Multiple Events: WJEC GCSE Higher June 2022 Unit 1 Q17

Higher June 2022 Unit 1 Q17

WJECHigherCurrent spec5 marksMultiple Events

17. A bag contains 10 balls.
5 of the balls are blue, 4 of the balls are red and 1 ball is green.
Three balls are chosen at random, one at a time, without replacement.

(a) Calculate the probability that the first ball is blue, the second ball is red and the third ball is green.
You must show all your working. [2]

(b) Calculate the probability that at least one blue ball is chosen.
You must show all your working. [3]

Mark scheme (a)

Scheme	Marks	Comments
\(\frac{5}{10} \times \frac{4}{9} \times \frac{1}{8}\) or equivalent	M1	Accept e.g. \(\frac{5 \times 4 \times 1}{10 \times 9 \times 8}\)
\(\frac{20}{720}\ \left( = \frac{1}{36}\right)\) or equivalent	A1	ISW

Mark scheme (b)

Scheme	Marks	Comments
1 – P(no blue)	S1	May be implied by subsequent working.
\(= 1 - \frac{5}{10} \times \frac{4}{9} \times \frac{3}{8}\)	M1	Complete method.
\(= \frac{660}{720}\ \left( = \frac{11}{12}\right)\) or equivalent	A1	ISW FT from part (a) consistent use of a wrongly calculated denominator. If no other marks awarded, SC1 for sight of \(\frac{875}{1000}\) or \(\frac{940}{1000}\) or equivalent.

Alternative method #1

Scheme	Marks	Comments
1 – P(three red) – P(two red, one green)	S1	May be implied by subsequent working.
\(= 1 - \frac{4}{10} \times \frac{3}{9} \times \frac{2}{8} - \frac{4}{10} \times \frac{3}{9} \times \frac{1}{8} \times 3\) \(\left(= 1 - \frac{24}{720} - \frac{36}{720}\quad\text{or}\quad 1 - \frac{1}{30} - \frac{1}{20}\right)\)	M1	Complete method. (Missing x3 is S1 M0 A0.)
\(= \frac{660}{720}\ \left( = \frac{11}{12}\right)\) or equivalent	A1	ISW FT from part (a) consistent use of a wrongly calculated denominator. If no other marks awarded, SC1 for sight of \(\frac{888}{1000}\) or \(\frac{940}{1000}\) or equivalent.

Alternative method #2

Scheme	Marks	Comments
P(one blue, two not blue OR two blue, one not blue OR three blue)	S1	May be implied by subsequent working.
\(= \frac{5}{10} \times \frac{5}{9} \times \frac{4}{8} \times 3 + \frac{5}{10} \times \frac{4}{9} \times \frac{5}{8} \times 3 + \frac{5}{10} \times \frac{4}{9} \times \frac{3}{8}\)	M1	Complete method. (Missing x3 is S1 M0 A0.)
\(= \frac{660}{720}\ \left( = \frac{11}{12}\right)\) or equivalent	A1	ISW FT from part (a) consistent use of a wrongly calculated denominator. If no other marks awarded, SC1 for sight of \(\frac{875}{1000}\) or \(\frac{660}{1000}\) or equivalent.

Alternative method #3

Scheme	Marks	Comments
P(two red, one blue OR one red, one green, one blue OR two blue, one red OR two blue, one green OR three blue)	S1	May be implied by subsequent working.
\(= \frac{4}{10} \times \frac{3}{9} \times \frac{5}{8} \times 3 + \frac{4}{10} \times \frac{1}{9} \times \frac{5}{8} \times 6\) \(\quad + \frac{5}{10} \times \frac{4}{9} \times \frac{4}{8} \times 3 + \frac{5}{10} \times \frac{4}{9} \times \frac{1}{8} \times 3 + \frac{5}{10} \times \frac{4}{9} \times \frac{3}{8}\)	M1	Complete method. (Missing x3 and / or x6 is S1 M0 A0.)
\(= \frac{660}{720}\ \left( = \frac{11}{12}\right)\) or equivalent	A1	ISW FT from part (a) consistent use of a wrongly calculated denominator. If no other marks awarded, SC1 for sight of \(\frac{860}{1000}\) or \(\frac{660}{1000}\) or equivalent.