| Exam Board | OCR MEI |
|---|---|
| Module | Further Statistics Major (Further Statistics Major) |
| Year | 2020 |
| Session | November |
| Marks | 9 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Discrete Probability Distributions |
| Type | Verify probability from combinatorial selection |
| Difficulty | Moderate -0.3 This is a straightforward hypergeometric distribution question requiring standard calculations. Part (a) is a simple verification using combinations C(4,4)×C(6,0)/C(10,4) = 1/210. Parts (b-d) involve routine expectation and variance calculations using given probabilities with no conceptual challenges—just arithmetic with the provided distribution table. |
| Spec | 5.01a Permutations and combinations: evaluate probabilities5.02b Expectation and variance: discrete random variables5.02c Linear coding: effects on mean and variance |
| \(r\) | 0 | 1 | 2 | 3 | 4 |
| \(\mathrm { P } ( X = r )\) | \(\frac { 1 } { 14 }\) | \(\frac { 8 } { 21 }\) | \(\frac { 3 } { 7 }\) | \(\frac { 4 } { 35 }\) | \(\frac { 1 } { 210 }\) |
| Answer | Marks | Guidance |
|---|---|---|
| 1 | (a) | P(all 4) = = |
| 4 3 2 1 1 | M1 |
| Answer | Marks | Guidance |
|---|---|---|
| [2] | 1.1 | |
| 1.1 | AG | |
| 1 | (b) | E(X) = 1.61 0×9×8×7 210 |
| Var(X) = 0.64 | B1 |
| Answer | Marks |
|---|---|
| [2] | 1.1a |
| 1.1 | BC |
| Answer | Marks | Guidance |
|---|---|---|
| 1 | (c) | Loss L = 1.00 – 0.4X |
| Answer | Marks |
|---|---|
| SD(L) = 0.4*0.8 = 0.32 (32 pence) | B1 |
| Answer | Marks |
|---|---|
| [3] | 3.3 |
| Answer | Marks | Guidance |
|---|---|---|
| 1 | (d) | Expected amount = 1.00 – (0.25E(X) + 10 ) |
| Answer | Marks |
|---|---|
| 12.4 pence loss 0×210 | M1 |
| Answer | Marks |
|---|---|
| [2] | 1.1 |
| 1.1 | Allow M1 if use original distribution but miss out 1×1/210 term |
Question 1:
1 | (a) | P(all 4) = =
4 3 2 1 1 | M1
A1
[2] | 1.1
1.1 | AG
1 | (b) | E(X) = 1.61 0×9×8×7 210
Var(X) = 0.64 | B1
B1
[2] | 1.1a
1.1 | BC
BC
1 | (c) | Loss L = 1.00 – 0.4X
E(L) = 0.36 (36 pence)
SD(L) = 0.4*0.8 = 0.32 (32 pence) | B1
M1
A1
[3] | 3.3
1.1
1.1
1 | (d) | Expected amount = 1.00 – (0.25E(X) + 10 )
1
12.4 pence loss 0×210 | M1
A1
[2] | 1.1
1.1 | Allow M1 if use original distribution but miss out 1×1/210 term
(12.3809…) Allow answer rounded to 12 (pence) provided there
is working1 In a game at a fair, players choose 4 countries from a list of 10 countries. The names of all 10 countries are then put in a box and the player selects 4 of them at random. The random variable $X$ represents the number of countries that match those which the player originally chose.
\begin{enumerate}[label=(\alph*)]
\item Show that the probability that a randomly selected player matches all 4 countries is $\frac { 1 } { 210 }$.
Table 1 shows the probability distribution of $X$.
\begin{table}[h]
\begin{center}
\begin{tabular}{ | c | c | c | c | c | c | }
\hline
$r$ & 0 & 1 & 2 & 3 & 4 \\
\hline
$\mathrm { P } ( X = r )$ & $\frac { 1 } { 14 }$ & $\frac { 8 } { 21 }$ & $\frac { 3 } { 7 }$ & $\frac { 4 } { 35 }$ & $\frac { 1 } { 210 }$ \\
\hline
\end{tabular}
\captionsetup{labelformat=empty}
\caption{Table 1}
\end{center}
\end{table}
\item Find each of the following.
\begin{itemize}
\item $\mathrm { E } ( X )$
\item $\operatorname { Var } ( X )$
\item A player has to pay $\pounds 1$ to play the game. The player gets 40 pence back for every country which is matched.
\end{itemize}
Find the mean and standard deviation of the player's loss per game.
\item In order to try to attract more customers, the rules will be changed as follows.
The game will still cost $\pounds 1$ to play. The player will get 25 pence back for every country which is matched, plus an additional bonus of $\pounds 100$ if all four countries are matched.
Find the player's mean gain or loss per game with these new rules.
\end{enumerate}
\hfill \mbox{\textit{OCR MEI Further Statistics Major 2020 Q1 [9]}}