| Exam Board | Edexcel |
|---|---|
| Module | AS Paper 2 (AS Paper 2) |
| Year | 2018 |
| Session | June |
| Marks | 8 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Discrete Probability Distributions |
| Type | Construct probability distribution from scenario |
| Difficulty | Moderate -0.3 This is a straightforward probability distribution question requiring basic probability axioms and binomial distribution application. Part (a) uses simple algebra with the constraint that probabilities sum to 1, part (b) is standard binomial calculation, and part (c) requires checking which X values satisfy an inequality. All techniques are routine for AS-level with no novel problem-solving required, making it slightly easier than average. |
| Spec | 2.04a Discrete probability distributions2.04b Binomial distribution: as model B(n,p)2.04c Calculate binomial probabilities |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(P(X=4) = P(X=2)\) so \(P(X=4) = 0.35\) | M1 | For using given information to obtain \(P(X=4)\); award for statement \(P(X=4)=P(X=2)\) or writing \(P(X=4)=0.35\) |
| \(P(X=1) = P(X=3)\) and \(P(X=1)+P(X=3) = 1-0.7\) | ||
| Table: \(x\): 1, 2, 3, 4 with \(P(X=x)\): 0.15, 0.35, 0.15, [0.35] | A1 | For fully correct distribution in any form; condone missing \(P(X=2)\) as given in QP |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| Let \(A\) = number of spins landing on 4, \(A \sim B(60, \text{"0.35"})\) | B1ft | For selecting suitable model, sight of \(B(60, \text{their } 0.35)\) |
| \(P(A>30) = 1 - P(A \leq 30)\) | M1 | For using model and interpreting "more than half"; need \(1-P(A\leq 30)\); can ignore incorrect LHS such as \(P(A\geq 30)\) |
| \(= 1 - 0.99411... = \textbf{awrt } 0.00589\) | A1 | For awrt 0.00589 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(Y-X \leq 4 \Rightarrow \frac{12}{X} - X \leq 4\) or \(12-X^2 \leq 4X\) (since \(X>0\)) | M1 | For translating prob. problem into a correct mathematical inequality; just an inequality in 1 variable |
| \(0 \leq X^2+4X-12 \Rightarrow 0 \leq (X+6)(X-2)\) so \(X \geq 2\) | M1 | For solving inequality leading to range of values; may be quadratic or cubic but must lead to set of values of \(X\) or \(Y-X\) |
| \(P(Y-X \leq 4) = P(X \geq 2) = 0.35+0.15+0.35 = \textbf{0.85}\) | A1 | For interpreting inequality and solving; ALT table: \(Y-X\): 11, 4, 1, \(-1\) |
# Question 5:
## Part (a):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $P(X=4) = P(X=2)$ so $P(X=4) = 0.35$ | M1 | For using given information to obtain $P(X=4)$; award for statement $P(X=4)=P(X=2)$ or writing $P(X=4)=0.35$ |
| $P(X=1) = P(X=3)$ and $P(X=1)+P(X=3) = 1-0.7$ | | |
| Table: $x$: 1, 2, 3, 4 with $P(X=x)$: 0.15, 0.35, 0.15, [0.35] | A1 | For fully correct distribution in any form; condone missing $P(X=2)$ as given in QP |
## Part (b):
| Answer/Working | Marks | Guidance |
|---|---|---|
| Let $A$ = number of spins landing on 4, $A \sim B(60, \text{"0.35"})$ | B1ft | For selecting suitable model, sight of $B(60, \text{their } 0.35)$ |
| $P(A>30) = 1 - P(A \leq 30)$ | M1 | For using model and interpreting "more than half"; need $1-P(A\leq 30)$; can ignore incorrect LHS such as $P(A\geq 30)$ |
| $= 1 - 0.99411... = \textbf{awrt } 0.00589$ | A1 | For awrt 0.00589 |
## Part (c):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $Y-X \leq 4 \Rightarrow \frac{12}{X} - X \leq 4$ or $12-X^2 \leq 4X$ (since $X>0$) | M1 | For translating prob. problem into a correct mathematical inequality; just an inequality in 1 variable |
| $0 \leq X^2+4X-12 \Rightarrow 0 \leq (X+6)(X-2)$ so $X \geq 2$ | M1 | For solving inequality leading to range of values; may be quadratic or cubic but must lead to set of values of $X$ or $Y-X$ |
| $P(Y-X \leq 4) = P(X \geq 2) = 0.35+0.15+0.35 = \textbf{0.85}$ | A1 | For interpreting inequality and solving; ALT table: $Y-X$: 11, 4, 1, $-1$ |
---5. A biased spinner can only land on one of the numbers $1,2,3$ or 4 . The random variable $X$ represents the number that the spinner lands on after a single spin and $\mathrm { P } ( X = r ) = \mathrm { P } ( X = r + 2 )$ for $r = 1,2$
Given that $\mathrm { P } ( X = 2 ) = 0.35$
\begin{enumerate}[label=(\alph*)]
\item find the complete probability distribution of $X$.
Ambroh spins the spinner 60 times.
\item Find the probability that more than half of the spins land on the number 4 Give your answer to 3 significant figures.
The random variable $Y = \frac { 12 } { X }$
\item Find $\mathrm { P } ( Y - X \leqslant 4 )$
\end{enumerate}
\hfill \mbox{\textit{Edexcel AS Paper 2 2018 Q5 [8]}}