| Exam Board | Edexcel |
|---|---|
| Module | S1 (Statistics 1) |
| Year | 2003 |
| Session | November |
| Marks | 18 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Discrete Probability Distributions |
| Type | Construct probability distribution from scenario |
| Difficulty | Standard +0.3 This is a straightforward S1 probability distribution question requiring construction of a simple discrete distribution from a scenario with independent events. Part (a) is basic probability multiplication (0.6³), part (b) requires systematic enumeration of outcomes (0, 10, 20, 30 points), and parts (c-d) use standard formulas. While multi-part, each step follows directly from the previous with no novel insight required—slightly easier than average A-level. |
| Spec | 5.02a Discrete probability distributions: general5.02b Expectation and variance: discrete random variables |
| Answer | Marks | Guidance |
|---|---|---|
| Part (a) | ||
| \(P(\text{scores 30 points}) = P(\text{hit, hit, hit.}) = 0.6^3 = 0.216\) | M1 | |
| \(0.6^3\) | A1 | (2 marks) |
| Part (b) | ||
| \(x\) | 0 | 10 |
| \(P(X=x)\) | 0.4 | 0.24 |
| \(0.6 \times 0.4\) | \(0.6^2 \times 0.4\) | |
| \(x = 0, 10, 20, 30\) | One correct \(P(X=x)\) | \(0.4; 0.24; 0.144\) |
| Part (c) | ||
| \(E[X] = (0 \times 0.4) + \ldots + (30 \times 0.216) = 11.76\) | M1 | |
| \(\sum xp[X = x]\) | ||
| \(11.8\) | A1 | |
| \(E[X^2] = [10^2 \times 0.24] + \ldots + [30^2 \times 0.216] = 276\) | B1 | |
| \(\text{Std Dev} = \sqrt{276 - 11.76^2} = 11.7346\ldots\) | M1 | |
| \(\sqrt{E[X^2] - [E[X]]^2}\) | ||
| \(11.7\) | A1 | (5 marks) |
| Part (d) | ||
| \(P(\text{Linda scores more in round 2 than in round 1})\) | ||
| \(= P[X_1 = 0 \text{ and } X_2 = 10, 20, 30] + X_2 > X_1\) | M1 | |
| \(+ P[X_1 = 0 \text{ and } X_2 = 10, 20, 30]\) | ||
| All possible | A1 | |
| \(+ P[X_1 = 20 \text{ and } X_2 = 30]\) | A1, \(\checkmark\) | |
| \(= 0.4 \times (0.24 + 0.144 + 0.216)\) | A1, \(\checkmark\) | |
| \(+ (0.24(0.144 + 0.216))\) | A1, \(\checkmark\) | |
| \(+ (0.144 \times 0.126)\) | A1, \(\checkmark\) | |
| \(= 0.357504\) | ||
| \(0.358\) | A1 | (6 marks) |
**Part (a)** | |
$P(\text{scores 30 points}) = P(\text{hit, hit, hit.}) = 0.6^3 = 0.216$ | M1 |
$0.6^3$ | A1 | (2 marks)
**Part (b)** | |
| $x$ | 0 | 10 | 20 | 30 |
| $P(X=x)$ | 0.4 | 0.24 | 0.144 | (0.216) |
| | | $0.6 \times 0.4$ | $0.6^2 \times 0.4$ | |
| | $x = 0, 10, 20, 30$ | One correct $P(X=x)$ | $0.4; 0.24; 0.144$ | A1, A1, A1 | (5 marks)
**Part (c)** | |
$E[X] = (0 \times 0.4) + \ldots + (30 \times 0.216) = 11.76$ | M1 |
$\sum xp[X = x]$ |
$11.8$ | A1 |
$E[X^2] = [10^2 \times 0.24] + \ldots + [30^2 \times 0.216] = 276$ | B1 |
$\text{Std Dev} = \sqrt{276 - 11.76^2} = 11.7346\ldots$ | M1 |
$\sqrt{E[X^2] - [E[X]]^2}$ |
$11.7$ | A1 | (5 marks)
**Part (d)** | |
$P(\text{Linda scores more in round 2 than in round 1})$ | |
$= P[X_1 = 0 \text{ and } X_2 = 10, 20, 30] + X_2 > X_1$ | M1 |
$+ P[X_1 = 0 \text{ and } X_2 = 10, 20, 30]$ | |
All possible | A1 |
$+ P[X_1 = 20 \text{ and } X_2 = 30]$ | A1, $\checkmark$ |
$= 0.4 \times (0.24 + 0.144 + 0.216)$ | A1, $\checkmark$ |
$+ (0.24(0.144 + 0.216))$ | A1, $\checkmark$ |
$+ (0.144 \times 0.126)$ | A1, $\checkmark$ |
$= 0.357504$ | |
$0.358$ | A1 | (6 marks)
---2. A fairground game involves trying to hit a moving target with a gunshot. A round consists of up to 3 shots. Ten points are scored if a player hits the target, but the round is over if the player misses. Linda has a constant probability of 0.6 of hitting the target and shots are independent of one another.
\begin{enumerate}[label=(\alph*)]
\item Find the probability that Linda scores 30 points in a round.
The random variable $X$ is the number of points Linda scores in a round.
\item Find the probability distribution of $X$.
\item Find the mean and the standard deviation of $X$.
A game consists of 2 rounds.
\item Find the probability that Linda scores more points in round 2 than in round 1.
\end{enumerate}
\hfill \mbox{\textit{Edexcel S1 2003 Q2 [18]}}