| Exam Board | CAIE |
|---|---|
| Module | FP2 (Further Pure Mathematics 2) |
| Year | 2014 |
| Session | June |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Binomial Distribution |
| Type | Direct binomial probability calculation |
| Difficulty | Easy -1.8 The question text is severely corrupted and unreadable, making it impossible to assess the actual mathematical content. However, based on the metadata indicating it's a 'Direct binomial probability calculation' from FP2, this would typically involve straightforward application of binomial probability formulas P(X=r) = nCr × p^r × (1-p)^(n-r), which is routine A-level material requiring only formula recall and arithmetic. |
| Spec | 5.02b Expectation and variance: discrete random variables5.02c Linear coding: effects on mean and variance |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(P(N=5) = \left(\frac{3}{4}\right)^4 \times \frac{1}{4} = \frac{81}{1024}\) or \(0.079[1]\) | B1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(P(N>8) = \left(\frac{3}{4}\right)^8 = \frac{6561}{65536}\) or \(0.1[00]\) | M1 A1 | |
| \(P_J = P(N \leq 6) = 1 - \left(\frac{3}{4}\right)^6\) or \(\{1 + \frac{3}{4} + \left(\frac{3}{4}\right)^2 + \left(\frac{3}{4}\right)^3 + \left(\frac{3}{4}\right)^4 + \left(\frac{3}{4}\right)^5\}\frac{1}{4} = \frac{3367}{4096}\) or \(0.822\) | M1 A1 | Find prob. \(P_J\) that James qualifies |
| \(P_C = 1 - \left(\frac{2}{3}\right)^6 = 0.9122\) | B1 | Find prob. \(P_C\) that Colin qualifies |
| \(P_J(1-P_C) + P_C(1-P_J) = \frac{3367}{4096} \cdot \frac{64}{729} + \frac{665}{729} \cdot \frac{729}{4096}\) | Find prob. exactly one qualifies | |
| \(= 0.0722 + 0.1624 = 0.235\) (allow \(0.234\)) | M1 A1 |
## Question 7:
**(i)**
| Answer/Working | Marks | Guidance |
|---|---|---|
| $P(N=5) = \left(\frac{3}{4}\right)^4 \times \frac{1}{4} = \frac{81}{1024}$ or $0.079[1]$ | B1 | |
**(ii)**
| Answer/Working | Marks | Guidance |
|---|---|---|
| $P(N>8) = \left(\frac{3}{4}\right)^8 = \frac{6561}{65536}$ or $0.1[00]$ | M1 A1 | |
| $P_J = P(N \leq 6) = 1 - \left(\frac{3}{4}\right)^6$ or $\{1 + \frac{3}{4} + \left(\frac{3}{4}\right)^2 + \left(\frac{3}{4}\right)^3 + \left(\frac{3}{4}\right)^4 + \left(\frac{3}{4}\right)^5\}\frac{1}{4} = \frac{3367}{4096}$ or $0.822$ | M1 A1 | Find prob. $P_J$ that James qualifies |
| $P_C = 1 - \left(\frac{2}{3}\right)^6 = 0.9122$ | B1 | Find prob. $P_C$ that Colin qualifies |
| $P_J(1-P_C) + P_C(1-P_J) = \frac{3367}{4096} \cdot \frac{64}{729} + \frac{665}{729} \cdot \frac{729}{4096}$ | | Find prob. exactly one qualifies |
| $= 0.0722 + 0.1624 = 0.235$ (allow $0.234$) | M1 A1 | |
---7 James throws a discus repeatedly in an attempt to achieve a successful throw. A throw is counted as successful if the distance achieved is over 40 metres. For each throw, the probability that James is successful is $\frac { 1 } { 4 }$, independently of all other throws. Find the probability that James takes\\
(i) exactly 5 throws to achieve the first successful throw,\\
(ii) more than 8 throws to achieve the first successful throw.
In order to qualify for a competition, a discus-thrower must throw over 40 metres within at most six attempts. When a successful throw is achieved, no further throws are taken. Find the probability that James qualifies for the competition.
Colin is another discus-thrower. For each throw, the probability that he will achieve a throw over 40 metres is $\frac { 1 } { 3 }$, independently of all other throws. Find the probability that exactly one of James and Colin qualifies for the competition.
\hfill \mbox{\textit{CAIE FP2 2014 Q7}}