| Exam Board | WJEC |
|---|---|
| Module | Further Unit 2 (Further Unit 2) |
| Year | 2018 |
| Session | June |
| Marks | 11 |
| 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 probability distribution question requiring systematic enumeration of outcomes, basic probability calculations (conditional probability with coin toss), and standard formulas for mean, variance, and expectation. While it has multiple parts and requires careful bookkeeping of the profit values (accounting for the 50p cost), it involves only routine A-level statistics techniques with no novel problem-solving or insight required. It's slightly easier than average due to its mechanical nature. |
| Spec | 5.02a Discrete probability distributions: general5.02b Expectation and variance: discrete random variables |
| Answer | Marks | Guidance |
|---|---|---|
| \(P(X = -50) = \frac{1}{2} + \frac{1}{2} \times \frac{39}{52}\left(= \frac{91}{104} = \frac{7}{8} = 0.875\right)\) | B1 M1 | B1 for all three values. M1 for correct working for \(P(X = 50)\) or \(P(X = 450)\) or \(P(X = -50)\) |
| OR \(P(X = 50) = \frac{1}{2} \times \frac{12}{52}\left(= \frac{104}{26} = 0.115 \ldots\right)\) | Accept answers in £ or pence for this question. | |
| OR \(P(X = 450) = \frac{1}{2} \times \frac{1}{52}\left(= \frac{1}{104} = 0.00961 \ldots\right)\) | ||
| \(x\) | -50 | 50 |
| \(P(X=x)\) | \(\frac{91}{104}\) | \(\frac{12}{104}\) |
| A1 A1 A1 | A1 for \(\frac{7}{8}\) oe, A1 for \(\frac{3}{26}\) oe, A1 for \(\frac{1}{104}\) oe. Only award final A1 if all correct probabilities are associated with the correct, corresponding values of \(x\). |
| Answer | Marks | Guidance |
|---|---|---|
| \(x\) | -50 | 50 |
| \(P(X=x)\) | \(\frac{7}{8}\) | \(\frac{3}{26}\) |
| Answer | Marks | Guidance |
|---|---|---|
| \(E(X) = -50 \times \frac{7}{8} + 50 \times \frac{3}{26} + 450 \times \frac{1}{104} = \frac{-875}{26} = -33.65\) (pence) OR \(\frac{-875}{26}\) awrt -33.7 | M1 A1 | FT their probability distribution for M1A1. \(\frac{-35}{104}\) if working in £ |
| \(E(X^2) = (-50)^2 \times \frac{7}{8} + 50^2 \times \frac{3}{26} + 450^2 \times \frac{1}{104}\) | ||
| \(Var(X) = (-50)^2 \times \frac{7}{8} + 50^2 \times \frac{3}{26} + 450^2 \times \frac{1}{104} - \left(\frac{-875}{26}\right)^2\) | M1 | |
| \(\sigma = \sqrt{3290.495562}\) | ||
| \(= 57.36(284\ldots \text{ pence})\) | A1 |
| Answer | Marks |
|---|---|
| \(\left(\frac{1}{8} \times 200\right) = 25\) players | B1 |
| Answer | Marks | Guidance |
|---|---|---|
| \(\left(\frac{875}{26} \times 200\right) = £67.31\) | B1 [11] | Accept £67.30 FT their \(-E(X)\) |
## 3(a)
Let the random variable X be the player's profit in pence. Values for $x$ are -50, 50 and 450
$P(X = -50) = \frac{1}{2} + \frac{1}{2} \times \frac{39}{52}\left(= \frac{91}{104} = \frac{7}{8} = 0.875\right)$ | B1 M1 | B1 for all three values. M1 for correct working for $P(X = 50)$ or $P(X = 450)$ or $P(X = -50)$
OR $P(X = 50) = \frac{1}{2} \times \frac{12}{52}\left(= \frac{104}{26} = 0.115 \ldots\right)$ | | Accept answers in £ or pence for this question.
OR $P(X = 450) = \frac{1}{2} \times \frac{1}{52}\left(= \frac{1}{104} = 0.00961 \ldots\right)$ | |
| $x$ | -50 | 50 | 450 |
|---|---|---|---|
| $P(X=x)$ | $\frac{91}{104}$ | $\frac{12}{104}$ | $\frac{1}{104}$ |
| A1 A1 A1 | A1 for $\frac{7}{8}$ oe, A1 for $\frac{3}{26}$ oe, A1 for $\frac{1}{104}$ oe. Only award final A1 if all correct probabilities are associated with the correct, corresponding values of $x$.
OR
| $x$ | -50 | 50 | 450 |
|---|---|---|---|
| $P(X=x)$ | $\frac{7}{8}$ | $\frac{3}{26}$ | $\frac{1}{104}$ |
## 3(b)
$E(X) = -50 \times \frac{7}{8} + 50 \times \frac{3}{26} + 450 \times \frac{1}{104} = \frac{-875}{26} = -33.65$ (pence) OR $\frac{-875}{26}$ awrt -33.7 | M1 A1 | FT their probability distribution for M1A1. $\frac{-35}{104}$ if working in £
$E(X^2) = (-50)^2 \times \frac{7}{8} + 50^2 \times \frac{3}{26} + 450^2 \times \frac{1}{104}$ | | |
$Var(X) = (-50)^2 \times \frac{7}{8} + 50^2 \times \frac{3}{26} + 450^2 \times \frac{1}{104} - \left(\frac{-875}{26}\right)^2$ | M1 |
$\sigma = \sqrt{3290.495562}$ | |
$= 57.36(284\ldots \text{ pence})$ | A1 |
## 3(c)(i)
$\left(\frac{1}{8} \times 200\right) = 25$ players | B1 |
## 3(c)(ii)
$\left(\frac{875}{26} \times 200\right) = £67.31$ | B1 [11] | Accept £67.30 FT their $-E(X)$
---A game at a school fete is played with a fair coin and a random number generator which generates random integers between 1 and 52 inclusive. It costs 50 pence to play the game. First, the player tosses the coin. If it lands on tails, the player loses. If it lands on heads, the player is allowed to generate a random number. If the number is 1, the player wins £5. If the number is between 2 and 13 inclusive, the player wins £1. If the number is greater than 13, the player loses.
\begin{enumerate}[label=(\alph*)]
\item Find the probability distribution of the player's profit. [5]
\item Find the mean and standard deviation of the player's profit. [4]
\item Given that 200 people play the game, calculate
\begin{enumerate}[label=(\roman*)]
\item the expected number of players who win some money,
\item the expected profit for the fete. [2]
\end{enumerate}
\end{enumerate}
\hfill \mbox{\textit{WJEC Further Unit 2 2018 Q3 [11]}}