| Exam Board | OCR MEI |
|---|---|
| Module | Further Statistics Major (Further Statistics Major) |
| Year | 2023 |
| Session | June |
| Marks | 10 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Binomial Distribution |
| Type | Direct binomial probability calculation |
| Difficulty | Moderate -0.3 This is a straightforward application of standard binomial and geometric probability formulas with clear setup. Parts (a)-(d) are direct calculations requiring only formula substitution, while part (e) involves solving a simple quadratic equation. The question is slightly easier than average because it requires no conceptual insight—just routine application of well-practiced techniques from the syllabus. |
| Spec | 5.02f Geometric distribution: conditions5.02g Geometric probabilities: P(X=r) = p(1-p)^(r-1) |
| Answer | Marks | Guidance |
|---|---|---|
| 3 | (a) | P(none in 3 serves) = 0.453 = 0.0911 |
| [1] | 1.1 | (0.091125…) |
| 3 | (b) | B(20, 0.55) |
| P(Ten or more) = 1 – 0.2493 = 0.7507 | M1 |
| Answer | Marks |
|---|---|
| [2] | 3.3 |
| 1.1 | s.o.i. |
| Answer | Marks | Guidance |
|---|---|---|
| 3 | (c) | P(fifth serve ) = 0.454 × 0.55 |
| = 0.0226 | M1 |
| Answer | Marks | Guidance |
|---|---|---|
| [2] | 3.3 | |
| 1.1 | (0.022553…) | |
| 3 | (d) | 4 successes in first nine so B(9, 0.55) then × 0.55 |
| P(fifth on tenth serve) = 0.2128 × 0.55 = 0.117 | M1 |
| Answer | Marks | Guidance |
|---|---|---|
| [2] | 3.1b | |
| 1.1 | BC (0.117016…) | |
| 3 | (e) | 𝑝(1−𝑝) = 0.2496 |
| Answer | Marks |
|---|---|
| 𝑝 = 0.48 or 𝑝 = 0.52, so answer 𝑝 = 0.48 | M1 |
| Answer | Marks |
|---|---|
| [3] | 3.1a |
| Answer | Marks |
|---|---|
| 2.2a | For expressing in the form 𝑎𝑥2+𝑏𝑥+𝑐 = 0 |
Question 3:
3 | (a) | P(none in 3 serves) = 0.453 = 0.0911 | B1
[1] | 1.1 | (0.091125…)
3 | (b) | B(20, 0.55)
P(Ten or more) = 1 – 0.2493 = 0.7507 | M1
A1
[2] | 3.3
1.1 | s.o.i.
BC
3 | (c) | P(fifth serve ) = 0.454 × 0.55
= 0.0226 | M1
A1
[2] | 3.3
1.1 | (0.022553…)
3 | (d) | 4 successes in first nine so B(9, 0.55) then × 0.55
P(fifth on tenth serve) = 0.2128 × 0.55 = 0.117 | M1
A1
[2] | 3.1b
1.1 | BC (0.117016…)
3 | (e) | 𝑝(1−𝑝) = 0.2496
𝑝2−𝑝+0.2496 = 0
𝑝 = 0.48 or 𝑝 = 0.52, so answer 𝑝 = 0.48 | M1
M1
A1
[3] | 3.1a
1.1
2.2a | For expressing in the form 𝑎𝑥2+𝑏𝑥+𝑐 = 03 A tennis player is practising her serve. Each time she serves, she has a $55 \%$ chance of being successful (getting the serve in the required area without hitting the net). You should assume that whether she is successful on any serve is independent of whether she is successful on any other serve.
\begin{enumerate}[label=(\alph*)]
\item Find the probability that the player is not successful in any of her first three serves.
\item Determine the probability that the player is successful at least 10 times in her first 20 serves.
\item Determine the probability that the player is successful for the first time on her fifth serve.
\item Determine the probability that the player is successful for the fifth time on her tenth serve.
Another player is also practising his serve. Each time he serves, he has a probability $p$ of being successful. You should assume that whether he is successful on any serve is independent of whether he is successful on any other serve.
The probability that he is successful for the first time on his second serve is 0.2496 and the probability that he is successful on both of his first two serves is less than 0.25 .
\item Determine the value of $p$.
\end{enumerate}
\hfill \mbox{\textit{OCR MEI Further Statistics Major 2023 Q3 [10]}}