| Exam Board | CAIE |
|---|---|
| Module | S1 (Statistics 1) |
| Year | 2016 |
| Session | November |
| Marks | 7 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Probability Definitions |
| Type | Two-way table probabilities |
| Difficulty | Easy -1.3 This is a straightforward two-way table probability question requiring only basic probability calculations (adding frequencies, dividing by total) and checking independence using P(X∩Y) = P(X)×P(Y). All values are given directly in the table with no problem-solving or conceptual insight needed—purely mechanical arithmetic. |
| Spec | 2.03a Mutually exclusive and independent events2.03b Probability diagrams: tree, Venn, sample space |
| Germany | Japan | Korea | |
| Silver | 40 | 26 | 34 |
| White | 32 | 22 | 26 |
| Red | 28 | 12 | 30 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(64/250\), \(0.256\) | B1 [1] | oe |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(190/250\), \(0.76(0)\) | B1 [1] | oe |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(P(X) = 80/250 = 8/25\) | M1 | attempt at \(P(X)\) |
| \(P(Y) = 100/250 = 2/5\) | M1 | attempt at \(P(Y)\) |
| \(P(X \cap Y) = 32/250 = 16/125\) | B1 | oe |
| \(P(X) \times P(Y) = \dfrac{8}{25} \times \dfrac{2}{5} = \dfrac{16}{125}\) | M1 | comparing \(P(X)\times P(Y)\) and \(P(X \cap Y)\) so long as independence has not been assumed |
| Since \(P(X)\times P(Y) = P(X \cap Y)\) therefore independent | A1 [5] | correct answer with all working correct |
## Question 4:
### Part (i):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $64/250$, $0.256$ | B1 [1] | oe |
### Part (ii):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $190/250$, $0.76(0)$ | B1 [1] | oe |
### Part (iii):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $P(X) = 80/250 = 8/25$ | M1 | attempt at $P(X)$ |
| $P(Y) = 100/250 = 2/5$ | M1 | attempt at $P(Y)$ |
| $P(X \cap Y) = 32/250 = 16/125$ | B1 | oe |
| $P(X) \times P(Y) = \dfrac{8}{25} \times \dfrac{2}{5} = \dfrac{16}{125}$ | M1 | comparing $P(X)\times P(Y)$ and $P(X \cap Y)$ so long as independence has not been assumed |
| Since $P(X)\times P(Y) = P(X \cap Y)$ therefore independent | A1 [5] | correct answer with all working correct |
---4 For a group of 250 cars the numbers, classified by colour and country of manufacture, are shown in the table.
\begin{center}
\begin{tabular}{ | l | c | c | c | }
\hline
& Germany & Japan & Korea \\
\hline
Silver & 40 & 26 & 34 \\
\hline
White & 32 & 22 & 26 \\
\hline
Red & 28 & 12 & 30 \\
\hline
\end{tabular}
\end{center}
One car is selected at random from this group. Find the probability that the selected car is\\
(i) a red or silver car manufactured in Korea,\\
(ii) not manufactured in Japan.\\
$X$ is the event that the selected car is white. $Y$ is the event that the selected car is manufactured in Germany.\\
(iii) By using appropriate probabilities, determine whether events $X$ and $Y$ are independent.
\hfill \mbox{\textit{CAIE S1 2016 Q4 [7]}}