| Exam Board | CAIE |
|---|---|
| Module | S1 (Statistics 1) |
| Year | 2022 |
| Session | November |
| Marks | 8 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Geometric Distribution |
| Type | Non-geometric distribution identification |
| Difficulty | Moderate -0.3 This question tests standard probability concepts: parts (a-b) involve calculating maximum of discrete uniform variables using systematic counting (P(X≤k) approach), while parts (c-d) are direct applications of geometric distribution formulas. All parts are routine textbook exercises requiring no novel insight, though the multi-step nature and need to recognize the geometric distribution structure place it slightly below average difficulty for A-level. |
| Spec | 2.04a Discrete probability distributions5.02g Geometric probabilities: P(X=r) = p(1-p)^(r-1)5.02h Geometric: mean 1/p and variance (1-p)/p^2 |
| \(x\) | 1 | 2 | 3 | 4 |
| \(\mathrm { P } ( X = x )\) | \(\frac { 7 } { 64 }\) | \(\frac { 19 } { 64 }\) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| Method 1: [no of ways for score of 2 are] 222, 211, 212, 221, 122, 112, 121. [Total options \(= 64\)] | B1 | 7 correct scenarios identified, no incorrect. |
| \([\text{So } P(X=2) =]\ \dfrac{7}{4\times4\times4} = \dfrac{7}{64}\) | M1 | \(\dfrac{a}{4\times4\times4}\), \(a =\) their number of correct identified scenarios \(> 4\) |
| A1 | Approach identified, WWW. | |
| Method 2: \(\left(\dfrac{1}{4}\right)^3 + {}^3C_2\left(\dfrac{1}{4}\times\dfrac{1}{4}\times\dfrac{1}{4}\right) + {}^3C_1\left(\dfrac{1}{4}\times\dfrac{1}{4}\times\dfrac{1}{4}\right)\) | B1 | \(\left(\dfrac{1}{4}\right)^3 + {}^3C_2\left(\text{or }{}^3C_1\right)\left(\dfrac{1}{4}\times\dfrac{1}{4}\times\dfrac{1}{4}\right) + d\), \(0 < d < 1\) |
| M1 | \(\left(\dfrac{1}{4}\right)^3 + e\left(\dfrac{1}{4}\right)^3 + f\left(\dfrac{1}{4}\right)^3\), \(1 < e < 5\) and \(1 < f < 5\) | |
| \([\text{So } P(X=2) =]\ \dfrac{7}{64}\) | A1 | Approach identified, WWW. |
| Method 3: \(\left(\dfrac{1}{2}\right)^3 - \left(\dfrac{1}{4}\right)^3\) | B1 | \(\left(\dfrac{1}{2}\right)^3 - b\) seen, \(0 < b < 1\) |
| M1 | \(\left(\dfrac{1}{2}\right)^3 - c^3\), \(0 < c < \frac{1}{2}\) | |
| \([\text{So } P(X=2) =]\ \dfrac{7}{64}\) | A1 | Approach identified, WWW. |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(P(X=1) = \dfrac{1}{64}\) | B1 | \(P(X=1)\) or \(P(X=4)\) correct. Condone answers not in probability distribution table if clearly identified. |
| \(P(X=4) = \left[1 - \dfrac{1}{64} - \dfrac{7}{64} - \dfrac{19}{64} =\right] \dfrac{37}{64}\) | B1 FT | All 4 probabilities summing to 1. |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(P(Y=6) = \left[\left(\dfrac{3}{4}\right)^5 \times \dfrac{1}{4} =\right] 0.0593,\ \dfrac{243}{4096}\) | B1 | Accept \(0.059326\ldots\) to 4 or more SF. |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(\left(\dfrac{3}{4}\right)^4\) | M1 | \(\left(\dfrac{3}{4}\right)^g\), \(g = 4, 5\) or \(p^4\) where \(0 < p < 1\) |
| \(= \dfrac{81}{256},\ 0.316\) | A1 | Accept \(0.316406\ldots\) to 4 or more SF. |
| Alternative: \(P(Y>4) = 1 - P(Y\leqslant4) = 1 - \left(\dfrac{1}{4} + \dfrac{3}{4}\times\dfrac{1}{4} + \left(\dfrac{3}{4}\right)^2\times\dfrac{1}{4} + \left(\dfrac{3}{4}\right)^3\times\dfrac{1}{4}\right)\) | M1 | Correct or \(1-\left(\dfrac{1}{4}+\dfrac{3}{4}\times\dfrac{1}{4}+\left(\dfrac{3}{4}\right)^2\times\dfrac{1}{4}+\left(\dfrac{3}{4}\right)^3\times\dfrac{1}{4}+\left(\dfrac{3}{4}\right)^4\right)\) or \(1-(p+qp+q^2p+q^3p)\) where \(0 < p < 1\) and \(q = 1-p\) |
| \(\left[= 1 - \dfrac{175}{256}\right] = \dfrac{81}{256},\ 0.316\) | A1 | Accept \(0.316406\ldots\) to 4 or more SF. |
## Question 4(a):
| Answer | Marks | Guidance |
|--------|-------|----------|
| **Method 1:** [no of ways for score of 2 are] 222, 211, 212, 221, 122, 112, 121. [Total options $= 64$] | B1 | 7 correct scenarios identified, no incorrect. |
| $[\text{So } P(X=2) =]\ \dfrac{7}{4\times4\times4} = \dfrac{7}{64}$ | M1 | $\dfrac{a}{4\times4\times4}$, $a =$ their number of correct identified scenarios $> 4$ |
| | A1 | Approach identified, WWW. |
| **Method 2:** $\left(\dfrac{1}{4}\right)^3 + {}^3C_2\left(\dfrac{1}{4}\times\dfrac{1}{4}\times\dfrac{1}{4}\right) + {}^3C_1\left(\dfrac{1}{4}\times\dfrac{1}{4}\times\dfrac{1}{4}\right)$ | B1 | $\left(\dfrac{1}{4}\right)^3 + {}^3C_2\left(\text{or }{}^3C_1\right)\left(\dfrac{1}{4}\times\dfrac{1}{4}\times\dfrac{1}{4}\right) + d$, $0 < d < 1$ |
| | M1 | $\left(\dfrac{1}{4}\right)^3 + e\left(\dfrac{1}{4}\right)^3 + f\left(\dfrac{1}{4}\right)^3$, $1 < e < 5$ and $1 < f < 5$ |
| $[\text{So } P(X=2) =]\ \dfrac{7}{64}$ | A1 | Approach identified, WWW. |
| **Method 3:** $\left(\dfrac{1}{2}\right)^3 - \left(\dfrac{1}{4}\right)^3$ | B1 | $\left(\dfrac{1}{2}\right)^3 - b$ seen, $0 < b < 1$ |
| | M1 | $\left(\dfrac{1}{2}\right)^3 - c^3$, $0 < c < \frac{1}{2}$ |
| $[\text{So } P(X=2) =]\ \dfrac{7}{64}$ | A1 | Approach identified, WWW. |
**Total: 3 marks**
---
## Question 4(b):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $P(X=1) = \dfrac{1}{64}$ | B1 | $P(X=1)$ or $P(X=4)$ correct. Condone answers not in probability distribution table if clearly identified. |
| $P(X=4) = \left[1 - \dfrac{1}{64} - \dfrac{7}{64} - \dfrac{19}{64} =\right] \dfrac{37}{64}$ | B1 FT | All 4 probabilities summing to 1. |
**Total: 2 marks**
---
## Question 4(c):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $P(Y=6) = \left[\left(\dfrac{3}{4}\right)^5 \times \dfrac{1}{4} =\right] 0.0593,\ \dfrac{243}{4096}$ | B1 | Accept $0.059326\ldots$ to 4 or more SF. |
**Total: 1 mark**
---
## Question 4(d):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $\left(\dfrac{3}{4}\right)^4$ | M1 | $\left(\dfrac{3}{4}\right)^g$, $g = 4, 5$ **or** $p^4$ where $0 < p < 1$ |
| $= \dfrac{81}{256},\ 0.316$ | A1 | Accept $0.316406\ldots$ to 4 or more SF. |
| **Alternative:** $P(Y>4) = 1 - P(Y\leqslant4) = 1 - \left(\dfrac{1}{4} + \dfrac{3}{4}\times\dfrac{1}{4} + \left(\dfrac{3}{4}\right)^2\times\dfrac{1}{4} + \left(\dfrac{3}{4}\right)^3\times\dfrac{1}{4}\right)$ | M1 | Correct or $1-\left(\dfrac{1}{4}+\dfrac{3}{4}\times\dfrac{1}{4}+\left(\dfrac{3}{4}\right)^2\times\dfrac{1}{4}+\left(\dfrac{3}{4}\right)^3\times\dfrac{1}{4}+\left(\dfrac{3}{4}\right)^4\right)$ or $1-(p+qp+q^2p+q^3p)$ where $0 < p < 1$ and $q = 1-p$ |
| $\left[= 1 - \dfrac{175}{256}\right] = \dfrac{81}{256},\ 0.316$ | A1 | Accept $0.316406\ldots$ to 4 or more SF. |
**Total: 2 marks**4 Three fair 4-sided spinners each have sides labelled 1,2,3,4. The spinners are spun at the same time and the number on the side on which each spinner lands is recorded. The random variable $X$ denotes the highest number recorded.
\begin{enumerate}[label=(\alph*)]
\item Show that $\mathrm { P } ( X = 2 ) = \frac { 7 } { 64 }$.
\item Complete the probability distribution table for $X$.
\begin{center}
\begin{tabular}{ | c | c | c | c | c | }
\hline
$x$ & 1 & 2 & 3 & 4 \\
\hline
$\mathrm { P } ( X = x )$ & & $\frac { 7 } { 64 }$ & $\frac { 19 } { 64 }$ & \\
\hline
\end{tabular}
\end{center}
On another occasion, one of the fair 4 -sided spinners is spun repeatedly until a 3 is obtained. The random variable $Y$ is the number of spins required to obtain a 3 .
\item Find $\mathrm { P } ( Y = 6 )$.
\item Find $\mathrm { P } ( Y > 4 )$.
\end{enumerate}
\hfill \mbox{\textit{CAIE S1 2022 Q4 [8]}}