| Exam Board | CAIE |
|---|---|
| Module | S1 (Statistics 1) |
| Year | 2020 |
| Session | March |
| Marks | 8 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Binomial Distribution |
| Type | Geometric distribution (first success) |
| Difficulty | Moderate -0.8 This is a straightforward application of geometric and binomial distributions with simple probability calculations. Part (a) requires basic geometric probability (3 terms), part (b) is a standard binomial distribution table with n=3, and part (c) is direct recall of E(X)=np. All calculations involve simple fractions with p=1/3, requiring no novel insight or complex multi-step reasoning. |
| Spec | 2.04a Discrete probability distributions2.04b Binomial distribution: as model B(n,p)5.02f Geometric distribution: conditions5.02g Geometric probabilities: P(X=r) = p(1-p)^(r-1) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(\left(\frac{1}{3}\right)\left(\frac{2}{3}\right)^2 + \left(\frac{1}{3}\right)\left(\frac{2}{3}\right)^3 + \left(\frac{1}{3}\right)\left(\frac{2}{3}\right)^4\) | M1 | One correct term with \(0 < p < 1\) |
| \(= \frac{4}{27} + \frac{8}{81} + \frac{16}{243}\left(= \frac{2432}{7776}\right)\) | A1 | Correct expression, accept unsimplified |
| \(= \frac{76}{243}\) or \(0.313\) | A1 | |
| 3 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| Probability distribution table with \(x\): 0, 1, 2, 3 and \(P(x)\): \(\frac{8}{27}\), \(\frac{12}{27}\), \(\frac{6}{27}\), \(\frac{1}{27}\) | B1 | Table with correct values of \(x\), no additional values unless probability 0 stated, at least one non-zero probability included |
| \(P(0) = \left(\frac{2}{3}\right)^3\); \(P(1) = \left(\frac{1}{3}\right)\left(\frac{2}{3}\right)^2 \times 3\); \(P(2) = \left(\frac{2}{3}\right)\left(\frac{1}{3}\right)^2 \times 3\); \(P(3) = \left(\frac{1}{3}\right)^3\) | B1 | 1 correct probability seen (may not be in table) or 3 or 4 non-zero probabilities summing to 1 |
| All probabilities correct | B1 | All probabilities correct |
| Total: 3 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(E(X) = \left[0 \times \frac{8}{27}\right] + 1 \times \frac{12}{27} + 2 \times \frac{6}{27} + 3 \times \frac{1}{27}\) | M1 | Correct method from *their* probability distribution table with at least 3 terms, \(0 \leqslant \text{their } P(x) \leqslant 1\), accept unsimplified |
| \(= \left[\frac{0}{27}\right] + \frac{12}{27} + \frac{12}{27} + \frac{3}{27}\) | ||
| \(= 1\) | A1 | |
| Total: 2 |
## Question 2(a):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $\left(\frac{1}{3}\right)\left(\frac{2}{3}\right)^2 + \left(\frac{1}{3}\right)\left(\frac{2}{3}\right)^3 + \left(\frac{1}{3}\right)\left(\frac{2}{3}\right)^4$ | M1 | One correct term with $0 < p < 1$ |
| $= \frac{4}{27} + \frac{8}{81} + \frac{16}{243}\left(= \frac{2432}{7776}\right)$ | A1 | Correct expression, accept unsimplified |
| $= \frac{76}{243}$ or $0.313$ | A1 | |
| | **3** | |
## Question 2(b):
| Answer | Marks | Guidance |
|--------|-------|----------|
| Probability distribution table with $x$: 0, 1, 2, 3 and $P(x)$: $\frac{8}{27}$, $\frac{12}{27}$, $\frac{6}{27}$, $\frac{1}{27}$ | B1 | Table with correct values of $x$, no additional values unless probability 0 stated, at least one non-zero probability included |
| $P(0) = \left(\frac{2}{3}\right)^3$; $P(1) = \left(\frac{1}{3}\right)\left(\frac{2}{3}\right)^2 \times 3$; $P(2) = \left(\frac{2}{3}\right)\left(\frac{1}{3}\right)^2 \times 3$; $P(3) = \left(\frac{1}{3}\right)^3$ | B1 | 1 correct probability seen (may not be in table) **or** 3 or 4 non-zero probabilities summing to 1 |
| All probabilities correct | B1 | All probabilities correct |
| **Total: 3** | | |
## Question 2(c):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $E(X) = \left[0 \times \frac{8}{27}\right] + 1 \times \frac{12}{27} + 2 \times \frac{6}{27} + 3 \times \frac{1}{27}$ | M1 | Correct method from *their* probability distribution table with at least 3 terms, $0 \leqslant \text{their } P(x) \leqslant 1$, accept unsimplified |
| $= \left[\frac{0}{27}\right] + \frac{12}{27} + \frac{12}{27} + \frac{3}{27}$ | | |
| $= 1$ | A1 | |
| **Total: 2** | | |2 An ordinary fair die is thrown repeatedly until a 1 or a 6 is obtained.
\begin{enumerate}[label=(\alph*)]
\item Find the probability that it takes at least 3 throws but no more than 5 throws to obtain a 1 or a 6 .\\
On another occasion, this die is thrown 3 times. The random variable $X$ is the number of times that a 1 or a 6 is obtained.
\item Draw up the probability distribution table for $X$.
\item Find $\mathrm { E } ( X )$.
\end{enumerate}
\hfill \mbox{\textit{CAIE S1 2020 Q2 [8]}}