| Exam Board | CAIE |
|---|---|
| Module | FP2 (Further Pure Mathematics 2) |
| Year | 2018 |
| Session | June |
| Marks | 7 |
| 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) and (iii) use the standard geometric probability formula P(X=k) = (1-p)^(k-1)p. All steps are routine with no conceptual challenges beyond recalling the correct formulas, making it slightly easier than average. |
| Spec | 5.02g Geometric probabilities: P(X=r) = p(1-p)^(r-1) |
| Answer | Marks | Guidance |
|---|---|---|
| \((1-p)/p^2 = 3.75\), \(15p^2 + 4p - 4 = 0\) | M1 A1 AG | Find given eqn. for \(p\) using \(\text{Var}(X) = (1-p)/p^2\) |
| \((5p-2)(3p+2) = 0\), \(p = 2/5\) or \(0.4\) | M1 A1 | Solve quadratic for \(p\) (A0 if \(p = -\frac{2}{3}\) not [implicitly] rejected) |
| Answer | Marks | Guidance |
|---|---|---|
| \(P(X=5) = (1-p)^4 p = 0.6^4 \times 0.4 = 0.0518\) or \(162/3125\) | B1 | Find \(P(X=5)\) |
| Answer | Marks | Guidance |
|---|---|---|
| \(P(3 \leqslant X \leqslant 7) = (1-p)^2 - (1-p)^7\) | M1 | Find \(P(3 \leqslant X \leqslant 7)\) |
| \(= 0.6^2 - 0.6^7 = 0.36 - 0.028 = 0.332\) | A1 | M0 for \(P(X \leqslant 7) - P(X \leqslant 3)\) [= 0.188] or similar error |
| OR: \(P(3 \leqslant X \leqslant 7) = \sum_{i=3}^{7}(1-p)^{i-1}p\) | (M1) | |
| \(= (0.6^2 + 0.6^3 + 0.6^4 + 0.6^5 + 0.6^6) \times 0.4 = 0.830016 \times 0.4 = 0.332\) | (A1) |
## Question 7:
### Part 7(i):
$(1-p)/p^2 = 3.75$, $15p^2 + 4p - 4 = 0$ | M1 A1 AG | Find given eqn. for $p$ using $\text{Var}(X) = (1-p)/p^2$
$(5p-2)(3p+2) = 0$, $p = 2/5$ or $0.4$ | M1 A1 | Solve quadratic for $p$ (A0 if $p = -\frac{2}{3}$ not [implicitly] rejected)
### Part 7(ii):
$P(X=5) = (1-p)^4 p = 0.6^4 \times 0.4 = 0.0518$ or $162/3125$ | B1 | Find $P(X=5)$
### Part 7(iii):
$P(3 \leqslant X \leqslant 7) = (1-p)^2 - (1-p)^7$ | M1 | Find $P(3 \leqslant X \leqslant 7)$
$= 0.6^2 - 0.6^7 = 0.36 - 0.028 = 0.332$ | A1 | M0 for $P(X \leqslant 7) - P(X \leqslant 3)$ [= 0.188] or similar error
OR: $P(3 \leqslant X \leqslant 7) = \sum_{i=3}^{7}(1-p)^{i-1}p$ | (M1) |
$= (0.6^2 + 0.6^3 + 0.6^4 + 0.6^5 + 0.6^6) \times 0.4 = 0.830016 \times 0.4 = 0.332$ | (A1) |7 The probability that a driver passes an advanced driving test has a fixed value $p$ for each attempt. A driver keeps taking the test until he passes. The random variable $X$ denotes the number of attempts required for the driver to pass. The variance of $X$ is 3.75 .\\
(i) Show that $15 p ^ { 2 } + 4 p - 4 = 0$ and hence find the value of $p$.\\
(ii) Find $\mathrm { P } ( X = 5 )$.\\
(iii) Find $\mathrm { P } ( 3 \leqslant X \leqslant 7 )$.\\
\hfill \mbox{\textit{CAIE FP2 2018 Q7 [7]}}