CAIE Further Paper 4 2020 November — Question 5 10 marks

Exam BoardCAIE
ModuleFurther Paper 4 (Further Paper 4)
Year2020
SessionNovember
Marks10
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicProbability Generating Functions
TypeMultiple independent coins/dice
DifficultyStandard +0.3 This is a straightforward application of PGF theory with standard coin tosses. Part (a) is routine binomial, parts (b-c) use the independence property of PGFs (multiply them together), part (d) applies the standard derivative formula for expectation, and part (e) requires finding the maximum coefficient. All techniques are direct applications of core PGF methods with no novel insight required, making it slightly easier than average.
Spec5.02a Discrete probability distributions: general

5 Keira has two unbiased coins. She tosses both coins. The number of heads obtained by Keira is denoted by \(X\).
  1. Find the probability generating function \(\mathrm { G } _ { \mathrm { X } } ( \mathrm { t } )\) of \(X\).
    Hassan has three coins, two of which are biased so that the probability of obtaining a head when the coin is tossed is \(\frac { 1 } { 3 }\). The corresponding probability for the third coin is \(\frac { 1 } { 4 }\). The number of heads obtained by Hassan when he tosses these three coins is denoted by \(Y\).
  2. Find the probability generating function \(\mathrm { G } _ { Y } ( \mathrm { t } )\) of \(Y\).
    The random variable \(Z\) is the total number of heads obtained by Keira and Hassan.
  3. Find the probability generating function of \(Z\), expressing your answer as a polynomial.
  4. Use the probability generating function of \(Z\) to find \(\mathrm { E } ( Z )\).
  5. Use the probability generating function of \(Z\) to find the most probable value of \(Z\).

Question 5(a):
AnswerMarks Guidance
AnswerMarks Guidance
\(G_X(t) = \frac{1}{4} + \frac{1}{2}t + \frac{1}{4}t^2\)B1 Accept \((0.5 + 0.5t)^2\)
Question 5(b):
AnswerMarks Guidance
AnswerMarks Guidance
\(P(0H) = \frac{12}{36}\), \(P(1H) = \frac{16}{36}\), \(P(2H) = \frac{7}{36}\), \(P(3H) = \frac{1}{36}\)M1 A1 Attempt at probs, at least 2 correct; All correct
\(G_Y(t) = \frac{12}{36} + \frac{16}{36}t + \frac{7}{36}t^2 + \frac{1}{36}t^3\)B1 FT FT their probabilities, must be cubic with 4 non-zero terms
Question 5(c):
AnswerMarks Guidance
AnswerMarks Guidance
\(G_Z(t) = \left(\frac{1}{4} + \frac{1}{2}t + \frac{1}{4}t^2\right)\left(\frac{12}{36} + \frac{16}{36}t + \frac{7}{36}t^2 + \frac{1}{36}t^3\right)\)M1 Attempt to multiply their two PGFs
\(= \frac{1}{144}\left(12 + 40t + 51t^2 + 31t^3 + 9t^4 + t^5\right)\)M1 A1 Obtain quintic expression and collect terms
Question 5(d):
AnswerMarks Guidance
AnswerMarks Guidance
\(G'_Z(t) = \frac{1}{144}\left(40 + 102t + 93t^2 + 36t^3 + 5t^4\right)\)M1 Differentiate
\(E(Z) = G'_Z(1) = \frac{23}{12}\ (= 1.92)\)A1
Question 5(e):
AnswerMarks Guidance
AnswerMarks Guidance
2B1 FT FT power of term with largest coefficient in their \(G_Z(t)\)
## Question 5(a):

| Answer | Marks | Guidance |
|--------|-------|----------|
| $G_X(t) = \frac{1}{4} + \frac{1}{2}t + \frac{1}{4}t^2$ | B1 | Accept $(0.5 + 0.5t)^2$ |

## Question 5(b):

| Answer | Marks | Guidance |
|--------|-------|----------|
| $P(0H) = \frac{12}{36}$, $P(1H) = \frac{16}{36}$, $P(2H) = \frac{7}{36}$, $P(3H) = \frac{1}{36}$ | M1 A1 | Attempt at probs, at least 2 correct; All correct |
| $G_Y(t) = \frac{12}{36} + \frac{16}{36}t + \frac{7}{36}t^2 + \frac{1}{36}t^3$ | B1 FT | FT their probabilities, must be cubic with 4 non-zero terms |

## Question 5(c):

| Answer | Marks | Guidance |
|--------|-------|----------|
| $G_Z(t) = \left(\frac{1}{4} + \frac{1}{2}t + \frac{1}{4}t^2\right)\left(\frac{12}{36} + \frac{16}{36}t + \frac{7}{36}t^2 + \frac{1}{36}t^3\right)$ | M1 | Attempt to multiply their two PGFs |
| $= \frac{1}{144}\left(12 + 40t + 51t^2 + 31t^3 + 9t^4 + t^5\right)$ | M1 A1 | Obtain quintic expression and collect terms |

## Question 5(d):

| Answer | Marks | Guidance |
|--------|-------|----------|
| $G'_Z(t) = \frac{1}{144}\left(40 + 102t + 93t^2 + 36t^3 + 5t^4\right)$ | M1 | Differentiate |
| $E(Z) = G'_Z(1) = \frac{23}{12}\ (= 1.92)$ | A1 | |

## Question 5(e):

| Answer | Marks | Guidance |
|--------|-------|----------|
| 2 | B1 FT | FT power of term with largest coefficient in their $G_Z(t)$ |
5 Keira has two unbiased coins. She tosses both coins. The number of heads obtained by Keira is denoted by $X$.
\begin{enumerate}[label=(\alph*)]
\item Find the probability generating function $\mathrm { G } _ { \mathrm { X } } ( \mathrm { t } )$ of $X$.\\

Hassan has three coins, two of which are biased so that the probability of obtaining a head when the coin is tossed is $\frac { 1 } { 3 }$. The corresponding probability for the third coin is $\frac { 1 } { 4 }$. The number of heads obtained by Hassan when he tosses these three coins is denoted by $Y$.
\item Find the probability generating function $\mathrm { G } _ { Y } ( \mathrm { t } )$ of $Y$.\\

The random variable $Z$ is the total number of heads obtained by Keira and Hassan.
\item Find the probability generating function of $Z$, expressing your answer as a polynomial.
\item Use the probability generating function of $Z$ to find $\mathrm { E } ( Z )$.
\item Use the probability generating function of $Z$ to find the most probable value of $Z$.
\end{enumerate}

\hfill \mbox{\textit{CAIE Further Paper 4 2020 Q5 [10]}}