Write down the probability that the first marble Anna selects is red.

Find the probability that Anna selects two red marbles.

Find the probability that one marble is red and one marble is blue.

Note: In this question, method marks may be awarded for selecting without replacement, as noted in the examples.

\({\rm{P}}(R) = \frac{6}{8}\left( { = \frac{3}{4}} \right)\)     A1     N1

[1 mark]

attempt to find \({\rm{P(Red)}} \times {\rm{P(Red)}}\)     (M1)

e.g. \({\rm{P(}}R{\rm{)}} \times {\rm{P(}}R{\rm{)}}\) , \(\frac{3}{4} \times \frac{3}{4}\) , \(\frac{6}{8} \times \frac{5}{7}\)

\({\rm{P}}(2R) = \frac{{36}}{{64}}\left( { = \frac{9}{{16}}} \right)\)     A1     N2

[2 marks]

METHOD 1

attempt to find \({\rm{P(Red)}} \times {\rm{P(Blue)}}\)     (M1)

e.g. \({\rm{P(}}R{\rm{)}} \times {\rm{P(}}B{\rm{)}}\) , \(\frac{6}{8} \times \frac{2}{8}\) , \(\frac{6}{8} \times \frac{2}{7}\)

recognizing two ways to get one red, one blue     (M1)

e.g. \({\rm{P}}(RB) + {\rm{P}}(BR)\) , \(2\left( {\frac{{12}}{{64}}} \right)\) , \(\frac{6}{8} \times \frac{2}{7} + \frac{2}{8} \times \frac{6}{7}\)

\({\rm{P}}(1R,1B) = \frac{{24}}{{64}}\left( { = \frac{3}{8}} \right)\)     A1     N2

[3 marks]

METHOD 2

recognizing that \({\rm{P}}(1R,1B)\) is \(1 - {\rm{P}}(2B) - {\rm{P}}(2R)\)     (M1)

attempt to find \({\rm{P}}(2R)\) and \({\rm{P}}(2B)\)     (M1)

e.g. \({\rm{P}}(2R) = \frac{3}{4} \times \frac{3}{4}\) , \(\frac{6}{8} \times \frac{5}{7}\) ; \({\rm{P}}(2B) = \frac{1}{4} \times \frac{1}{4}\) , \(\frac{2}{8} \times \frac{1}{7}\)

\({\rm{P}}(1R,1B) = \frac{{24}}{{64}}\left( { = \frac{3}{8}} \right)\)     A1     N2

[3 marks]

Candidates did very well on parts (a) and (b) of this probability question.

Candidates did very well on parts (a) and (b) of this probability question, and knew to multiply the probabilities of independent events in part (b). However, in part (c), very few candidates considered that there are two ways to draw one red and one blue marble, and therefore did not earn full marks on this question. There were also some candidates who tried to add, rather than multiply, the probabilities in parts (b) and (c).

Candidates did very well on parts (a) and (b) of this probability question, and knew to multiply the probabilities of independent events in part (b). However, in part (c), very few candidates considered that there are two ways to draw one red and one blue marble, and therefore did not earn full marks on this question. There were also some candidates who tried to add, rather than multiply, the probabilities in parts (b) and (c).