| Exam Board | Edexcel |
|---|---|
| Module | S1 (Statistics 1) |
| Marks | 17 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Uniform Distribution |
| Type | Name the distribution |
| Difficulty | Easy -1.8 This is a very straightforward S1 question requiring only basic probability concepts: naming a discrete uniform distribution (trivial recognition), calculating expectations using the formula E(X) = Σxp(x), finding probabilities of sums by enumeration, and verifying the linearity of expectation. All parts are routine bookwork with no problem-solving or insight required. |
| Spec | 5.02a Discrete probability distributions: general5.02b Expectation and variance: discrete random variables5.02e Discrete uniform distribution5.04a Linear combinations: E(aX+bY), Var(aX+bY) |
| \(a\) | 1 | 2 | 3 |
| \(\mathrm { P } ( A = a )\) | \(\frac { 1 } { 3 }\) | \(\frac { 1 } { 3 }\) | \(\frac { 1 } { 3 }\) |
| \(b\) | 1 | 2 | 3 |
| \(\mathrm { P } ( B = b )\) | \(\frac { 1 } { 2 }\) | \(\frac { 1 } { 4 }\) | \(\frac { 1 } { 4 }\) |
| Answer | Marks | Guidance |
|---|---|---|
| (a) discrete uniform | B1 | |
| (b) \(2\) | A1 | |
| (c) \(\sum bP(b) = \frac{1}{2} + \frac{1}{2} + \frac{3}{4} = \frac{7}{4}\) | M1 A1 | |
| (d) \(P(C = 2) = P(A = 1 \text{ and } B = 1) = \frac{1}{4} \times \frac{1}{4} = \frac{1}{16}\) | M2 A1 | |
| (e) \(P(C = 3) = (\frac{1}{4} \times \frac{1}{4}) + (\frac{1}{4} \times \frac{1}{4}) = \frac{1}{8}\) | ||
| \(P(C = 4) = (\frac{1}{4} \times \frac{1}{4}) + (\frac{1}{4} \times \frac{1}{4}) + (\frac{1}{4} \times \frac{1}{4}) = \frac{1}{4}\) | ||
| \(P(C = 5) = (\frac{1}{4} \times \frac{1}{4}) + (\frac{1}{4} \times \frac{1}{4}) = \frac{1}{8}\) | ||
| \(P(C = 6) = (\frac{1}{4} \times \frac{1}{4}) = \frac{1}{12}\) | M3 A3 | |
| \(c\) | \(2\) | \(3\) |
| \(P(C = c)\) | \(\frac{1}{6}\) | \(\frac{1}{4}\) |
| (f) \(E(C) = \sum cP(c) = \frac{1}{3} + \frac{3}{4} + \frac{4}{3} + \frac{5}{6} + \frac{1}{2} = \frac{15}{4}\) | M1 A1 | |
| \(E(A) + E(B) = 2 + \frac{7}{4} = \frac{15}{4}\) \(\therefore E(C) = E(A) + E(B)\) | M1 A1 | (17) |
(a) discrete uniform | B1 |
(b) $2$ | A1 |
(c) $\sum bP(b) = \frac{1}{2} + \frac{1}{2} + \frac{3}{4} = \frac{7}{4}$ | M1 A1 |
(d) $P(C = 2) = P(A = 1 \text{ and } B = 1) = \frac{1}{4} \times \frac{1}{4} = \frac{1}{16}$ | M2 A1 |
(e) $P(C = 3) = (\frac{1}{4} \times \frac{1}{4}) + (\frac{1}{4} \times \frac{1}{4}) = \frac{1}{8}$ | |
$P(C = 4) = (\frac{1}{4} \times \frac{1}{4}) + (\frac{1}{4} \times \frac{1}{4}) + (\frac{1}{4} \times \frac{1}{4}) = \frac{1}{4}$ | |
$P(C = 5) = (\frac{1}{4} \times \frac{1}{4}) + (\frac{1}{4} \times \frac{1}{4}) = \frac{1}{8}$ | |
$P(C = 6) = (\frac{1}{4} \times \frac{1}{4}) = \frac{1}{12}$ | M3 A3 |
| $c$ | $2$ | $3$ | $4$ | $5$ | $6$ |
|-----|-----|-----|-----|-----|--------|
| $P(C = c)$ | $\frac{1}{6}$ | $\frac{1}{4}$ | $\frac{1}{3}$ | $\frac{1}{6}$ | $\frac{1}{12}$ |
(f) $E(C) = \sum cP(c) = \frac{1}{3} + \frac{3}{4} + \frac{4}{3} + \frac{5}{6} + \frac{1}{2} = \frac{15}{4}$ | M1 A1 |
$E(A) + E(B) = 2 + \frac{7}{4} = \frac{15}{4}$ $\therefore E(C) = E(A) + E(B)$ | M1 A1 | (17)6. In a game two spinners are used. The score on the first spinner is given by the random variable $A$, which has the following probability distribution:
\begin{center}
\begin{tabular}{ | c | c | c | c | }
\hline
$a$ & 1 & 2 & 3 \\
\hline
$\mathrm { P } ( A = a )$ & $\frac { 1 } { 3 }$ & $\frac { 1 } { 3 }$ & $\frac { 1 } { 3 }$ \\
\hline
\end{tabular}
\end{center}
\begin{enumerate}[label=(\alph*)]
\item State the name of this distribution.
\item Write down $\mathrm { E } ( A )$.
The score on the second spinner is given by the random variable $B$, which has the following probability distribution:
\begin{center}
\begin{tabular}{ | c | c | c | c | }
\hline
$b$ & 1 & 2 & 3 \\
\hline
$\mathrm { P } ( B = b )$ & $\frac { 1 } { 2 }$ & $\frac { 1 } { 4 }$ & $\frac { 1 } { 4 }$ \\
\hline
\end{tabular}
\end{center}
\item Find $\mathrm { E } ( B )$.
On each player's turn in the game, both spinners are used and the scores on the two spinners are added together. The total score on the two spinners is given by the random variable $C$.
\item Show that $\mathrm { P } ( C = 2 ) = \frac { 1 } { 6 }$.
\item Find the probability distribution of $C$.
\item Show that $\mathrm { E } ( C ) = \mathrm { E } ( A ) + \mathrm { E } ( B )$.
\end{enumerate}
\hfill \mbox{\textit{Edexcel S1 Q6 [17]}}