Selection without replacement scenarios

5 Nine balls labelled \(1,2,3,4,5,6,7,8,9\) are placed in a bag. Kai selects three balls at random from the bag, without replacement. The random variable \(X\) is the number of balls selected by Kai that are labelled with a multiple of 3 .

1. Find the probability generating function \(\mathrm { G } _ { \mathrm { X } } ( \mathrm { t } )\) of \(X\).
   The balls are replaced in the bag.
   Jacob now selects two balls at random from the bag, without replacement. The random variable \(Y\) is the number of balls selected by Jacob that are labelled with an even number.
2. Find the probability generating function \(\mathrm { G } _ { Y } ( \mathrm { t } )\) of \(Y\).
   The random variable \(Z\) is the sum of the number of balls that are labelled with a multiple of 3 selected by Kai and the number of balls that are labelled with an even number selected by Jacob.
3. Find the probability generating function of \(Z\), expressing your answer as a polynomial.
4. Use the probability generating function of \(Z\) to find \(\mathrm { E } ( Z )\).

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 \(\operatorname { Var } ( Z )\).