Exam-Style Problem

9231 P42 - Nov 2023 - Q3 - 10 marks

6699

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.
(a) 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.
(b) Find the probability generating function of \(Z\), expressing your answer as a polynomial in \(t\).
(c) Use the probability generating function of \(Z\) to find \(\operatorname{Var}(Z)\).

Solution

Answer: (a) \(G_X(t)=\frac{1}{84}\left(1+18t+45t^2+20t^3\right)\).

(b) \(G_Z(t)=\frac{1}{1008}\left(1+24t+158t^2+380t^3+345t^4+100t^5\right)\).

(a) Toby chooses 3 marbles from 6 red and 3 green, without replacement. If \(X\) is the number of red marbles chosen, then \(X\) can take the values \(0,1,2,3\).

The total number of ways to choose 3 marbles from 9 is \(\binom{9}{3}=84\).

\(P(X=0)=\dfrac{\binom{6}{0}\binom{3}{3}}{\binom{9}{3}}=\dfrac{1}{84}\)
\(P(X=1)=\dfrac{\binom{6}{1}\binom{3}{2}}{\binom{9}{3}}=\dfrac{18}{84}\)
\(P(X=2)=\dfrac{\binom{6}{2}\binom{3}{1}}{\binom{9}{3}}=\dfrac{45}{84}\)
\(P(X=3)=\dfrac{\binom{6}{3}\binom{3}{0}}{\binom{9}{3}}=\dfrac{20}{84}\)

So the probability generating function is

\(G_X(t)=E(t^X)=\sum P(X=r)t^r\)

\(\displaystyle G_X(t)=\frac{1}{84}+\frac{18}{84}t+\frac{45}{84}t^2+\frac{20}{84}t^3=\frac{1}{84}\left(1+18t+45t^2+20t^3\right).\)

(b) We are given

\(\displaystyle G_Y(t)=\frac{1}{12}(1+6t+5t^2).\)

Since \(Z=X+Y\), and Toby's and Ling's choices are from different bags, \(X\) and \(Y\) are independent. Therefore

\(G_Z(t)=G_X(t)G_Y(t).\)

\(\displaystyle G_Z(t)=\frac{1}{84}(1+18t+45t^2+20t^3)\cdot \frac{1}{12}(1+6t+5t^2).\)

Now expand:

\((1+18t+45t^2+20t^3)(1+6t+5t^2)\)

\(=1+6t+5t^2+18t+108t^2+90t^3+45t^2+270t^3+225t^4+20t^3+120t^4+100t^5\)

\(=1+24t+158t^2+380t^3+345t^4+100t^5.\)

Hence

\(\displaystyle G_Z(t)=\frac{1}{1008}\left(1+24t+158t^2+380t^3+345t^4+100t^5\right).\)

(c) For a probability generating function \(G_Z(t)\),

\(E(Z)=G_Z'(1), \qquad E(Z(Z-1))=G_Z''(1).\)

Differentiate:

\(\displaystyle G_Z'(t)=\frac{1}{1008}\left(24+316t+1140t^2+1380t^3+500t^4\right),\)

\(\displaystyle G_Z''(t)=\frac{1}{1008}\left(316+2280t+4140t^2+2000t^3\right).\)

Now substitute \(t=1\):

\(\displaystyle E(Z)=G_Z'(1)=\frac{24+316+1140+1380+500}{1008}=\frac{3360}{1008}=\frac{10}{3}.\)

Also,

\(\displaystyle G_Z''(1)=\frac{316+2280+4140+2000}{1008}=\frac{8736}{1008}=\frac{26}{3}.\)

Since

\(E(Z^2)=E(Z(Z-1))+E(Z)=G_Z''(1)+G_Z'(1),\)

we get

\(\displaystyle E(Z^2)=\frac{26}{3}+\frac{10}{3}=12.\)

Therefore

\(\displaystyle \operatorname{Var}(Z)=E(Z^2)-[E(Z)]^2=12-\left(\frac{10}{3}\right)^2=12-\frac{100}{9}=\frac{8}{9}.\)