| Exam Board | CAIE |
|---|---|
| Module | S1 (Statistics 1) |
| Year | 2012 |
| Session | November |
| Marks | 11 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Discrete Probability Distributions |
| Type | Verify probability from independent trials |
| Difficulty | Standard +0.3 This is a straightforward discrete probability question requiring systematic enumeration of outcomes. Part (i) involves counting favorable outcomes (standard technique), part (ii) extends this to complete a table, and part (iii) tests independence using P(R∩S) = P(R)P(S). While tedious, it requires only basic counting and probability rules with no novel insight—slightly easier than average A-level. |
| Spec | 2.03a Mutually exclusive and independent events2.04a Discrete probability distributions5.02a Discrete probability distributions: general |
| \(x\) | 3 | 4 | 5 | 6 | 7 | 8 | 9 | 10 | 11 | 12 |
| \(\mathrm { P } ( X = x )\) | \(\frac { 1 } { 64 }\) | \(\frac { 3 } { 64 }\) | \(\frac { 12 } { 64 }\) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(P(9) = P(1,4,4)\times3 + P(2,3,4)\times6 + P(3,3,3)\) | M1 | Listing at least 2 different options |
| M1 | Multiplying \(P(4,3,2)\) by 6 or \(P(1,4,4)\) by 3 | |
| \(= \frac{10}{64}\ \left(\frac{5}{32}\right)\ (0.156)\) AG | A1 [3] | Correct answer — must see numerical justification |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| Probs: \(\frac{1}{64}, \frac{3}{64}, \frac{6}{64}, \frac{10}{64}, \frac{12}{64}, \frac{12}{64}, \frac{10}{64}, \frac{6}{64}, \frac{3}{64}, \frac{1}{64}\) | B1 | 3 or more additional correct probabilities |
| B1 | 5 or more correct | |
| B1 [3] | All correct |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(P(S) = \frac{6}{64}\ \left(\frac{3}{32}\right)\) | M1 | An attempt at \(P(S)\) for \(4,4,1\) or \(4,2,2\) |
| A1 | Correct \(P(S)\) | |
| \(P(R \cap S) = \frac{3}{64} \neq \frac{15}{1024}\) ie \(P(R) \times P(S)\) | B1 | Correct \(P(R \cap S)\) in either intersection or conditional probability cases |
| OR \(P(R | S) = \frac{3/64}{6/64} = \frac{1}{2} \neq \frac{10}{64}\) ie \(P(R)\) | M1 |
| Not independent | A1ft [5] | Correct conclusion ft wrong \(P(S)\) or \(P(R \cap S)\) only |
## Question 6:
**(i)**
| Answer | Mark | Guidance |
|--------|------|----------|
| $P(9) = P(1,4,4)\times3 + P(2,3,4)\times6 + P(3,3,3)$ | M1 | Listing at least 2 different options |
| | M1 | Multiplying $P(4,3,2)$ by 6 or $P(1,4,4)$ by 3 |
| $= \frac{10}{64}\ \left(\frac{5}{32}\right)\ (0.156)$ AG | A1 [3] | Correct answer — must see numerical justification |
**(ii)**
| Answer | Mark | Guidance |
|--------|------|----------|
| Probs: $\frac{1}{64}, \frac{3}{64}, \frac{6}{64}, \frac{10}{64}, \frac{12}{64}, \frac{12}{64}, \frac{10}{64}, \frac{6}{64}, \frac{3}{64}, \frac{1}{64}$ | B1 | 3 or more additional correct probabilities |
| | B1 | 5 or more correct |
| | B1 [3] | All correct |
**(iii)**
| Answer | Mark | Guidance |
|--------|------|----------|
| $P(S) = \frac{6}{64}\ \left(\frac{3}{32}\right)$ | M1 | An attempt at $P(S)$ for $4,4,1$ or $4,2,2$ |
| | A1 | Correct $P(S)$ |
| $P(R \cap S) = \frac{3}{64} \neq \frac{15}{1024}$ ie $P(R) \times P(S)$ | B1 | Correct $P(R \cap S)$ in either intersection or conditional probability cases |
| OR $P(R|S) = \frac{3/64}{6/64} = \frac{1}{2} \neq \frac{10}{64}$ ie $P(R)$ | M1 | Comparing their $P(R \cap S)$ with their $P(R) \times P(S)$, or their $P(R|S)$ with their $P(R)$ — need numerical values |
| Not independent | A1ft [5] | Correct conclusion ft wrong $P(S)$ or $P(R \cap S)$ only |6 A fair tetrahedral die has four triangular faces, numbered $1,2,3$ and 4 . The score when this die is thrown is the number on the face that the die lands on. This die is thrown three times. The random variable $X$ is the sum of the three scores.\\
(i) Show that $\mathrm { P } ( X = 9 ) = \frac { 10 } { 64 }$.\\
(ii) Copy and complete the probability distribution table for $X$.
\begin{center}
\begin{tabular}{ | c | c | c | c | c | c | c | c | c | c | c | }
\hline
$x$ & 3 & 4 & 5 & 6 & 7 & 8 & 9 & 10 & 11 & 12 \\
\hline
$\mathrm { P } ( X = x )$ & $\frac { 1 } { 64 }$ & $\frac { 3 } { 64 }$ & & & $\frac { 12 } { 64 }$ & & & & & \\
\hline
\end{tabular}
\end{center}
(iii) Event $R$ is 'the sum of the three scores is 9 '. Event $S$ is 'the product of the three scores is 16 '. Determine whether events $R$ and $S$ are independent, showing your working.
\hfill \mbox{\textit{CAIE S1 2012 Q6 [11]}}