June 2025 Paper 3 Q1

EdexcelCurrent spec9 marksProbability

1. Bag A contains 5 red, 4 yellow and 3 green beads.

Bag B only contains red and yellow beads.

A bead is selected at random from bag A and a second bead is selected at random from bag B

Given that the probability that both beads selected are yellow is \(\dfrac{3}{16}\)

(a)

(i) find the probability of selecting a yellow bead from bag B (2)

(ii) Hence complete the tree diagram below by finding the probability on each branch.

tree diagram: first branches from bag A to red, yellow and green; from each, branches from bag B to red and yellow; every branch has a dotted space for its probability

(2)

(b) Find the exact probability that at least one of the two beads selected is yellow. (3)

The event \(X\) is that at least one of the beads selected is yellow.

The event \(W\) is that a green bead is selected.

Mark scheme (a)(i)

Scheme	Marks	AO
[ Let \(p\) = probability of yellow from bag B] \(\mathrm{P}(YY) = \dfrac{4}{12} \times p\)	M1	1.1b
\(\left[\dfrac{4}{12} \times p = \dfrac{3}{16}\right]\) so \(p = \dfrac{9}{16}\) (Allow 0.5625 and condone 0.563)	A1	1.1b
	(2)

Notes

M1: for a suitable expression (in their \(p\)) for the probability both are yellow.

A1: for correct probability of yellow from B. Can be implied by values on tree diagram.

Handwritten tree diagram: bag A branches R 5/12, Y 4/12, G 3/12; from each, bag B branches R 7/16 and Y 9/16 — Mark scheme (a)(ii)

Scheme	Marks	AO
	B1 B1ft	1.1b 1.1b
	(2)

Mark scheme (b)

Scheme	Marks	AO
e.g. \(\mathrm{P}(RY) + \mathrm{P}(Y) + \mathrm{P}(GY)\) or \(1 - \mathrm{P}(RR) - \mathrm{P}(GR)\) or \(\mathrm{P}(\underline{\ldots}Y) + \mathrm{P}(YR)\)	M1	2.1
\(\dfrac{5}{12} \times \dfrac{\text{``}9\text{''}}{\text{``}16\text{''}} + \dfrac{4}{12} + \dfrac{3}{12} \times \dfrac{\text{``}9\text{''}}{\text{``}16\text{''}}\) or \(1 - \left[\dfrac{5}{12} \times \dfrac{\text{``}7\text{''}}{\text{``}16\text{''}} + \dfrac{3}{12} \times \dfrac{\text{``}7\text{''}}{\text{``}16\text{''}}\right]\) or \(\dfrac{\text{``}9\text{''}}{\text{``}16\text{''}} + \dfrac{1}{3} \times \dfrac{\text{``}7\text{''}}{\text{``}16\text{''}}\)	A1ft	1.1b
\(= \dfrac{136}{192} = \dfrac{17}{24} = 0.708\dot{3}\)	A1	1.1b
	(3)

Notes

M1: for selection of all correct cases (including the \(1 -\) if required). Don’t need P(… may be implied by correct probability products seen (ft their tree diagram)

A1ft: for a correct complete probability expression (ft their tree diagram) or awrt 0.708
Must see product pairs (for ft) or implied by correct values e.g. \(\frac{15}{64} + \frac{7}{48} + \frac{3}{16} + \frac{9}{64}\)
May see \(1 - \mathrm{P}\left(\bar{Y}\,\bar{Y}\right)\) (M1) and \(1 - \dfrac{2}{3} \times \dfrac{\text{``}7\text{''}}{\text{``}16\text{''}}\) (A1ft)

A1: for an exact answer …fraction needn’t be simplified but decimal must have \(\ldots\dot{3}\)

Acc: If scored 2^nd A0 in (b) for 0.708 (or better) then allow A1 in (c) for awrt 0.199

Mark scheme (c)

Scheme	Marks	AO
\(\left[\mathrm{P}(W \mid X) =\right] \dfrac{\mathrm{P}(W \cap X)}{\mathrm{P}(X)}\) or \(\dfrac{\frac{3}{12} \times \frac{\text{``}9\text{''}}{\text{``}16\text{''}}}{\text{``}(\text{b})\text{''}}\)	M1	1.1b
\(= \dfrac{27}{136}\)	A1	1.1b
	(2)
	(9 marks)

Notes

M1: for a correct ratio of probabilities expression (in symbols) or ft their values seen

A1: for an exact answer NB an answer of awrt 0.199 (0.198529411…) implies M1

Acc: If scored 2^nd A0 in (b) for 0.708 (or better) then allow A1 in (c) for awrt 0.199