| Exam Board | Edexcel |
|---|---|
| Module | S1 (Statistics 1) |
| Year | 2017 |
| Session | October |
| Marks | 17 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Discrete Probability Distributions |
| Type | Construct probability distribution from scenario |
| Difficulty | Moderate -0.3 This is a multi-part S1 question involving constructing and analyzing a probability distribution from a game scenario. While it requires careful tracking of cases (parts b-c) and standard calculations (E(X), Var(X)), all techniques are routine for S1. The most challenging aspect is part (h) comparing X and Y, but this still only requires systematic enumeration of cases. Overall, slightly easier than average due to being a standard S1 exercise with clear structure and no novel insights required. |
| Spec | 2.04a Discrete probability distributions5.02a Discrete probability distributions: general5.02b Expectation and variance: discrete random variables5.02e Discrete uniform distribution |
| \(d\) | 1 | 2 | 3 | 4 |
| \(\mathrm { P } ( D = d )\) | \(\frac { 1 } { 4 }\) | \(\frac { 1 } { 4 }\) | \(\frac { 1 } { 4 }\) | \(\frac { 1 } { 4 }\) |
| \(x\) | 0 | 2 | 3 | 4 | 5 |
| \(\mathrm { P } ( X = x )\) | \(\frac { 1 } { 4 }\) | \(\frac { 5 } { 16 }\) | \(\frac { 1 } { 16 }\) |
| Answer | Marks | Guidance |
|---|---|---|
| (a) | [Discrete uniform (BUT continuous uniform is B0) | B1 |
| (b) | \(P(D = 3) + P(D = 1) \times P(D = 2) = \frac{1}{4} + \frac{1}{4} \times \frac{1}{4} = \frac{5}{16}\) (*) | M1A1cso |
| (c) | \([P(D = 1) \times P(D = 1)] = \frac{1}{4} \times \frac{1}{4}\) or \(1 - (\frac{1}{4} + \frac{5}{16} + \frac{1}{16} + \frac{1}{16}) = ] \frac{1}{16}\) | B1 |
| (d) | \(E(X) = 0 \times 2 \times \frac{1}{4} + 3 \times \frac{5}{16} + 4 \times \frac{5}{16} + 5 \times \frac{1}{16} = \mathbf{3}\) | M1A1 |
| (e) | \(E(X^2) = 0 \times 2^2 \times \frac{1}{4} + 3^2 \times \frac{5}{16} + 4^2 \times \frac{5}{16} + 5^2 \times \frac{1}{16} = [\frac{166}{16}\) or \(\frac{83}{8}\) or 10.375] | M1, dM1, A1 |
| \(\text{Var}(X) = \frac{166}{16} - 3^2\) | dM1 | A1 for 1.375 or an exact equivalent |
| \(\sigma_X^2 = 1.375\) or \(\frac{11}{8}\) | ||
| (f) | r | |
| \(P(R = r)\) | \(\frac{1}{4}\) | \(\frac{1}{4}\) |
| (g) | \(E(R) = 1 \times \frac{1}{4} + 2 \times \frac{1}{4} = [1.25\) o.e.] | M1, M1, A1cso |
| \(E(Y) = 2E(R) + 0.5\) | y | |
| \(= 2.5 + 0.5 = 3\) (*) | \(P(Y = y)\) | |
| M1, M1, A1cso | ||
| (h) | \(R = 1\) so \(Y = 2.5 \Rightarrow X = D = 2\) or 3 or 4 so \(D = 3\) or 4 work and prob \(= \frac{1}{4} + \frac{1}{4}\) | M1 |
| \(R = 2\) so \(Y = 4.5 \Rightarrow D = 1\) then \(X = 0, 3\) or 4 or 5 so \(X = 5\) only prob \(= \frac{1}{16}\) | M1 | 2nd M1 for cases where \(R = 2\) and prob |
| So \(P(X > Y) = \frac{1}{4} + \frac{1}{4} + \frac{1}{16} = \frac{9}{16}\) | A1 | A1 for \(\frac{9}{16}\) or exact equivalent |
| **(a)** | [**Discrete** uniform (BUT continuous uniform is B0) | B1 | |
|---|---|---|---|
| **(b)** | $P(D = 3) + P(D = 1) \times P(D = 2) = \frac{1}{4} + \frac{1}{4} \times \frac{1}{4} = \frac{5}{16}$ (*) | M1A1cso | M1 for a correct expression in terms of $P(D)$ or with $\frac{1}{4}$ s for $P(X = 3)$ A1cso M1 scored and no incorrect working seen $[P(X = 0) + P(X = 3)$ is M0A0 if identified!] |
| **(c)** | $[P(D = 1) \times P(D = 1)] = \frac{1}{4} \times \frac{1}{4}$ or $1 - (\frac{1}{4} + \frac{5}{16} + \frac{1}{16} + \frac{1}{16}) = ] \frac{1}{16}$ | B1 | |
| **(d)** | $E(X) = 0 \times 2 \times \frac{1}{4} + 3 \times \frac{5}{16} + 4 \times \frac{5}{16} + 5 \times \frac{1}{16} = \mathbf{3}$ | M1A1 | M1 for an attempt i.e. an expression with at least 3 correct products seen A1 for 3 or an exact equivalent e.g. $\frac{48}{16}$ |
| **(e)** | $E(X^2) = 0 \times 2^2 \times \frac{1}{4} + 3^2 \times \frac{5}{16} + 4^2 \times \frac{5}{16} + 5^2 \times \frac{1}{16} = [\frac{166}{16}$ or $\frac{83}{8}$ or 10.375] | M1, dM1, A1 | 1st M1 for an attempt i.e. an expression with at least 3 correct products seen[implied by $\frac{166}{16}$] 2nd dM1 dep on 1st M1 for use of $\text{Var}(X) = E(X^2) - [E(X)]^2$ must see values but ft their values |
| | $\text{Var}(X) = \frac{166}{16} - 3^2$ | dM1 | A1 for 1.375 or an exact equivalent |
| | $\sigma_X^2 = 1.375$ or $\frac{11}{8}$ | | |
| **(f)** | | r | 1 | 2 |
| | | $P(R = r)$ | $\frac{1}{4}$ | $\frac{1}{4}$ | M1, A1 | M1 for one correct value of $r$ and it's associated probability A1 for a fully correct probability distribution – needn't be in a table |
| **(g)** | $E(R) = 1 \times \frac{1}{4} + 2 \times \frac{1}{4} = [1.25$ o.e.] | M1, M1, A1cso | 1st M1 for correct expression for $E(R)$ [ft their (f)] or 1.25 or correct distribution for Y[ft (f)] 2nd M1 for correct use of $E(Y) = 2E(R) + 0.5$ or correct expression for $E(Y)$ [ft (f)] [⇒ 1st M1]] A1cso for 3 with no incorrect working seen provided both Ms are scored |
| | $E(Y) = 2E(R) + 0.5$ | | y | 2.5 | 4.5 |
| | $= 2.5 + 0.5 = 3$ (*) | | $P(Y = y)$ | $\frac{1}{4}$ | $\frac{1}{4}$ |
| | | M1, M1, A1cso | | | |
| **(h)** | $R = 1$ so $Y = 2.5 \Rightarrow X = D = 2$ or 3 or 4 so $D = 3$ or 4 work and prob $= \frac{1}{4} + \frac{1}{4}$ | M1 | 1st M1 for cases where $R = 1$ **and** prob. |
| | $R = 2$ so $Y = 4.5 \Rightarrow D = 1$ then $X = 0, 3$ or 4 or 5 so $X = 5$ only prob $= \frac{1}{16}$ | M1 | 2nd M1 for cases where $R = 2$ and prob |
| | So $P(X > Y) = \frac{1}{4} + \frac{1}{4} + \frac{1}{16} = \frac{9}{16}$ | A1 | A1 for $\frac{9}{16}$ or exact equivalent |
**[Total 17]**\begin{enumerate}
\item The discrete random variable $D$ with the following probability distribution represents the score when a 4-sided die is rolled.
\end{enumerate}
\begin{center}
\begin{tabular}{ | c | c | c | c | c | }
\hline
$d$ & 1 & 2 & 3 & 4 \\
\hline
$\mathrm { P } ( D = d )$ & $\frac { 1 } { 4 }$ & $\frac { 1 } { 4 }$ & $\frac { 1 } { 4 }$ & $\frac { 1 } { 4 }$ \\
\hline
\end{tabular}
\end{center}
(a) Write down the name of this distribution.
The die is used to play a game and the random variable $X$ represents the number of points scored. The die is rolled once and if $D = 2,3$ or 4 then $X = D$. If $D = 1$ the die is rolled a second time and $X = 0$ if $D = 1$ again, otherwise $X$ is the sum of the two scores on the die.\\
(b) Show that the probability of scoring 3 points in this game is $\frac { 5 } { 16 }$\\
(c) Find the probability of scoring 0 in this game.
The table below shows the probability distribution for the remaining values of $X$.
\begin{center}
\begin{tabular}{ | c | c | c | c | c | c | }
\hline
$x$ & 0 & 2 & 3 & 4 & 5 \\
\hline
$\mathrm { P } ( X = x )$ & & $\frac { 1 } { 4 }$ & & $\frac { 5 } { 16 }$ & $\frac { 1 } { 16 }$ \\
\hline
\end{tabular}
\end{center}
(d) Find $\mathrm { E } ( X )$\\
(e) Find $\operatorname { Var } ( X )$
The discrete random variable $R$ represents the number of times the die is rolled in the game.\\
(f) Write down the probability distribution of $R$.
The random variable $Y = 2 R + 0.5$\\
(g) Show that $\mathrm { E } ( Y ) = \mathrm { E } ( X )$
The game is played once.\\
(h) Find $\mathrm { P } ( X > Y )$
\hfill \mbox{\textit{Edexcel S1 2017 Q6 [17]}}