Basic Probability

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Higher June 2025 Paper 2 Q11

EdexcelCurrent spec3 marksBasic Probability

11 Khalid has a box of counters.

17 of the counters are red
28 of the counters are blue
the rest of the counters are orange

Khalid is going to take at random a counter from the box.

The probability that Khalid will take an orange counter is \(\dfrac{4}{9}\)

Work out the number of orange counters that are in the box.

(3)

Mark scheme

Scheme	Marks
(17 + 28) ÷ (9 – 4) or 45 ÷ 5 or \(1 - \dfrac{4}{9}\left(= \dfrac{5}{9}\right)\) or \((17 + 28 =)\dfrac{5}{9}\) oe or \(\dfrac{p}{p + 28 + 17} = \dfrac{4}{9}\) oe or \(\dfrac{p}{p + 45} = \dfrac{4}{9}\) or \(\dfrac{m - 45}{m} = \dfrac{4}{9}\)	M1
(17 + 28) ÷ 5 × 4 or 45 ÷ 5 × 4 or \((17 + 28) \div \text{``}{\dfrac{5}{9}}\text{''}\) (= 81) or \(45 \div \text{``}{\dfrac{5}{9}}\text{''}\) (= 81) or \((17 + 28) \times \text{``}{\dfrac{9}{5}}\text{''}\) (= 81) or \(45 \times \text{``}{\dfrac{9}{5}}\text{''}\) (= 81) or (17 + 28) ÷ 5 × 4 or 45 ÷ 5 × 4 or \(\text{``}{\dfrac{5}{9}}\text{''} = \dfrac{17 + 28}{n}\) oe or \(n = 81\) or \(9p = 4(p + 28 + 17)\) or \(9p - 4p = 180\) oe or \(5p = 180\) oe or \(9(m - 45) = 4m\) or \(9m - 4m = 405\) oe or \(5m = 405\) oe or \(m = 81\)	M1
Working not required, so correct answer scores full marks (unless from obvious incorrect working) Answer: 36	A1
	(3)
	(3 marks)

Notes

M1: Allow 0.55(555…) or 55(.555…)% truncated or rounded

M1: for the correct calculation for the total number of counters
or
for the correct calculation for the number of orange counters
or
for the correct equation for the total number of counters (removing the denominators)
or
for the correct equation for the number of orange counters (removing the denominators)

A1: cao

Higher June 2025 Paper 1R Q4

EdexcelCurrent spec5 marksBasic Probability

4 Here is a biased 5-sided spinner.
When the spinner is spun, it can land on blue or on green or on purple or on red or on yellow.

Pentagonal spinner with sections blue, green, purple, red and yellow

The table gives information about the probability of the spinner landing on each colour.

Colour	blue	green	purple	red	yellow
Probability	0.12	0.20	0.38	\(4x\)	\(x\)

Sophie spins the spinner once.

(a) Work out the probability that the spinner lands on blue or on green or on purple. (1)

Max spins the spinner 350 times.

(b) Work out an estimate for the number of times the spinner lands on red. (4)

Scheme	Marks
0.7	B1
	(1)

Mark scheme (b)

Scheme	Marks
eg 1 – (0.12 + 0.2 + 0.38) (= 0.3) oe or 1 – “0.7” (= 0.3) oe or 0.12 + 0.20 + 0.38 + 4\(x\) + \(x\) = 1 oe or “0.7” × 350 (= 245) oe or 0.12 × 350 (= 42) or 0.38 × 350 (= 133)	M1
eg “0.3” ÷ 5 (= 0.06) or “0.3” ÷ 5 × 4 (= 0.24) or 0.24 or (\(x\) =) 0.06 or (4\(x\) =) 0.24 or “0.3” × 350 (= 105) oe or 350 – “245” (= 105) oe or 350 – “42” – 0.2 × 350 – “133” (=105) oe	M1
eg “0.06” × 350 (= 21) oe or “105” ÷ 5 (= 21) oe or “0.06” × 4 × 350 oe or “0.24” × 350	M1
Correct answer scores full marks (unless from obvious incorrect working) Answer: 84	A1
	(4)
	(5 marks)

Notes

M1: ft their “0.7”

If probabilities are given as percentages then % sign must be seen

M1: or for \(\dfrac{21}{350}\) or \(\dfrac{84}{350}\)

A1: cao

Higher November 2024 Paper 2 Q5

EdexcelCurrent spec4 marksBasic Probability

5 Cody has a biased 5-sided spinner, numbered 1, 2, 3, 4, 5

The table gives the probabilities that when the spinner is spun it will land on 2 or on 3 or on 5

Number	1	2	3	4	5
Probability		0.14	0.17		0.21

The probability that the spinner will land on 1 is the same as the probability that the spinner will land on 4

Cody is going to spin the spinner 400 times.

Work out an estimate for the number of times the spinner will land on 4

(4)

Mark scheme

Scheme	Marks
1 – (0.14 + 0.17 + 0.21) (= 0.48) or \(0.14 + 0.17 + 0.21 + x + x = 1\) oe or 0.14 × 400 (= 56) or 0.17 × 400 (= 68) or 0.21 × 400 (= 84) or (0.14 + 0.17 + 0.21) × 400 (= 208) oe eg 0.52 × 400 (= 208)	M1
“0.48” ÷ 2 (= 0.24) [could be in table] or 400 – 56 – 68 – 84 (= 192) oe eg 400 – 208 (= 192) or 0.48 × 400 (= 192)	M1
“0.24” × 400 oe or “192” ÷ 2	M1
Correct answer scores full marks (unless from obvious incorrect working) Answer: 96	A1
	(4)
	(4 marks)

Notes

M1: Correct use of probabilities total 1
or
correct calculation for an estimate for number of times the spinner will land on 2 or on 3 or on 5

M1: A completely correct method to find the probability that the spinner will land on 4
or
a completely correct method to find the number of times the spinner will land on 1 or on 4

M1: a correct calculation to find the estimate required

an answer leading from 96 seen

eg \(\dfrac{96}{400}\)

A1: cao
SCB1 for 104 if no other marks have been awarded