| Exam Board | CAIE |
|---|---|
| Module | S1 (Statistics 1) |
| Year | 2022 |
| Session | March |
| Marks | 6 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Binomial Distribution |
| Type | Independent binomial samples with compound probability |
| Difficulty | Standard +0.3 This is a straightforward two-part binomial distribution question requiring standard probability calculations. Part (a) involves computing P(X > 2) for X ~ B(7, 0.18) using complement rule. Part (b) requires recognizing a binomial-of-binomial structure: first finding P(at least one day) = 1 - (0.82)^7, then using this in a second binomial B(3, p). Both parts are routine applications of standard techniques with no novel insight required, making it slightly easier than average. |
| Spec | 2.04b Binomial distribution: as model B(n,p)2.04c Calculate binomial probabilities |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \([P(>2) = 1 - P(0,1,2) =]\) \(1 - (^7C_0\ 0.18^0\ 0.82^7 + ^7C_1\ 0.18^1\ 0.82^6 + ^7C_2\ 0.18^2\ 0.82^5)\) | M1 | One term \(^7C_x\ p^x(1-p)^{7-x}\), \(0 < p < 1\), \(0 < x < 7\) |
| \(= 1 - (0.249285 + 0.383048 + 0.252251) = 1 - 0.88458\) | A1 | Correct unsimplified expression or better. Condone omission of brackets if recovered |
| \(0.115\) | B1 | WWW. \(0.115 \leqslant p < 0.1155\) not from wrong working |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \([P(\text{at least 1 day of rain}) = 1 - P(0) = 1-(0.82)^7 =]\ 0.7507\) | B1 | AWRT 0.751 seen |
| \([P(\text{exactly 2 periods}) =]\ 0.7507^2 \times (1-0.7507) \times 3\) | M1 | FT *their* \(1-p^7\) or *their* 0.7507 if identified, not 0.18, 0.82. Accept \(^x C_r\), r=1,2 or \(^x P_1\) for \(\times 3\). Condone \(\times 2\) |
| \(0.421\) | A1 | Accept \(0.421 \leqslant p \leqslant 0.4215\). SC B1 if 0/3 scored for final answer only \(0.421 \leqslant p \leqslant 0.4215\) |
## Question 2(a):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $[P(>2) = 1 - P(0,1,2) =]$ $1 - (^7C_0\ 0.18^0\ 0.82^7 + ^7C_1\ 0.18^1\ 0.82^6 + ^7C_2\ 0.18^2\ 0.82^5)$ | M1 | One term $^7C_x\ p^x(1-p)^{7-x}$, $0 < p < 1$, $0 < x < 7$ |
| $= 1 - (0.249285 + 0.383048 + 0.252251) = 1 - 0.88458$ | A1 | Correct unsimplified expression or better. Condone omission of brackets if recovered |
| $0.115$ | B1 | WWW. $0.115 \leqslant p < 0.1155$ not from wrong working |
## Question 2(b):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $[P(\text{at least 1 day of rain}) = 1 - P(0) = 1-(0.82)^7 =]\ 0.7507$ | B1 | AWRT 0.751 seen |
| $[P(\text{exactly 2 periods}) =]\ 0.7507^2 \times (1-0.7507) \times 3$ | M1 | FT *their* $1-p^7$ or *their* 0.7507 if identified, not 0.18, 0.82. Accept $^x C_r$, r=1,2 or $^x P_1$ for $\times 3$. Condone $\times 2$ |
| $0.421$ | A1 | Accept $0.421 \leqslant p \leqslant 0.4215$. **SC B1** if 0/3 scored for final answer only $0.421 \leqslant p \leqslant 0.4215$ |2 In a certain country, the probability of more than 10 cm of rain on any particular day is 0.18 , independently of the weather on any other day.
\begin{enumerate}[label=(\alph*)]
\item Find the probability that in any randomly chosen 7-day period, more than 2 days have more than 10 cm of rain.
\item For 3 randomly chosen 7-day periods, find the probability that exactly two of these periods have at least one day with more than 10 cm of rain.
\end{enumerate}
\hfill \mbox{\textit{CAIE S1 2022 Q2 [6]}}