| Exam Board | OCR MEI |
|---|---|
| Module | Further Statistics A AS (Further Statistics A AS) |
| Year | 2021 |
| Session | November |
| Marks | 10 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Geometric Distribution |
| Type | First success on specific trial |
| Difficulty | Moderate -0.3 This is a straightforward application of geometric distribution formulas with standard calculations. Parts (a)-(b) require direct formula recall, (c) uses negative binomial, and (d) involves routine inequality solving with logarithms. While multi-part, each component is a textbook exercise requiring no novel insight, making it slightly easier than average for A-level Further Maths statistics. |
| Spec | 2.04a Discrete probability distributions2.04b Binomial distribution: as model B(n,p)2.04c Calculate binomial probabilities5.02f Geometric distribution: conditions5.02g Geometric probabilities: P(X=r) = p(1-p)^(r-1)5.02h Geometric: mean 1/p and variance (1-p)/p^2 |
| Answer | Marks | Guidance |
|---|---|---|
| 2 | (a) | Geometric(0.7) |
| Answer | Marks |
|---|---|
| = 0.0189 | M1 |
| Answer | Marks |
|---|---|
| [2] | 3.3 |
| Answer | Marks | Guidance |
|---|---|---|
| 2 | (b) | E(X) = 1.43 |
| Var(X) = 0.612 | B1 |
| Answer | Marks |
|---|---|
| [2] | 1.1a |
| Answer | Marks | Guidance |
|---|---|---|
| 2 | (c) | P(second time fourth attempt)=3×(0.3)2×(0.7)2 |
| = 0.1323 | M1 |
| Answer | Marks | Guidance |
|---|---|---|
| [2] | 3.1b | |
| 1.1 | M1 for n(1-p)2p2 n ≠ 1 | |
| 2 | (d) | (i) |
| Least n = 7 | M1 |
| Answer | Marks |
|---|---|
| [2] | 3.1b |
| Answer | Marks | Guidance |
|---|---|---|
| 2 | (d) | (ii) |
| Least n = 6 | M1 |
| Answer | Marks |
|---|---|
| [2] | 3.1b |
Question 2:
2 | (a) | Geometric(0.7)
P(Fourth attempt)=(0.3)3×0.7
= 0.0189 | M1
A1
[2] | 3.3
1.1
2 | (b) | E(X) = 1.43
Var(X) = 0.612 | B1
B1
[2] | 1.1a
1.1
2 | (c) | P(second time fourth attempt)=3×(0.3)2×(0.7)2
= 0.1323 | M1
A1
[2] | 3.1b
1.1 | M1 for n(1-p)2p2 n ≠ 1
2 | (d) | (i) | (0.3)n−1×0.7<0.001
Least n = 7 | M1
A1
[2] | 3.1b
1.1
2 | (d) | (ii) | 1−(0.3)n ≥0.999 (⇒(0.3)n ≤0.001)
Least n = 6 | M1
A1
[2] | 3.1b
1.12 A football player is practising taking penalties. On each attempt the player has a $70 \%$ chance of scoring a goal. The random variable $X$ represents the number of attempts that it takes for the player to score a goal.
\begin{enumerate}[label=(\alph*)]
\item Determine $\mathrm { P } ( X = 4 )$.
\item Find each of the following.
\begin{itemize}
\item $\mathrm { E } ( X )$
\item $\operatorname { Var } ( X )$
\item Determine the probability that the player needs exactly 4 attempts to score 2 goals.
\item The player has $n$ attempts to score a goal.
\begin{enumerate}[label=(\roman*)]
\item Determine the least value of $n$ for which the probability that the player first scores a goal on the $n$th attempt is less than 0.001 .
\item Determine the least value of $n$ for which the probability that the player scores at least one goal in $n$ attempts is at least 0.999.
\end{itemize}
\end{enumerate}\end{enumerate}
\hfill \mbox{\textit{OCR MEI Further Statistics A AS 2021 Q2 [10]}}