1. The diagram showed a scale marked at 0, 0.5 and 1. You could develop the idea that all probabilities can be marked on such a scale, with students suggesting examples that lie at different points on the scale. You could also discuss the use of both fractions and decimals to represent probabilities.

2. Lynne entered 500 competitions last year and won 10 prizes. We saw how to estimate her probability of winning from last year’s results. Students could comment on what they think this probability represents and how it can be interpreted.

3. The slogan is ‘Grab a Grand’. As Lynne points out, the chances of winning this prize would probably be relatively small. Why?

4. There are 7 unlabelled tins: 4 containing baked beans, 2 containing semolina and 1 containing dog food. Ben and Katie demonstrate how to calculate the probability of randomly selecting one that would not go well with toast. The ‘or rule’ for addition of probabilities is stated here, and could be developed further.

5. 4 pink balls are placed in the bag. There are also 2 blue, 1 yellow and 3 green. The problem is to find the probability of choosing either pink or blue at random. Many further examples could be generated from this context. What if the yellow ball was removed? What if one ball of another colour was added? What if two draws were made?

6. They draw 4 branches, representing each of the 4 colours. You could discuss more fully here the construction of tree diagrams and the confusion highlighted by Katie’s mistake (of stating the probabilities as quarters because there were four colours). You could also develop the idea of checking that the sum of probabilities of mutually exclusive outcomes equals one.

7. Ben suggests placing £10 notes in the packets. We see how much the campaign would cost, and the probability of winning for one, one million and one thousand £10 notes. Students could comment on why one thousand notes was suggested as an acceptable number. Can they give other examples of similar promotions and comment on the probability of winning in each case?

8. The chance of winning the mobile phone is given as 1/25000. Ben calculates his chance of winning both the phone and the £10 using multiplication. The idea that winning both would be far more unlikely than winning one or the other can be linked to the result of multiplying the probabilities. Students could also explore the nature of independent events. Is there any reason to think that if you win one prize you are less likely to win a second?

9. You could look in more detail at the construction of the diagram for the two events. Katie used multiplication to find the probability of selecting beans and bread fit to eat. Students could find all the other possible outcomes and discuss how to record them on the diagram.

10. Katie selects semolina and mouldy bread. The viewer is asked to calculate the probability of obtaining this outcome. Students could calculate this, and the probabilities of the other possible outcomes.

Worksheet 2: Tick or Trash

Question 1

Student A is correct in both (i) and (ii).

In (i), Student B has made a mistake in adding the decimals. The total is 0.9, not 0.36. This was frequently seen by the examiners, although the question appeared on a paper permitting calculator use.

In (ii), Student B has ignored the context of the question and assumed that the colours can be mixed like paint. Again this was often seen by the examiners.

In (ii), Student A should have written ‘0’, not ‘nought’, but the response would have been accepted by the examiners.

The phrase ‘write down’ implies that a calculation is not required.

Question 2

Student A is correct. Student B has treated the question as if the discs were not replaced.

In (b), Student B failed to recognise that there are two ways of getting discs of a different colour. Student B would have gained credit in (b) if the results of (a) had been correctly applied.

Student A completed columns for outcome and probability on the tree diagram, which helped with the calculation.

Worksheet 3: Exam Practise Questions (Edexcel)

This is a selection of past exam questions from Edexcel, targeted at the middle ability range. You can use the sequence as it stands or select individual questions to suit your needs.

Given below is the mark scheme for each question, which will help you to see how marks are allocated and what the examiner is especially looking out for. The mark schemes use the following abbreviations:
oe
or equivalent
cao
correct answer only
ft
follow-through marks
dep
dependent
indep
independent
M
method marks
A
accuracy marks
B
benefit-of-doubt mark
SC
special case

Question 1

(a) Red

(b)

(i) 1 — (0.6 + 0.15 + 0.15) = 0.1

(ii) 0

Notes:

(a) B1 cao.

(b)

(i) M1 for 1 — (0.6 + 0.15 + 0.15).

(ii) A1 for 0.1. B1 cao.

Question 2

1 — ¹/₁₂ =

accept ⁴/₁₂ + ²/₁₂ + ³/₁₂ + ²/₁₂ = ¹¹/₁₂

Question 3

1 — (0.31 + 0.28 + 0.24) = 0.17

Question 4

(a)

(i) ⁸/₁₃ or 0.62...

(ii) 1 — ^‘8’/₁₃ = ⁵/₁₃ (or 0.38...)

(b) (y,y) (y,g) (g,y) (g,g)

Question 5

(i) ¹⁵/₂₂

(ii) ⁷/₂₂

Question 6

(a) 1 — 0.995 = 0.005 (C)

(b) 10 000 x (a) — 50 (B)

Question 7

(a) ¹⁰/₂₀ or ¹/₂

(b) ⁷/₂₀

Notes:

(a) B1 cao oe 0.5, 50% only.

(b) B1 cao oe 0.35, 35% only.

Question 8

Working 1 — (0.9 + 0.03)

Answer 0.07

Notes:

M1. A1 cao.

Question 9

(i) 0.2 x 0.3 = 0.06

(ii) 1 — 0.3 or 1 — 0.2 (oe)

0.2 x ‘0.7’ + ‘0.8’ x 0.3 = 0.38

Notes:

(i) M1 for 0.2 x 0.3 seen on its own. A1 cao.

(ii) M1. M1 (dep on previous M1). A1 cao.

Question 10

(i) ¹/₄ x ¹/₄ = ¹/₁₆

(ii) 2 x ¹/₄ x ¹/₃ = ¹/₆

Notes:

(i) M1 ¹/₄ x ¹/₄. A1 cao [0.06 or better].

(ii) M1 ¹/₄ x ¹/₃. M1 x #2 (dep). A1 cao [0.17, 0.16 or better].

Question 11

(a)

3p + p = 1 or 1:3

p = 0.25

‘0.25’ x 232

= 58

(b)

PC ‘a’ x ⁴⁸/₈ = 348 pkts

COC (232 — ‘a’) x ⁴⁸/₁₆

= 522 pkts

(PC + COC) ÷ (232 x 48) = 870 ÷ 11136 = ⁵/₆₄

= 0.07812

Alternative

p x ¹/₈ + (1 — p) x ¹/₁₆

Notes:

(a) M1 (oe method to find probs). A1 cao for p = 0.25. A1 cao.

(b) M1 PC ‘a’ x 48 ÷ 8 or ‘a’ x 6 (oe for COC). M1 dep. A1 cao anything rounding up to 0.08. Alternative: M1 for p x ¹/₈. M1 dep full expression. A1 cao.

Question 12

(a) 1 — 0.17 = 0.83

(b) 0.27 x 0.17 = 0.0459

Notes:

(a)

M1 for 1 — 0.17. A1 cao.

SC: 0.73 B1

(b) M1 for 0.27 x 0.17. A1 cao exactly.

Question 13

(i) 0.0625 x 0.032 x 0.044 = 0.0000915

(ii) 0.065 x 0.968 x 0.954 = 0.06

(iii) 0.935 x 0.968 x 0.954 = 0.863

Question 14

(i) 0.65 x 0.8 = 0.52

(ii)

1 — 0.65 (= 0.35)

1 — 0.8 (= 0.2)

0.35 x 0.2 = 0.07

Question 15

(a) [AW]

(b)

(i) ⁵/₁₂ x ⁵/₁₂ = ²⁵/₁₄₄

(ii) (⁵/₁₂ x ⁷/₁₂) + (⁷/₁₂ x ⁵/₁₂) = ³⁵/₇₂

Notes:

(a) B1 LHS (oe). B1 RHS (oe) ft on probability values used on LHS. In each case (oe) is a % or a decimal to a minimum of 2 dp on LHS. NB: Probabilities must add to 1 on each pair of branches. ⁵/₁₂ = 0.416, ⁷/₁₂ = 0.583.

(b)

(i) M1 ⁵/₁₂ x ⁵/₁₂. A1 cao (oe) 0.17(3611...).

(ii)

M1 (⁵/₁₂ x ⁷/₁₂). M1 ( ) + ( ) (dep); OR 2 x ( ). A1 cao (oe) 0.48(611...) or 0.49.

SC: if without replacement, then allow B1 on (a), M1 x 3 for (b), B1 for both ⁵/₃₃, ³⁵/₆₆ (0.15..., 0.53030...).