Exam-Style Problem

9231 P41 - Nov 2024 - Q5 - 9 marks

6843

Nikita has three coins. One coin is fair, one coin is biased so that the probability of obtaining a head is \(\frac{1}{3}\) and the third coin is biased so that the probability of obtaining a head is \(\frac{1}{5}\). The random variable \(X\) is the number of heads that Nikita obtains when he throws all three coins at the same time.
(a) Find the probability generating function of \(X\).

Rajesh has two fair six-sided dice with faces labelled 1, 2, 3, 4, 5, 6. The random variable \(Y\) is the number of 4 s that Rajesh obtains when he throws the two dice.

The random variable \(Z\) is the sum of the number of heads obtained by Nikita and the number of 4 s obtained by Rajesh.
(b) Find the probability generating function of \(Z\), expressing your answer as a polynomial.
(c) Use your answer to part (b) to find \(\mathrm{E}(Z)\).

Solution

Answer:

(a) \(G_X(t)=\frac{8}{30}+\frac{14}{30}t+\frac{7}{30}t^2+\frac{1}{30}t^3\).

(b) \(G_Y(t)=\frac{25}{36}+\frac{10}{36}t+\frac{1}{36}t^2\), so

\(G_Z(t)=G_X(t)G_Y(t)=\frac{1}{1080}\left(200+430t+323t^2+109t^3+17t^4+t^5\right)\).

(a) Let the three coins have head probabilities \(\frac12\), \(\frac13\) and \(\frac15\).

For \(X\), the number of heads:

\(P(X=0)=\left(\frac12\right)\left(\frac23\right)\left(\frac45\right)=\frac{8}{30}\).
\(P(X=3)=\left(\frac12\right)\left(\frac13\right)\left(\frac15\right)=\frac{1}{30}\).
\(P(X=1)\) is the sum of the three cases where exactly one coin shows heads:
\(\left(\frac12\right)\left(\frac23\right)\left(\frac45\right)+\left(\frac12\right)\left(\frac13\right)\left(\frac45\right)+\left(\frac12\right)\left(\frac23\right)\left(\frac15\right)=\frac{8}{30}+\frac{4}{30}+\frac{2}{30}=\frac{14}{30}\).
\(P(X=2)\) is the sum of the three cases where exactly two coins show heads:
\(\left(\frac12\right)\left(\frac13\right)\left(\frac45\right)+\left(\frac12\right)\left(\frac23\right)\left(\frac15\right)+\left(\frac12\right)\left(\frac13\right)\left(\frac45\right)\) is better written carefully as
\(\left(\frac12\right)\left(\frac13\right)\left(\frac45\right)+\left(\frac12\right)\left(\frac23\right)\left(\frac15\right)+\left(\frac12\right)\left(\frac13\right)\left(\frac45\right)\) from the combinations \((H,H,T),(H,T,H),(T,H,H)\), giving
\(\frac{4}{30}+\frac{2}{30}+\frac{1}{30}=\frac{7}{30}\).

So the probability generating function is

\(G_X(t)=\sum P(X=r)t^r=\frac{8}{30}+\frac{14}{30}t+\frac{7}{30}t^2+\frac{1}{30}t^3\).

(b) For each die, the probability of a 4 is \(\frac16\), so \(Y\sim \mathrm{Bin}(2,\frac16)\).

Hence

\(P(Y=0)=\left(\frac56\right)^2=\frac{25}{36},\quad P(Y=1)=2\left(\frac16\right)\left(\frac56\right)=\frac{10}{36},\quad P(Y=2)=\left(\frac16\right)^2=\frac{1}{36}\).

Therefore

\(G_Y(t)=\frac{25}{36}+\frac{10}{36}t+\frac{1}{36}t^2\).

Since \(Z=X+Y\) and the coin throws are independent of the dice throws,

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

\(G_Z(t)=\left(\frac{8}{30}+\frac{14}{30}t+\frac{7}{30}t^2+\frac{1}{30}t^3\right)\left(\frac{25}{36}+\frac{10}{36}t+\frac{1}{36}t^2\right)\).

Take out the factor \(\frac{1}{1080}\):

\(G_Z(t)=\frac{1}{1080}(8+14t+7t^2+t^3)(25+10t+t^2)\).

Now expand:

\((8+14t+7t^2+t^3)(25+10t+t^2)\)

\(=200+80t+8t^2+350t+140t^2+14t^3+175t^2+70t^3+7t^4+25t^3+10t^4+t^5\)

\(=200+430t+323t^2+109t^3+17t^4+t^5\).

Hence

\(G_Z(t)=\frac{1}{1080}\left(200+430t+323t^2+109t^3+17t^4+t^5\right)\).

Differentiate:

\(G_Z'(t)=\frac{1}{1080}\left(430+646t+327t^2+68t^3+5t^4\right)\).

\(G_Z'(1)=\frac{1}{1080}(430+646+327+68+5)=\frac{1476}{1080}=\frac{41}{30}\).

Therefore \(\mathrm{E}(Z)=\frac{41}{30}\).