Basic Probability: Eduqas GCSE Foundation June 2025 Paper 1 Q3

Foundation June 2025 Paper 1 Q3

EduqasFoundationCurrent spec5 marksBasic Probability

(a)

Kali has a bag containing 20 marbles.
She chooses one marble from the bag at random.

(i) It is impossible for Kali to choose a blue marble.
How many blue marbles are there in Kali’s bag? [1]

(ii) It is likely that Kali chooses a red marble.
What is the greatest number of red marbles there could be in Kali’s bag? [1]

(b)

William has a bag containing 40 counters.
20 of the counters are purple, 12 of the counters are black, and the remaining counters are green.

A counter is taken at random from the bag.
William says that there is an even chance of choosing a purple counter.

Kali says that William cannot be correct because there are 3 different colours of counter in the bag.

Who is correct?

William ☐ Kali ☐ Neither ☐

Explain your reasoning. [1]

(c)

Lauren has some pens in a drawer.

A pen is selected at random from the drawer.
The table shows the probability of choosing a blue pen, a black pen or a red pen.

Colour	Blue	Black	Red
Probability	\(\dfrac{7}{20}\)	\(\dfrac{1}{4}\)	\(\dfrac{3}{20}\)

Explain how you know that the drawer must contain pens that are not blue, black or red.

You must show all your working. [2]

Corrected by Westie's Workshop

The board's question paper printed “William has a box containing 40 counters”. Changed to “William has a bag containing 40 counters” because the question calls the same container a “box” once and a “bag” twice.

Original: see the board's question paper PDF (link above).

Mark scheme (a)

Scheme	Marks	Comments
(i) 0	B1	Allow in words e.g. ‘zero’ or ‘none’. Accept ‘no blue marbles’ or ‘0 out of 20’.
(ii) 19	B1	Allow in words e.g. ‘nineteen’. Accept ’19 out of 20’. B0 for 19/20.

Corrected by Westie's Workshop

The board's mark scheme printed “Accept ‘no blue counters’”. Changed to “Accept ‘no blue marbles’” because this part of the question is about marbles, not counters.

Original: see the board's mark scheme PDF (link above).

Mark scheme (b)

Scheme	Marks	Comments
William indicated, with a suitable reason: ‘20 is half/50% of 40 (so there’s an even chance of getting a purple counter)’ ‘Half/50% of the counters are purple (so there is an even chance of getting a purple counter)’	E1	Accept equivalent suitable reasons. Allow 20/40 = ½ oe E0 for either: 12 + 8 = 20 20 out of 40 Do not allow ‘there is an even chance as most of the counters are purple’ ‘there are 20 purple counters and 20 counters which are black or green’.

Scheme

Marks

Comments

William indicated, with a suitable reason:

‘Half/50% of the counters are purple (so there is an even chance of getting a purple counter)’

Accept equivalent suitable reasons.
Allow 20/40 = ½ oe

E0 for either:

Do not allow

Mark scheme (c)

Scheme	Marks	Comments
¼ = 5/20 OR 7/20 = 35% AND 1/4 = 25% AND 3/20 = 15% OR 7/20 = 0.35 AND 1/4 = 0.25 AND 3/20 = 0.15	B1	Accept conversion to decimals or percentages providing all fractions are converted correctly. If using a common denominator other than 20, all fractions must be converted correctly. Accept ‘7 + 5 + 3 = 15’ for B1.
7/20 + 5/20 + 3/20 = 15/20 oe (≠ 1) OR 35% + 25% + 15% = 75% (≠ 100%) OR 0.35 + 0.25 + 0.15 = 0.75 (≠ 1) (so, there must be pens of other colours in the drawer)	E1	Dependent upon B1 being awarded. Accept ‘probabilities do not add to 1’ with supporting working e.g. ‘7/20 + 5/20 + 3/20. Does not add up to 1’. Allow B1 E1 for example ‘7 + 5 + 3 = 15 so there are 5 pens of other colours’ ‘7 + 5 + 3 does not add up to 20’.
	(5)

Alternative method

Scheme	Marks	Comments
Three correctly evaluated calculations for a common amount (e.g. 40) (7/20 x 40) = 14 (1/4 x 40) = 10 (3/20 x 40) = 6	B1
Showing the sum of the three values does not equal the common amount e.g. 14 + 10 + 6 = 30 (≠ 40) (so, there must be pens of other colours in the drawer)	E1