| Exam Board | Edexcel |
|---|---|
| Module | FS1 AS (Further Statistics 1 AS) |
| Year | 2022 |
| Session | June |
| Marks | 14 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Discrete Random Variables |
| Type | Probability distributions with parameters |
| Difficulty | Challenging +1.2 This is a multi-part Further Statistics question requiring probability distribution properties, variance manipulation, and conditional probability across multiple games. While it involves several steps and careful bookkeeping (parts c and d require considering multiple cases), the techniques are standard for FS1: using ΣP=1, calculating E(X) and Var(X) from definitions, and systematic enumeration of outcomes. The conceptual demand is moderate—higher than typical A-level stats but routine for Further Maths students who have practiced these methods. |
| Spec | 5.01a Permutations and combinations: evaluate probabilities5.02a Discrete probability distributions: general5.02b Expectation and variance: discrete random variables5.02c Linear coding: effects on mean and variance |
| \(x\) | 0 | 2 | 3 | 6 |
| \(\mathrm { P } ( X = x )\) | \(p\) | 0.25 | \(q\) | 0.4 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(E(X) = [0 \times p] + (2 \times 0.25) + 3q + (6 \times 0.4) [= 2.9 + 3q]\) | B1 | Correct expression for \(E(X)\), need not be simplified |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(E(X^2) = [0 \times p] + (2^2 \times 0.25) + 3^2q + (6^2 \times 0.4) [= 15.4 + 9q]\) | B1 | Correct expression for \(E(X^2)\), need not be simplified |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(("15.4 + 9q") - ("2.9 + 3q")^2 = 3.66\) | M1 | Using "their \(E(X^2)\)" \(-\) "their \((E(X))^2\)" \(= 3.66\) |
| \(9q^2 + 8.4q - 3.33 = 0 \Rightarrow q = 0.3\) and \(-\dfrac{37}{30}\) | M1 | Rearranging to get a correct 3 term quadratic (condone missing \(= 0\)) leading to \(0.3\) and \(-37/30\) (awrt \(-1.23\)) or \((10q-3)(10q+37)\) |
| \(q = 0.3^*\) since \(q\) cannot be negative | A1cso* | cso with a comment why \(-37/30\) is eliminated. Minimum required is \(q > 0\) or they say it is impossible. |
| SC \(("15.4 + 9 \times 0.3") - ("2.9 + 3 \times 0.3")^2\) can get M1M0A0 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(P(x_1 + x_2 + x_3 + x_4 = 20) = P(6,6,6,2\ \text{or}\ 6,6,2,6\ \text{or}\ 6,2,6,6\ \text{or}\ 2,6,6,6)\) | M1 | Realising that combination is 6662, any order. Implied by \(0.4^3 \times 0.25\) |
| \(= 4 \times 0.4^3 \times 0.25\) | M1 | Correct calculation |
| \(= 0.064\) oe | A1 | 0.064 oe only e.g. 8/125 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(P(x_5 + x_6 \geqslant 7) = P(6,6\ \text{or}\ 6,3\ \text{or}\ 3,6)\) | M1 | Realising all the different combinations 7 or more can be scored from 2 games (no need for arrangements). Implied by \((0.4^2)\) and \((0.4 \times 0.3)\) and \((0.4 \times 0.25)\) |
| \(= (0.4^2) + 2 \times (0.4 \times 0.3) + 2 \times 0.4 \times 0.25\ [= 0.6]\) | M1 | Fully correct method |
| \(P(\text{score} \geqslant 27) = "0.064" \times "0.6"\ [= 24/625 = 0.0384]\) | M1 | For multiplying "their (c)" with "their \(P(x_5 + x_6 \geqslant 7)\)" providing at least 2 combinations are used to find \(P(x_5 + x_6 \geqslant 7)\) |
| \(Y \sim \text{B}(3,\ "0.0384")\) | dM1 | Dependent on 3rd M1 being awarded, for using or writing \(\text{B}(3,\ \text{"their } P(x_1+x_2+x_3+x_4+x_5+x_6 \geqslant 27)\text{"})\ (1-"0.0384")^3\) or similar |
| \(P(Y \geqslant 1) = 1 - P(Y = 0)\) | M1 | For writing or using \(1 - P(Y=0)\), e.g. \(1-(1-"0.0384")^3\) |
| \(= 0.1108...\) | A1cso | awrt 0.111 from correct working |
# Question 4:
## Part (a)(i)
| Answer/Working | Mark | Guidance |
|---|---|---|
| $E(X) = [0 \times p] + (2 \times 0.25) + 3q + (6 \times 0.4) [= 2.9 + 3q]$ | B1 | Correct expression for $E(X)$, need not be simplified |
## Part (a)(ii)
| Answer/Working | Mark | Guidance |
|---|---|---|
| $E(X^2) = [0 \times p] + (2^2 \times 0.25) + 3^2q + (6^2 \times 0.4) [= 15.4 + 9q]$ | B1 | Correct expression for $E(X^2)$, need not be simplified |
## Part (b)
| Answer/Working | Mark | Guidance |
|---|---|---|
| $("15.4 + 9q") - ("2.9 + 3q")^2 = 3.66$ | M1 | Using "their $E(X^2)$" $-$ "their $(E(X))^2$" $= 3.66$ |
| $9q^2 + 8.4q - 3.33 = 0 \Rightarrow q = 0.3$ and $-\dfrac{37}{30}$ | M1 | Rearranging to get a correct 3 term quadratic (condone missing $= 0$) leading to $0.3$ and $-37/30$ (awrt $-1.23$) or $(10q-3)(10q+37)$ |
| $q = 0.3^*$ since $q$ cannot be negative | A1cso* | cso with a comment why $-37/30$ is eliminated. Minimum required is $q > 0$ or they say it is impossible. |
| **SC** $("15.4 + 9 \times 0.3") - ("2.9 + 3 \times 0.3")^2$ can get M1M0A0 | | |
## Part (c)
| Answer/Working | Mark | Guidance |
|---|---|---|
| $P(x_1 + x_2 + x_3 + x_4 = 20) = P(6,6,6,2\ \text{or}\ 6,6,2,6\ \text{or}\ 6,2,6,6\ \text{or}\ 2,6,6,6)$ | M1 | Realising that combination is 6662, any order. Implied by $0.4^3 \times 0.25$ |
| $= 4 \times 0.4^3 \times 0.25$ | M1 | Correct calculation |
| $= 0.064$ oe | A1 | 0.064 oe only e.g. 8/125 |
## Part (d)
| Answer/Working | Mark | Guidance |
|---|---|---|
| $P(x_5 + x_6 \geqslant 7) = P(6,6\ \text{or}\ 6,3\ \text{or}\ 3,6)$ | M1 | Realising all the different combinations 7 or more can be scored from 2 games (no need for arrangements). Implied by $(0.4^2)$ and $(0.4 \times 0.3)$ and $(0.4 \times 0.25)$ |
| $= (0.4^2) + 2 \times (0.4 \times 0.3) + 2 \times 0.4 \times 0.25\ [= 0.6]$ | M1 | Fully correct method |
| $P(\text{score} \geqslant 27) = "0.064" \times "0.6"\ [= 24/625 = 0.0384]$ | M1 | For multiplying "their (c)" with "their $P(x_5 + x_6 \geqslant 7)$" providing at least 2 combinations are used to find $P(x_5 + x_6 \geqslant 7)$ |
| $Y \sim \text{B}(3,\ "0.0384")$ | dM1 | Dependent on 3rd M1 being awarded, for using or writing $\text{B}(3,\ \text{"their } P(x_1+x_2+x_3+x_4+x_5+x_6 \geqslant 27)\text{"})\ (1-"0.0384")^3$ or similar |
| $P(Y \geqslant 1) = 1 - P(Y = 0)$ | M1 | For writing or using $1 - P(Y=0)$, e.g. $1-(1-"0.0384")^3$ |
| $= 0.1108...$ | A1cso | awrt 0.111 from correct working |\begin{enumerate}
\item The discrete random variable $X$ has the following probability distribution
\end{enumerate}
\begin{center}
\begin{tabular}{ | c | c | c | c | c | }
\hline
$x$ & 0 & 2 & 3 & 6 \\
\hline
$\mathrm { P } ( X = x )$ & $p$ & 0.25 & $q$ & 0.4 \\
\hline
\end{tabular}
\end{center}
(a) Find in terms of $q$\\
(i) $\mathrm { E } ( X )$\\
(ii) $\mathrm { E } \left( X ^ { 2 } \right)$
Given that $\operatorname { Var } ( X ) = 3.66$\\
(b) show that $q = 0.3$
In a game, the score is given by the discrete random variable $X$\\
Given that games are independent,\\
(c) calculate the probability that after the 4th game has been played, the total score is exactly 20
A round consists of 4 games plus 2 bonus games. The bonus games are only played if after the 4th game has been played the total score is exactly 20
A prize of $\pounds 10$ is awarded if 6 games are played in a round and the total score for the round is at least 27
Bobby plays 3 rounds.\\
(d) Find the probability that Bobby wins at least $\pounds 10$
\hfill \mbox{\textit{Edexcel FS1 AS 2022 Q4 [14]}}