CAIE Further Paper 4 2023 November — Question 3

Exam BoardCAIE
ModuleFurther Paper 4 (Further Paper 4)
Year2023
SessionNovember
TopicProbability Generating Functions
TypeSelection without replacement scenarios

3 Toby has a bag which contains 6 red marbles and 3 green marbles. He randomly chooses 3 marbles from the bag, without replacement. The random variable \(X\) is the number of red marbles that Toby obtains.
  1. Find the probability generating function of \(X\).
    Ling also has a bag which contains 6 red marbles and 3 green marbles. He randomly chooses 2 marbles from his bag, without replacement. The random variable \(Y\) is the number of red marbles that Ling obtains. It is given that the probability generating function of \(Y\) is \(\frac { 1 } { 12 } \left( 1 + 6 t + 5 t ^ { 2 } \right)\). The random variable \(Z\) is the total number of red marbles that Toby and Ling obtain.
  2. Find the probability generating function of \(Z\), expressing your answer as a polynomial in \(t\).
  3. Use the probability generating function of \(Z\) to find \(\operatorname { Var } ( Z )\).