (a).

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

(A1)(ft)(G2)

Note: Follow through from part (a). (iii) (A1)(ft)(A1)(ft)

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

(A1)(ft)(G2)

[8 marks]

0.7 × 0.85

= 0.595 ( , 59.5% )119200

0.3 × 0.2 + 0.7 × 0.85

= 0.655 ( , 65.5% )131200

0.5950.655

= 0.908 (0.90839… , , 90,8% )119131

9c. [2 marks]The conference lasts for two days.

Calculate the probability that Tomek wears trousers on both days.

Markscheme (M1)

(A1)(G2)

[2 marks]

0.3 × 0.3

= 0.09( ,9%)9100

9d. [3 marks]The conference lasts for two days.

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

Markscheme (M1)

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

(A1)(ft)(G2)

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

0.3 × 0.7

0.3 × 0.7 × 2 (0.3 × 0.7) + (0.7 × 0.3)

= 0.42( , ,42%)42100

2150

10a. [3 marks]

Ramzi travels to work each day, either by bus or by train. The probability that he travels by bus is . If he travels by bus, the

probability that he buys a magazine is . If he travels by train, the probability that he buys a magazine is .

Complete the tree diagram.

Markscheme

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

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

35

23

34

[3 marks]10b. Find the probability that Ramzi buys a magazine when he travels to work.


Notes: Award (A1)(ft) for two consistent products from tree diagram, (M1) for addition of their products.

Follow through from their tree diagram provided all probabilities are between and .

(A1)(ft) (C3)

[3 marks]

× + ×35

23

25

34

0 1

(0.7, 70% , )710

4260

[5 marks]11a.

A group of women in the USA were asked whether they had visited the continents of Europe ( ) or South America ( ) or Asia (

). had visited all three continents

had visited Europe only

had visited South America only

had visited Asia only

had visited Europe and South America but had not visited Asia

had visited South America and Asia but had not visited Europe

had visited Europe and Asia but had not visited South America

had not visited any of these continents

Draw a Venn diagram, using sets labelled , and , to show this information.

Markscheme

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

Notes: Award (A1) for rectangle and three labelled intersecting circles.

Award (A1) for in correct place.

Award (A1) for , and in the correct places.

Award (A1) for , and in the correct places.

Award (A1) for in the correct place.

Accept and instead of and .

Do not penalize if is omitted from the diagram.

[5 marks]

120 E S

A

7

28

22

16

15

x

2x

20

E S A

7

28 22 16

15 x 2x

20

4 8 x 2x

U

[2 marks]11b. Calculate the value of .

Markscheme (M1)

Note: Award (M1) for setting up a correct equation involving , the and values from their diagram.

(A1)(ft)(G2)

Note: Follow through from part (a). For the follow through to be awarded must be a positive integer.

[2 marks]

x

3x = 120 − (20 + 28 + 15 + 22 + 7 + 16)

x 120

x = 4

x

[2 marks]11c. Explain, in words, the meaning of .(E ∪ S) ∩ A′

Markscheme(Women who had visited) Europe or South America and (but had) not (visited) Asia (A1)(A1) Notes: Award (A1) for “(visited) Europe or South America” (or both).

Award (A1) for “and (but) had not visited Asia”.

(urope) union (outh America) intersected with not (sia) earns no marks, (A0). [2 marks]

E S A

[1 mark]11d. Write down .

Markscheme (A1)

Note: Award (A0) for the embedded answer of .

[1 mark]

n ((E ∪ S ∪ A ))′

20

n(20)

[2 marks]11e. Find the probability that a woman selected at random from the group had visited Europe.

Markscheme (A1)(ft)(A1)(G2)

Note: Award (A1)(ft) for numerator, follow through from their value of , or their diagram, (A1) for denominator.

[2 marks]

( , 0.483, 48.3% ) (0.48333 … )58120

2960

x

[2 marks]11f. Find the probability that a woman selected at random from the group had visited Europe, given that she had visited Asia.

Markscheme (A1)(ft)(A1)(ft)(G2)

Note: Award (A1)(ft) for numerator, (A1)(ft) for denominator, follow through from their value of or their diagram.

[2 marks]

( , 0.429, 42.9% ) (0.428571 … )1535

37

x

11g. [3 marks]Two women from the group are selected at random.

Find the probability that both women selected had visited South America.


Notes: Award (A1)(ft) for two correct fractions, follow through from their denominator in part (e), follow through the numerator from

their answer to part (b) or from their diagram, (M1) for multiplication of their two fractions.

(A1)(ft)(G2)

Notes: Award (A1)(M1)(A1) for correct fractions, correctly multiplied together with an answer of .

Award (A0)(M1)(A0) for .

Award (G1) for an answer of with no working seen.

[3 marks]

×48120

47119

= ( , 0.158, 15,8% ) (0.157983 … )225614 280

94595

0.16

× = 0.1648120

48120

0.16

12a. [3 marks]

The probability that it snows today is 0.2. If it does snow today, the probability that it will snow tomorrow is 0.6. If it does not snow

today, the probability that it will not snow tomorrow is 0.9.

Using the information given, complete the following tree diagram.

Markscheme

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

Note: Award (A1) for each correct pair of probabilities.

[3 marks]

[3 marks]12b. Calculate the probability that it will snow tomorrow.


Note: Award (A1)(ft) for two correct products of probabilities taken from their diagram, (M1) for the addition of their products.

(A1)(ft) (C3)

Note: Accept any equivalent correct fraction.

Follow through from their tree diagram.

[3 marks]

0.2 × 0.6 + 0.8 × 0.1

= 0.2 ( , 20% )15

[4 marks]13a.

A group of tourists went on safari to a game reserve. The game warden wanted to know how many of the tourists saw Leopard ( ),

Cheetah ( ) or Rhino ( ). 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.

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]

L

C R

13b. [2 marks]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.

Markscheme (M1)

Notes: Award (M1) for their seen even if total is greater than .

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

(A1)(ft)(G2)

[2 marks]

25 − (5 + 2 + 1 + 4 + 3 + 9)

5 + 2 + 1 + 4 + 3 + 9 25

25

= 1

13c. [6 marks]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.

Markscheme(i) (A1)(ft)(A1)(G2)

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

Follow through from Venn diagram.

(ii) (A1)(A1)(G2)

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

There is no follow through; all information is given.

(iii) ) (A1)(A1)(ft)(G2)

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

Follow through from part (c)(ii) only.

[6 marks]

(0.48, 48%)1225

(0.64, 64%)1625

(0.25, 25%)416

13d. [2 marks]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.

Markscheme (A1)(A1)(ft)(G2)

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

Follow through from Venn diagram.

[2 marks]

(0.5, 50%)612

14a. [3 marks]

Maria travels to school either by walking or by bicycle. The probability she cycles to school is 0.75.

If she walks, the probability that she is late for school is 0.1.If she cycles, the probability that she is late for school is 0.05.

Complete the tree diagram below, showing the appropriate probabilities.

Markscheme

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

Note: Award (A1) for 0.25, (A1) for 0.1 and 0.9, (A1) for 0.05 and 0.95

[3 marks]

[3 marks]14b. Find the probability that Maria is late for school.


Note: Award (A1)(ft) for two correct products from their diagram and award (M1) for addition of their two products.

(A1)(ft) (C3)

[3 marks]

P(late) = 0.25 × 0.1 + 0.75 × 0.05

= 0.0625( , 6.25%)116

[2 marks]15a.

In a research project on the relation between the gender of 150 science students at college and their degree subject, the following setof data is collected.

Find the probability that a student chosen at random is male.

Markscheme (A1)(A1) (C2)


[2 marks]

= (0.607, 60.6%, 60.7%)91150

[2 marks]15b. Find the probability that a student chosen at random is either male or studies Chemistry.

Markscheme (A1)(ft)(A1) (C2)

Note: Award (A1)(ft) for their numerator in (a) +20 provided the final answer is not greater than 1. (A1) for denominator.

[2 marks]

= ( , 0.74, 74 %)111150

3750

[2 marks]15c. Find the probability that a student chosen at random studies Physics, given that the student is male.

Markscheme (A1)(A1)(ft) (C2)

Note: Award (A1) for numerator and (A1)(ft) for denominator. Follow through from their numerator in (a) provided answer is notgreater than 1.

[2 marks]

(0.176, 17.6%)1691

16a. [2 marks]

A group of 30 students were asked about their favourite topping for toast.

18 liked peanut butter (A)

10 liked jam (B)

6 liked neither

Show this information on the Venn diagram below.

Markscheme

OR (A2) (C2)

Note: Award (A2) for 3 correctly placed values, and no extras (4 need not be seen), (A1) for 2 correctly placed values, (A0) for 1 orno correctly placed values.

[2 marks]

[2 marks]16b. Find the number of students who like both peanut butter and jam.

Markscheme18 + 10 + 6 = 30 (M1)= 4 (A1) (C2)

[2 marks]

[2 marks]16c. Find the probability that a randomly chosen student from the group likes peanut butter, given that they like jam.

Markscheme (A1)(ft)(A1) (C2)

Note: Award (A1)(ft) for their numerator from part (b), (A1) for denominator.

[2 marks]

P(A|B) = ( , 0.4, 40 %)410

25

[2 marks]17a.

Let , and .

Find .

Markscheme (M1)

(A1) (C2)

Note: Award (M1) for correct substitution, (A1) for correct answer.

[2 marks]

P(A) = 0.5 P(B) = 0.6 P(A ∪ B) = 0.8

P(A ∩ B)

0.8 = 0.5 + 0.6 − P(A ∩ B)P(A ∩ B) = 0.3

[2 marks]17b. Find .

Markscheme (M1)

= 0.5 (A1)(ft) (C2)

Note: Award (M1) for correct substitution in conditional probability formula. Follow through from their answer to part (a), providedprobability is not greater than one.

[2 marks]

P(A|B)

P(A|B) = 0.30.6

[2 marks]17c. Decide whether A and B are independent events. Give a reason for your answer.

Markscheme or 0.3 = 0.5 × 0.6 (R1)

OR

(R1)

they are independent. (Yes) (A1)(ft) (C2)

Note: Follow through from their answers to parts (a) or (b).

Do not award (R0)(A1).

[2 marks]

P(A ∩ B) = P(A) × P(B)

P(A|B) = P(A)

18a. [3 marks]

A bag contains 7 red discs and 4 blue discs. Ju Shen chooses a disc at random from the bag and removes it. Ramón then chooses adisc from those left in the bag.

Write down the probability that

(i) Ju Shen chooses a red disc from the bag;

(ii) Ramón chooses a blue disc from the bag, given that Ju Shen has chosen a red disc;

(iii) Ju Shen chooses a red disc and Ramón chooses a blue disc from the bag.

Markscheme(i) ( , ) ( ) (A1) (C1)

(ii) (A1) (C1)

(iii) (A1)(ft) (C1)

Note: Follow through from the product of their answers to parts (a) (i) and (ii).

[3 marks]

711

0.636 63.6% 0.636363…

410

( , 0.4, 40%)25

28110

( , 0.255, 25.5%)1455

0.254545…

[3 marks]18b. Find the probability that Ju Shen and Ramón choose different coloured discs from the bag.

Markscheme OR (M1)(M1)

Notes: Award (M1) for using their as part of a combined probability expression. (M1) for either adding or formultiplying by 2.

( ) (A1)(ft) (C3)

Note: Follow through applies from their answer to part (a) (iii) and only when their answer is between 0 and 1.

[3 marks]

+ ( × )28110

411

710

2 × 28110

28110

×411

710

= 56110

( , 0.509, 50.9%)2855

0.509090…

[2 marks]19a.

A survey was carried out in a group of 200 people. They were asked whether they smoke or not. The collected information wasorganized in the following table.

One person from this group is chosen at random.

Write down the probability that this person is a smoker.



[2 marks]

(0.45, 45 %)90200

[2 marks]19b. Write down the probability that this person is male given that they are a smoker.

Markscheme (A1)(A1)(ft) (C2)

Notes: Award (A1) for numerator, (A1)(ft) for denominator, follow through from their numerator in part (a). Last mark is lost ifanswer is not a probability.

[2 marks]

(0. , 0.667, 66. %, 66.6… %, 66.7 %)6090

6̄ 6̄

[2 marks]19c. Find the probability that this person is a smoker or is male.

Markscheme (M1)

Note: Award (M1) for correct substitution in the combined events formula. Follow through from their answer to part (a).

(A1)(ft)

OR

(M1)

Note: Award (M1) for adding the correct fractions.

(A1)

OR

(M1)

Note: Award (M1) for subtraction of correct fraction from 1.

(A1) (C2)

[2 marks]

+ −90200

100200

60200

= (0.65, 65 %)130200

+ +60200

40200

30200

= (0.65, 65 %)130200

1 − 70200

= (0.65, 65 %)130200

[1 mark]20a.

Given the set .

List the elements of the set .

Markscheme, , , , , , (A1) (C1)

Note: Award (A1) for correct numbers, do not penalise if braces, brackets or parentheses seen.

[1 mark]

A = {x − 4 ⩽ x ⩽ 2, x is an integer}

A

−4 −3 −2 −1 0 1 2

[2 marks]20b. A number is chosen at random from set . Write down the probability that the number chosen is a negative integer.A

Markscheme (A1)(ft)(A1)(ft) (C2)

Notes: Award (A1)(ft) for numerator, (A1)(ft) for denominator. Follow through from part (a).

Note: There is no further penalty in parts (c) and (d) for use of denominator consistent with that in part (b).

[2 marks]

(0.571, 57.1%)47

[1 mark]20c. A number is chosen at random from set . Write down the probability that the number chosen is a positive even integer.

Markscheme (A1)(ft) (C1)

Note: Follow through from part (a).

[1 mark]

A

(0.143, 14.3%)17

[2 marks]20d. A number is chosen at random from set . Write down the probability that the number chosen is an odd integer less than .

Markscheme (A1)(ft)(A1)(ft) (C2)

Note: Award (A1)(ft) for numerator, (A1)(ft) for denominator. Follow through from part (a).

[2 marks]

A −1

(0.143, 14.3%)17

21a. [3 marks]

The probability that it rains today is . If it rains today, the probability that it will rain tomorrow is . If it does not rain today, theprobability that it will rain tomorrow is .


0.4 0.80.7

Markscheme

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

Note: Award (A1) for each correct pair.

[3 marks]

[3 marks]21b. Calculate the probability of rain tomorrow.


Notes: Award (A1)(ft) for two consistent products from tree diagram, (M1) for addition of their products. Follow through from theirtree diagram provided all probabilities are between 0 and 1.

(A1)(ft) (C3)

[3 marks]

0.4 × 0.8 + 0.6 × 0.7

= 0.74

22a. [2 marks]

A survey was carried out at an international airport. A number of travellers were interviewed and asked for their flight destinations.The results are shown in the table below.

One traveller is to be chosen at random from all those interviewed.

Find the probability that this traveller was going to Africa.



[2 marks]

( , 0.432, 43.2%)108250

54125

22b. [2 marks]One female traveller is to be chosen at random from all those interviewed.

Find the probability that this female traveller was going to Asia.



[2 marks]

(0.236, 23.6%)25106

22c. [2 marks]One traveller is to be chosen at random from those not going to America.

Find the probability that the chosen traveller is female.



[2 marks]

(0.418, 41.8%)71170

23a. [3 marks]

Neil has three dogs. Two are brown and one is grey. When he feeds the dogs, Neil uses three bowls and gives them out randomly.There are two red bowls and one yellow bowl. This information is shown on the tree diagram below.

One of the dogs is chosen at random.

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

(ii) Find P (the dog does not get the yellow bowl).

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

(M1)(A1)(G2)

The (M1) is for multiplying two values along any branch of the tree.

(ii) P (dog does not get yellow bowl) ( = 0.667 (3sf) or 0.6) (A1)

[3 marks]

= × = (= 0.111)13

13

19

= 23

23b. [9 marks]Neil often takes the dogs to the park after they have eaten. He has noticed that the grey dog plays with a stick for a quarter ofthe time and both brown dogs play with sticks for half of the time. This information is shown on the tree diagram below.

(i) Copy the tree diagram and add the four missing probability values on the branches that refer to playing with a stick.

During a trip to the park, one of the dogs is chosen at random.

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

(iii) Find P (the dog is grey given that the dog is playing with a stick).

(iv) Find P (the dog is grey and was fed from the yellow bowl and is not playing with a stick).

Markscheme(i) The tree diagram should show the values for the brown branch and in the correct positions for the grey branch. (A1)(A1)(ft)

Follow through if the branches are interchanged.

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

(M1)

( = 0.583) (A1)(ft)(G1)

The (M1) is for showing two correct products (whether added or not). Follow through from b(i). Award (M1) for \( \frac{1}{3} +\frac{1}{4}\) (must be a sum).

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

or (M1)(A1)(ft)

(M1) for substituted conditional probability formula, (A1) for correct substitutions.

( = 0.2) (A1)(ft)(G2)

(iv) P (grey and fed from yellow bowl and not playing with stick) ( = = 0.0833 3sf). (M1)(A1)(ft)(G1)

(M1) is for product of 3 reasonable probability values.

[9 marks]

,12

12

,14

34

= × + ×13

34

23

12

= 712

=P(G∩S)

P(S)

×13

14

( × )+( × )23

12

13

14

/112

512

= 15

= × × =13

13

34

112

336

[4 marks]23c.

There are 49 mice in a pet shop.

30 mice are white.

27 mice are male.

18 mice have short tails.

8 mice are white and have short tails.

11 mice are male and have short tails.

7 mice are male but neither white nor short-tailed.

5 mice have all three characteristics and

2 have none.

Copy the diagram below to your examination script.

Complete the diagram, using the information given in the question.

Markscheme

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

Award (A1) for 2 (must be in a box), (A1) for 7, (A1)(ft) for 6 and 4, (A1)(ft) for 9 and 13. Observe the assignment of (ft) marksstrictly here. Example A common error is likely to be 11 instead of 6 (A0). In this case follow through to 4 and 18 (A1)(ft) for thefinal pair. Here the 4 follows from the total of 27 for n(M).

[4 marks]

23d. [3 marks]Find (i)

(ii)

n(M ∩ W)

n(M ' ∪ S)

Markscheme(i) (A1)(ft)

(ii) OR (A1)(ft)

= 33 (A1)(ft)

Award (A2) if answer 33 is seen. Award (A1) for any of 22, 11, 15 or 18 seen but 33 absent.

[3 marks]

n(M ∩ W) = 14

n( ∪ S) = 22 + 11M ′ 15 + 18

23e. [2 marks]Two mice are chosen without replacement.

Find P (both mice are short-tailed).

MarkschemeP (both mice short-tailed) (= 0.130). (M1)(A1)(ft)(G1)

(Allow alternatives such as 153/1176 or 51/392.) Award (M1) for any of and or or seen.

[2 marks]

= × =1849

1748

306352

1849

1748

×1849

1749

+1849

1748

24a. [5 marks]

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 usesthe 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 0.08respectively.

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

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]

[1 mark]24b. Find the probability that Geraldine travels by bus and is late.

Markscheme0.06 (accept or 6%) (A1)(ft)

[1 mark]

0.5 × 0.12

[3 marks]24c. Find the probability that Geraldine is late.

Markschemefor a relevant two-factor product, either or (M1)

for summing three two-factor products (M1)

0.094 (A1)(ft)(G2)

[3 marks]

C × L T × L

(0.2 × 0.05 + 0.06 + 0.3 × 0.08)

[3 marks]24d. Find the probability that Geraldine travelled by train, given that she is late.

Markscheme (M1)(A1)(ft)

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

= 0.255 (A1)(ft)(G2)

[3 marks]

0.3×0.080.094

[3 marks]24e.

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

Sketch the curve of the function for values of from −2 to 4, giving the intercepts with both axes.

Markscheme

(G3)

[3 marks]

f(x) = − 2 + x − 3x3 x2 x

[3 marks]24f. On the same diagram, sketch the line and find the coordinates of the point of intersection of the line with the curve.y = 7 − 2x

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

(2.45, 2.11) (G1)(G1)

[3 marks]

[2 marks]24g. Find the value of the gradient of the curve where .

Markscheme (M1)

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

2.87 (A1)(G2)

[2 marks]

x = 1.7

(1.7) = 3(1.7 − 4(1.7) + 1f ′ )2

[4 marks]25a.

students at Rambling High School were asked how they travelled to school yesterday. All of the students travelled by bus, by caror walked.

students travelled by car only students travelled by bus only students travelled by car and walked, but did not use a bus students travelled by bus and walked, but did not use a car students used all three forms of travel.

Represent this information on a Venn Diagram.

Markscheme

(A4)

Note: Award (A1) for rectangle and three labelled intersecting circles, (A1) for , (A1) for and , (A1) for and .

50

1275103

3 5 10 7 12

25b. [4 marks]There were students who used a bus to travel to school. Calculate the number of students(i) who travelled by car and by bus but did not walk;

(ii) who travelled by car.

28

Printed for Victoria Shanghai Academy

© International Baccalaureate Organization 2016 International Baccalaureate® - Baccalauréat International® - Bachillerato Internacional®

Markscheme(i) (M1)(A1)(ft)(G2)

Note: Follow through from their Venn diagram.

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

Note: Follow through from part (b)(i) and their Venn diagram.

28 − (10 + 3 + 7) = 8

5 + 3 + 8 + 12 = 28

25c. [2 marks]Tomoko used a bus to travel to school yesterday.

Find the probability that she also walked.

Markscheme ( ) (A1)(A1)(ft)(G2)

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

P(walk|bus) = 1328

(0.464, 46.4%) 0.464285…

25d. [7 marks]Two students are chosen at random from all students.

Find the probability that(i) both students walked;(ii) only one of the students walked.

Markscheme(i) (A1)(M1)(M1)

Note: Award (A1) for seen, (M1) for non replacement, (M1) for multiplying their fractions.

( ) (A1)(G3)

(ii) (A1)(ft)(M1)

Notes: Award (A1)(ft) for two products, (M1) for adding two products. Do not penalise in (ii) for consistent use of with replacement.

( ) (A1)(ft)(G2)

50

×2350

2249

23

= 5062450

(0.207, 20.7%) 0.206530…

× + ×2350

2749

2750

2349

= 12422450

(0.507, 50.7%) 0.509638…

practice: probability 2 - mr kenneth...

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