| Exam Board | CAIE |
|---|---|
| Module | S1 (Statistics 1) |
| Year | 2024 |
| Session | June |
| Marks | 11 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Geometric Distribution |
| Type | Second success on trial n |
| Difficulty | Standard +0.3 This is a straightforward application of standard probability distributions (geometric and binomial with normal approximation). Parts (a)-(c) require direct formula application with minimal problem-solving, while part (d) is a routine normal approximation with continuity correction. All techniques are standard S1 content with no novel insights required. |
| Spec | 2.04b Binomial distribution: as model B(n,p)2.04c Calculate binomial probabilities2.04d Normal approximation to binomial5.02f Geometric distribution: conditions5.02g Geometric probabilities: P(X=r) = p(1-p)^(r-1)5.02h Geometric: mean 1/p and variance (1-p)/p^2 |
| Answer | Marks | Guidance |
|---|---|---|
| \(\left[(0.35)^4 \times 0.65 =\right] 0.00975\) | B1 | AWRT |
| Answer | Marks | Guidance |
|---|---|---|
| \((0.35)^3 \times (0.65)^2 \times 4\) | M1 | \((0.35)^3 \times (0.65)^2 \times k\), where \(k\) is an integer, \(1 \leqslant k \leqslant 5\), no \(+\) or \(-\) |
| \(= 0.0725\) | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| \([1 - P(5,6,7) =]\) \(1 - (^7C_5\ 0.65^5\ 0.35^2 + ^7C_6\ 0.65^6\ 0.35^1 + 0.65^7)\) | M1 | One term \(^7C_x(p)^x(1-p)^{7-x}\), \(0 < p < 1\), \(0 < x < 7\) |
| \([= 1-(0.29848 + 0.18478 + 0.049022)]\) | A1 | Correct expression, accept unsimplified, no terms omitted leading to final answer |
| \(= 0.468\) | B1 | AWRT |
| Answer | Marks | Guidance |
|---|---|---|
| \([P(0,1,2,3,4) =]\) \(0.35^7 + ^7C_1\ 0.65^1\ 0.35^6 + ^7C_2\ 0.65^2\ 0.35^5 + ^7C_3\ 0.65^3\ 0.35^4 + ^7C_4\ 0.65^4\ 0.35^3\) | (M1) | One term \(^7C_x(p)^x(1-p)^{7-x}\), \(0 < p < 1\), \(0 < x < 7\) |
| \([0.00064 + 0.00836 + 0.04660 + 0.14424 + 0.26787]\) | (A1) | Correct expression, accept unsimplified, no terms omitted leading to final answer |
| \(= 0.468\) | (B1) | AWRT |
| Answer | Marks | Guidance |
|---|---|---|
| \([\text{Mean} =]\ 84 \times 0.65 = 54.6\) \([\text{Var} =]\ 84 \times 0.65 \times 0.35 = 19.11\) | B1 | 54.6 and 19.11 seen, allow unsimplified. \([\sigma =]\ 4.371\) or \(\frac{7\sqrt{39}}{10}\) implies correct variance. Incorrect notation penalised, but condone use of values in standardisation formula |
| \(P(X > 50) = P\!\left(Z > \dfrac{50.5 - 54.6}{\sqrt{19.11}}\right)\) | M1 | Substituting *their* \(\mu\) and *their* positive \(\sigma\) into the \(\pm\)standardising formula (any number), not *their* \(\sigma^2\), or \(\sqrt{\textit{their}\ \sigma}\) |
| M1 | Use continuity correction 49.5 or 50.5 in *their* standardisation formula. Note: \(\frac{\pm 4.1}{\sqrt{19.11}}\) or \(\pm\frac{4.1}{4.371}\) seen gains M2 BOD | |
| \(P(Z > -0.9379) = \Phi(0.9379)\) | M1 | Appropriate area \(\Phi\), from final process, must be a probability |
| \(0.826\) | A1 | \(0.8255 < p \leq 0.826\) |
## Question 5(a):
| $\left[(0.35)^4 \times 0.65 =\right] 0.00975$ | B1 | AWRT |
**Total: 1 mark**
---
## Question 5(b):
| $(0.35)^3 \times (0.65)^2 \times 4$ | M1 | $(0.35)^3 \times (0.65)^2 \times k$, where $k$ is an integer, $1 \leqslant k \leqslant 5$, no $+$ or $-$ |
|---|---|---|
| $= 0.0725$ | A1 | |
**Total: 2 marks**
---
## Question 5(c):
**Method 1:**
| $[1 - P(5,6,7) =]$ $1 - (^7C_5\ 0.65^5\ 0.35^2 + ^7C_6\ 0.65^6\ 0.35^1 + 0.65^7)$ | M1 | One term $^7C_x(p)^x(1-p)^{7-x}$, $0 < p < 1$, $0 < x < 7$ |
|---|---|---|
| $[= 1-(0.29848 + 0.18478 + 0.049022)]$ | A1 | Correct expression, accept unsimplified, no terms omitted leading to final answer |
| $= 0.468$ | B1 | AWRT |
**Method 2:**
| $[P(0,1,2,3,4) =]$ $0.35^7 + ^7C_1\ 0.65^1\ 0.35^6 + ^7C_2\ 0.65^2\ 0.35^5 + ^7C_3\ 0.65^3\ 0.35^4 + ^7C_4\ 0.65^4\ 0.35^3$ | (M1) | One term $^7C_x(p)^x(1-p)^{7-x}$, $0 < p < 1$, $0 < x < 7$ |
|---|---|---|
| $[0.00064 + 0.00836 + 0.04660 + 0.14424 + 0.26787]$ | (A1) | Correct expression, accept unsimplified, no terms omitted leading to final answer |
| $= 0.468$ | (B1) | AWRT |
**Total: 3 marks**
---
## Question 5(d):
| $[\text{Mean} =]\ 84 \times 0.65 = 54.6$ $[\text{Var} =]\ 84 \times 0.65 \times 0.35 = 19.11$ | B1 | 54.6 and 19.11 seen, allow unsimplified. $[\sigma =]\ 4.371$ or $\frac{7\sqrt{39}}{10}$ implies correct variance. Incorrect notation penalised, but condone use of values in standardisation formula |
|---|---|---|
| $P(X > 50) = P\!\left(Z > \dfrac{50.5 - 54.6}{\sqrt{19.11}}\right)$ | M1 | Substituting *their* $\mu$ and *their* positive $\sigma$ into the $\pm$standardising formula (any number), not *their* $\sigma^2$, or $\sqrt{\textit{their}\ \sigma}$ |
| | M1 | Use continuity correction 49.5 or 50.5 in *their* standardisation formula. Note: $\frac{\pm 4.1}{\sqrt{19.11}}$ or $\pm\frac{4.1}{4.371}$ seen gains M2 BOD |
| $P(Z > -0.9379) = \Phi(0.9379)$ | M1 | Appropriate area $\Phi$, from final process, must be a probability |
| $0.826$ | A1 | $0.8255 < p \leq 0.826$ |
**Total: 5 marks**
---5 Salah decides to attempt the crossword puzzle in his newspaper each day. The probability that he will complete the puzzle on any given day is 0.65 , independent of other days.\\[0pt]
\begin{enumerate}[label=(\alph*)]
\item Find the probability that Salah completes the puzzle for the first time on the 5th day. [1]
\item Find the probability that Salah completes the puzzle for the second time on the 5th day.
\item Find the probability that Salah completes the puzzle fewer than 5 times in a week (7 days). [3]\\
\includegraphics[max width=\textwidth, alt={}, center]{9b21cc0f-b043-4251-8aa9-cb1e5c2fb5d0-10_2713_31_145_2014}
\item Use a suitable approximation to find the probability that Salah completes the puzzle more than 50 times in a period of 84 days.
\end{enumerate}
\hfill \mbox{\textit{CAIE S1 2024 Q5 [11]}}