Foundation June 2021 Paper 1 Q21

EdexcelCurrent spec5 marksBasic Probability

21 Grace has a biased 5-sided spinner.

Grace is going to spin the arrow on the spinner once.

The table below gives the probabilities that the spinner will land on red or on blue or on green.

Colour	Red	Blue	Green	Orange	Pink
Probability	0.20	0.12	0.08

The probability that the spinner will land on orange is 3 times the probability that the spinner will land on pink.

(a) Work out the probability that the spinner will land on orange. (3)

Grace spins the arrow on the spinner 150 times.

(b) Work out an estimate for the number of times the spinner lands on blue. (2)

Mark scheme (a)

Scheme	Marks
eg 1 – (0.2 + 0.12 + 0.08) (= 0.6) or \(1 - \left(\dfrac{20}{100} + \dfrac{12}{100} + \dfrac{8}{100}\right)\left(= \dfrac{60}{100}\right)\) oe or 100(%) – (20(%) + 12(%) + 8(%)) (= 60(%)) or \(0.2 + 0.12 + 0.08 + 3x + x = 1\) oe	M1
“0.6” ÷ 4 (= 0.15) oe or “0.6” ÷ 4 × 3 or “0.6” × 0.75 oe (Sight of 0.15 in the table for Orange or Pink or 0.45 for Pink gains M2)	M1
Working not required, so correct answer scores full marks (unless from obvious incorrect working) Answer: 0.45	A1
	(3)

M1: for a correct calculation for the remaining probabilities or a correct equation for the remaining probabilities

M1: For dividing the remaining probability by 4 or finding ¾ of the remaining probability
NB “0.6” means 0.6 must come from correct working

A1: or \(\dfrac{9}{20}\) oe or 45% (if working in % final answer must have % sign). Allow \(\dfrac{0.45}{1}\)

If no answer on answer line, check in the correct space in table above.

Mark scheme (b)

Scheme	Marks
0.12 × 150 oe eg 12 + 6	M1
Working not required, so correct answer scores full marks (unless from obvious incorrect working) Answer: 18	A1
	(2)
	(5 marks)

M1: for a correct calculation to find the number of times the spinner lands on blue

A1: (an answer of \(\dfrac{18}{150}\) scores M1A0 as this is a probability not a number of times)