Find the probability that

  (i)     none of the marbles are green;

  (ii)     exactly one marble is green.

Find the expected number of green marbles drawn from the jar.

(i)     Write down the probability that the marble is drawn from jar B.

(ii)     Given that the marble was drawn from jar B, write down the probability that it is red.

Given that the marble is red, find the probability that it was drawn from jar A.

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

eg   \(\frac{3}{8} \times \frac{2}{7}\) , \(\frac{3}{8} \times \frac{3}{8}\) , \(\frac{3}{8} \times \frac{2}{8}\)

\({\text{P(none green)}} = \frac{6}{{56}}\) \(\left( { = \frac{3}{{28}}} \right)\)     A1     N2

(ii)     attempt to find \({\rm{P(red)}} \times {\rm{P(green)}}\)     (M1)

eg   \(\frac{5}{8} \times \frac{3}{7}\) , \(\frac{3}{8} \times \frac{5}{8}\) , \(\frac{{15}}{{56}}\)

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

eg   \(2{\rm{P}}(R) \times {\rm{P}}(G)\) , \(\frac{5}{8} \times \frac{3}{7} + \frac{3}{8} \times \frac{5}{7}\) , \(\frac{3}{8} \times \frac{5}{8} \times 2\)

\({\text{P(exactly one green)}} = \frac{{30}}{{56}}\) \(\left( { = \frac{{15}}{{28}}} \right)\)     A1     N2

[5 marks]

\({\text{P(both green)}} = \frac{{20}}{{56}}\) (seen anywhere)     (A1)

correct substitution into formula for \({\rm{E}}(X)\)     A1

eg   \(0 \times \frac{6}{{56}} + 1 \times \frac{{30}}{{56}} + 2 \times \frac{{20}}{{56}}\) , \(\frac{{30}}{{64}} + \frac{{50}}{{64}}\)

expected number of green marbles is \(\frac{{70}}{{56}}\) \(\left( { = \frac{5}{4}} \right)\)     A1     N2

[3 marks]

(i)     \({\text{P(jar B)}} = \frac{4}{6}\) \(\left( { = \frac{2}{3}} \right)\)     A1     N1

(ii)     \({\text{P(red| jar B)}} = \frac{6}{8}\) \(\left( { = \frac{3}{4}} \right)\)     A1     N1

[2 marks]

recognizing conditional probability     (M1)

eg   \({\rm{P}}(A|R)\) , \(\frac{{{\text{P(jar A and red)}}}}{{{\rm{P(red)}}}}\) , tree diagram

attempt to multiply along either branch (may be seen on diagram)     (M1)

eg   \({\text{P(jar A and red)}} = \frac{1}{3} \times \frac{3}{8}\) \(\left( { = \frac{1}{8}} \right)\)

attempt to multiply along other branch     (M1)

eg   \({\text{P(jar B and red)}} = \frac{2}{3} \times \frac{6}{8}\) \(\left( { = \frac{1}{2}} \right)\)

adding the probabilities of two mutually exclusive paths     (A1)

eg   \({\rm{P(red)}} = \frac{1}{3} \times \frac{3}{8} + \frac{2}{3} \times \frac{6}{8}\)

correct substitution

eg   \({\text{P(jar A|red)}} = \frac{{\frac{1}{3} \times \frac{3}{8}}}{{\frac{1}{3} \times \frac{3}{8} + \frac{2}{3} \times \frac{6}{8}}}\) , \(\frac{{\frac{1}{8}}}{{\frac{5}{8}}}\)     A1

\({\text{P(jar A|red)}} = \frac{1}{5}\)     A1     N3

[6 marks]

Many candidates correctly found the probability of selecting no green marbles in two draws, although some candidates treated the second draw as if replacing the first. When finding the probability for exactly one green marble, candidates often failed to recognize two pathways for selecting one of each color.

Few candidates understood the concept of expected value in this context, often leaving this blank or treating as if a binomial experiment. Successful candidates often made a distribution table before making the final calculation.

Most candidates answered part (c) correctly. However, many overcomplicated (c)(ii) by using the conditional probability formula. Those with a clear understanding of the concept easily followed the “write down” instruction.

Only a handful of candidates correctly applied conditional probability to find \(P(A|R)\) in part (d). While some wrote down the formula, or drew a tree diagram, few correctly calculated \(P(red) = \frac{5}{8}\) . A common error was to combine the marbles in the two jars to give \(P(red) = \frac{9}{16}\) .