| Exam Board | CAIE |
|---|---|
| Module | S1 (Statistics 1) |
| Year | 2010 |
| Session | June |
| Marks | 10 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Discrete Probability Distributions |
| Type | Sampling without replacement |
| Difficulty | Moderate -0.3 This is a standard S1 question covering hypergeometric distribution and Bayes' theorem. Part (i) requires systematic enumeration of a small sample space, part (ii) involves routine expectation/variance calculations with given answer to verify, and part (iii) is a textbook conditional probability problem using the law of total probability. All techniques are standard for this specification with no novel insight required, making it slightly easier than average. |
| Spec | 2.03c Conditional probability: using diagrams/tables2.03d Calculate conditional probability: from first principles2.04a Discrete probability distributions |
| Answer | Marks | Guidance |
|---|---|---|
| (i) | x | 0 |
| P(X = x) | 1/7 | 4/7 |
| B1 | 3, 4 … if in table must have blank or 0 for prob. One correct probability | |
| B1 [3] | All correct | |
| (ii) \(E(X) = 8/7 \text{ (1.14)}\) AG | B1 | Legitimate correct given answer rounding to 1.14 |
| \(\text{Var}(X) = 12/7 - (8/7)^2 = 20/49 \text{ (0.408)}\) | M1, A1 [3] | Correct method with mean\(^2\) subt numerically no dividing by anything. Correct final answer |
| (iii) \(P(G \mid NA) = \frac{P(G \cap NA)}{P(NA)} = \frac{2/5 \times 1/4}{2/5 \times 1/4 + 3/5 \times 9/10}\) | M1, M1 | Attempt at \(P(G \cap NA)\) or \(P(G \cap A)\) as numerator of a fraction. Attempt at P(NA) or P(A) in form of summing two 2-factor products, seen anywhere |
| \(= \frac{5}{32} \text{ (0.156)}\) | A1 [4] | Correct unsimplified denominator of a fraction. Correct answer |
(i) | x | 0 | 1 | 2 |
| P(X = x) | 1/7 | 4/7 | 2/7 | | B1 | 0, 1, 2 only in table or listed with some prob
| | | B1 | 3, 4 … if in table must have blank or 0 for prob. One correct probability
| | | B1 [3] | All correct
(ii) $E(X) = 8/7 \text{ (1.14)}$ AG | B1 | Legitimate correct given answer rounding to 1.14
$\text{Var}(X) = 12/7 - (8/7)^2 = 20/49 \text{ (0.408)}$ | M1, A1 [3] | Correct method with mean$^2$ subt numerically no dividing by anything. Correct final answer
(iii) $P(G \mid NA) = \frac{P(G \cap NA)}{P(NA)} = \frac{2/5 \times 1/4}{2/5 \times 1/4 + 3/5 \times 9/10}$ | M1, M1 | Attempt at $P(G \cap NA)$ or $P(G \cap A)$ as numerator of a fraction. Attempt at P(NA) or P(A) in form of summing two 2-factor products, seen anywhere
$= \frac{5}{32} \text{ (0.156)}$ | A1 [4] | Correct unsimplified denominator of a fraction. Correct answer
---A small farm has 5 ducks and 2 geese. Four of these birds are to be chosen at random. The random variable $X$ represents the number of geese chosen.
\begin{enumerate}[label=(\roman*)]
\item Draw up the probability distribution of $X$. [3]
\item Show that $\mathrm{E}(X) = \frac{8}{7}$ and calculate $\mathrm{Var}(X)$. [3]
\item When the farmer's dog is let loose, it chases either the ducks with probability $\frac{3}{5}$ or the geese with probability $\frac{2}{5}$. If the dog chases the ducks there is a probability of $\frac{1}{10}$ that they will attack the dog. If the dog chases the geese there is a probability of $\frac{1}{4}$ that they will attack the dog. Given that the dog is not attacked, find the probability that it was chasing the geese. [4]
\end{enumerate}
\hfill \mbox{\textit{CAIE S1 2010 Q6 [10]}}