| Exam Board | OCR MEI |
|---|---|
| Module | Further Statistics Major (Further Statistics Major) |
| Year | 2022 |
| Session | June |
| Marks | 8 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Geometric Distribution |
| Type | First success on specific trial |
| Difficulty | Standard +0.8 This is a multi-part Further Maths statistics question requiring geometric distribution for part (a), binomial distribution combined with geometric distribution for part (b), and solving a quadratic equation from probability theory in part (c). Part (b) requires recognizing that P(X>3) needs to be calculated first, then used in a binomial context, which demands careful multi-step reasoning. Part (c) involves algebraic manipulation of probability expressions. While the individual techniques are standard for Further Maths, the combination and the need to link different distributions makes this moderately challenging. |
| Spec | 5.02f 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 |
|---|---|---|
| 7 | (a) | P(X = 5) = 0.74 × 0.3 |
| =0.07203 | M1 |
| Answer | Marks | Guidance |
|---|---|---|
| [2] | 3.3 | |
| 1.1 | Allow 0.072 | |
| 7 | (b) | P(X > 3) = 0.73 = 0.343 |
| Answer | Marks |
|---|---|
| P(At least 4 out of 6) = 1 – 0.8900 = 0.1100 | B1 |
| Answer | Marks |
|---|---|
| [3] | 1.1 |
| Answer | Marks |
|---|---|
| 1.1 | With their 0.343 but not 0.3 or 0.7. Must be stated if not 0.343 |
| Answer | Marks | Guidance |
|---|---|---|
| 7 | (c) | (1− p)p= 28 |
| Answer | Marks |
|---|---|
| 4 7 | M1 |
| Answer | Marks |
|---|---|
| [3] | 3.1a |
| Answer | Marks |
|---|---|
| 1.1 | Allow sign errors |
Question 7:
7 | (a) | P(X = 5) = 0.74 × 0.3
=0.07203 | M1
A1
[2] | 3.3
1.1 | Allow 0.072
7 | (b) | P(X > 3) = 0.73 = 0.343
Use of binomial (6, their 0.343)
P(At least 4 out of 6) = 1 – 0.8900 = 0.1100 | B1
M1
A1
[3] | 1.1
3.3
1.1 | With their 0.343 but not 0.3 or 0.7. Must be stated if not 0.343
Allow 0.110
7 | (c) | (1− p)p= 28
121
121p2 – 121p + 28 = 0
p = or p =
⇒
4 7 | M1
M1
A1
[3] | 3.1a
2.1
1.1 | Allow sign errors
BC Allow 0.363… and 0.636…7 Amir is trying to thread a needle. On each attempt the probability that he is successful is 0.3 , independently of any other attempt. The random variable $X$ represents the number of attempts that he takes to thread the needle.
\begin{enumerate}[label=(\alph*)]
\item Find $\mathrm { P } ( X = 5 )$.
\item During the course of a day, Amir has to thread 6 needles.
Determine the probability that it takes him more than 3 attempts to be successful for at least 4 of the 6 needles.
\item Amaya is also trying to thread a needle. On each attempt the probability that she is successful is $p$, independently of any other attempt. The probability that Amaya takes 2 attempts to thread a particular needle is $\frac { 28 } { 121 }$.
Determine the possible values of $p$.
\end{enumerate}
\hfill \mbox{\textit{OCR MEI Further Statistics Major 2022 Q7 [8]}}