| Exam Board | CAIE |
|---|---|
| Module | S1 (Statistics 1) |
| Year | 2009 |
| Session | June |
| Marks | 6 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Discrete Probability Distributions |
| Type | Sequential trials until success |
| Difficulty | Moderate -0.3 This is a straightforward probability distribution question requiring calculation of probabilities using tree diagrams/sample spaces, then applying standard expectation and variance formulas. Part (i) is guided (showing a given answer), and part (ii) involves routine computation with the complete distribution provided. While it requires careful enumeration of cases, it's a standard textbook exercise with no novel problem-solving or conceptual challenges beyond basic S1 material. |
| Spec | 5.02a Discrete probability distributions: general5.02b Expectation and variance: discrete random variables |
| \(x\) | 2 | 3 | 4 | 5 | 6 | 7 |
| \(\mathrm { P } ( X = x )\) | \(\frac { 5 } { 16 }\) | \(\frac { 1 } { 16 }\) | \(\frac { 3 } { 8 }\) | \(\frac { 1 } { 8 }\) | \(\frac { 1 } { 16 }\) | \(\frac { 1 } { 16 }\) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(P(X=2) = \frac{1}{4} \times \frac{1}{4} + \frac{1}{4} = \frac{5}{16}\) AG | M1 | Considering cases \((1,1)\) and \((2)\) |
| OR can use a table: \(\begin{array}{c\ | cccc} & 1 & 2 & 3 & 4 \\ 1 & 2 & 2 & 4 & 4 \\ 2 & 3 & 2 & 5 & 4 \\ 3 & 4 & 2 & 6 & 4 \\ 4 & 5 & 2 & 7 & 4 \end{array}\) | A1 [2] |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(E(X) = \Sigma xp\) | M1 | Using correct formula for \(E(X)\), no extra division |
| \(= \frac{15}{4}\) (3.75) | A1 | Correct answer |
| \(\text{Var}(X) = 2^2 \times \frac{5}{16} + 3^2 \times \frac{1}{16} + 4^2 \times \frac{3}{8} + \ldots - \left(\frac{15}{4}\right)^2\) | M1 | Using a variance formula correctly with mean\(^2\) subtracted numerically, no extra division |
| \(= \frac{260}{16} - \frac{225}{16} = \frac{35}{16}\) (2.19) | A1 [4] | Correct final answer |
## Question 2:
### Part (i)
| Answer/Working | Marks | Guidance |
|---|---|---|
| $P(X=2) = \frac{1}{4} \times \frac{1}{4} + \frac{1}{4} = \frac{5}{16}$ AG | M1 | Considering cases $(1,1)$ and $(2)$ |
| OR can use a table: $\begin{array}{c\|cccc} & 1 & 2 & 3 & 4 \\ 1 & 2 & 2 & 4 & 4 \\ 2 & 3 & 2 & 5 & 4 \\ 3 & 4 & 2 & 6 & 4 \\ 4 & 5 & 2 & 7 & 4 \end{array}$ | A1 **[2]** | Correct given answer legitimately obtained ($\frac{1}{16} + \frac{4}{16}$ needs some justification but $\frac{1}{16} + \frac{1}{4}$ is acceptable) |
### Part (ii)
| Answer/Working | Marks | Guidance |
|---|---|---|
| $E(X) = \Sigma xp$ | M1 | Using correct formula for $E(X)$, no extra division |
| $= \frac{15}{4}$ (3.75) | A1 | Correct answer |
| $\text{Var}(X) = 2^2 \times \frac{5}{16} + 3^2 \times \frac{1}{16} + 4^2 \times \frac{3}{8} + \ldots - \left(\frac{15}{4}\right)^2$ | M1 | Using a variance formula correctly with mean$^2$ subtracted numerically, no extra division |
| $= \frac{260}{16} - \frac{225}{16} = \frac{35}{16}$ (2.19) | A1 **[4]** | Correct final answer |
---2 Gohan throws a fair tetrahedral die with faces numbered $1,2,3,4$. If she throws an even number then her score is the number thrown. If she throws an odd number then she throws again and her score is the sum of both numbers thrown. Let the random variable $X$ denote Gohan's score.\\
(i) Show that $\mathrm { P } ( X = 2 ) = \frac { 5 } { 16 }$.\\
(ii) The table below shows the probability distribution of $X$.
\begin{center}
\begin{tabular}{ | c | c | c | c | c | c | c | }
\hline
$x$ & 2 & 3 & 4 & 5 & 6 & 7 \\
\hline
$\mathrm { P } ( X = x )$ & $\frac { 5 } { 16 }$ & $\frac { 1 } { 16 }$ & $\frac { 3 } { 8 }$ & $\frac { 1 } { 8 }$ & $\frac { 1 } { 16 }$ & $\frac { 1 } { 16 }$ \\
\hline
\end{tabular}
\end{center}
Calculate $\mathrm { E } ( X )$ and $\operatorname { Var } ( X )$.
\hfill \mbox{\textit{CAIE S1 2009 Q2 [6]}}