| Exam Board | CAIE |
|---|---|
| Module | FP2 (Further Pure Mathematics 2) |
| Year | 2018 |
| Session | June |
| Marks | 11 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Geometric Distribution |
| Type | Determine p from given mean or variance |
| Difficulty | Standard +0.3 This is a straightforward application of geometric distribution formulas. Part (i) requires knowing Var(X) = (1-p)/p² and solving a quadratic equation. Parts (ii)-(iv) are routine probability calculations using the geometric distribution formula. The algebra is simple and all steps follow standard procedures with no novel insight required. |
| 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 |
|---|---|---|
| \((1-p)/p^2 = 4/9\), leading to \(4p^2 + 9p - 9 = 0\) | M1 A1 (AG) | Find given equation for \(p\) using \(\text{Var}(X) = (1-p)/p^2\) |
| \((4p-3)(p+3) = 0,\quad p = \tfrac{3}{4}\) | M1 A1 | Solve quadratic for \(p\) (A0 if \(p = -3\) not [implicitly] rejected) |
| Answer | Marks | Guidance |
|---|---|---|
| \(P(X = 3) = (1-p)^2 p = (\tfrac{1}{4})^2 \cdot \tfrac{3}{4} = 3/64\) or \(0.0469\) | B1 | Find \(P(X = 3)\) |
| Answer | Marks | Guidance |
|---|---|---|
| \(P(X \geqslant 4) = (1-p)^3 = (\tfrac{1}{4})^3\) or \(1 - (1 + \tfrac{1}{4} + (\tfrac{1}{4})^2)\tfrac{3}{4} = 1/64\) or \(0.0156\) | M1 A1 | Find \(P(X \geqslant 4)\) |
| Answer | Marks | Guidance |
|---|---|---|
| \(1 - (1-p)^N > 0.999\) | M1 | Formulate condition for \(N\) (\(1-(1-p)^{N-1}\) is M0) |
| \(0.001 > (\tfrac{1}{4})^N\) | A1 | (\(<\) or \(=\) can earn M1 M1 only, max 2/4) |
| \(N > \log 0.001 / \log 0.25\) | M1 | Rearrange and take logs (any base) to give bound |
| \(N > 4.98,\ N_{\min} = 5\) | A1 | Find \(N_{\min}\) |
## Question 9(i):
| $(1-p)/p^2 = 4/9$, leading to $4p^2 + 9p - 9 = 0$ | M1 A1 (AG) | Find given equation for $p$ using $\text{Var}(X) = (1-p)/p^2$ |
|---|---|---|
| $(4p-3)(p+3) = 0,\quad p = \tfrac{3}{4}$ | M1 A1 | Solve quadratic for $p$ (A0 if $p = -3$ not [implicitly] rejected) |
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## Question 9(ii):
| $P(X = 3) = (1-p)^2 p = (\tfrac{1}{4})^2 \cdot \tfrac{3}{4} = 3/64$ or $0.0469$ | B1 | Find $P(X = 3)$ |
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## Question 9(iii):
| $P(X \geqslant 4) = (1-p)^3 = (\tfrac{1}{4})^3$ or $1 - (1 + \tfrac{1}{4} + (\tfrac{1}{4})^2)\tfrac{3}{4} = 1/64$ or $0.0156$ | M1 A1 | Find $P(X \geqslant 4)$ |
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## Question 9(iv):
| $1 - (1-p)^N > 0.999$ | M1 | Formulate condition for $N$ ($1-(1-p)^{N-1}$ is M0) |
|---|---|---|
| $0.001 > (\tfrac{1}{4})^N$ | A1 | ($<$ or $=$ can earn M1 M1 only, max 2/4) |
| $N > \log 0.001 / \log 0.25$ | M1 | Rearrange and take logs (any base) to give bound |
| $N > 4.98,\ N_{\min} = 5$ | A1 | Find $N_{\min}$ |
---9 At a ski resort, the probability of snow on any particular day is constant and equal to $p$. The skiing season begins on 1 November. The random variable $X$ denotes the day of the skiing season on which the first snowfall occurs. (For example, if the first snowfall is on 5 November, then $X = 5$.) The variance of $X$ is $\frac { 4 } { 9 }$.\\
(i) Show that $4 p ^ { 2 } + 9 p - 9 = 0$ and hence find the value of $p$.\\
(ii) Find the probability that the first snowfall will be on 3 November.\\
(iii) Find the probability that the first snowfall will not be before 4 November.\\
(iv) Find the least integer $N$ so that the probability of the first snowfall being on or before the $N$ th day of November is more than 0.999 .\\
\hfill \mbox{\textit{CAIE FP2 2018 Q9 [11]}}