| Exam Board | CAIE |
|---|---|
| Module | S1 (Statistics 1) |
| Year | 2024 |
| Session | November |
| Marks | 10 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Geometric Distribution |
| Type | First success before/after trial n |
| Difficulty | Standard +0.8 Part (a) is a standard binomial calculation. Part (b) is straightforward geometric distribution (complement of first success on trial 7+). Part (c) requires conditional probability with geometric distributions and careful reasoning about the constraint that trials 1-4 must all be wrapped (gold or red) with specific success patterns—this elevates it above routine exercises but remains within standard A-level techniques. |
| Spec | 2.03d Calculate conditional probability: from first principles2.04b Binomial distribution: as model B(n,p)2.04c Calculate binomial probabilities5.02f Geometric distribution: conditions5.02g Geometric probabilities: P(X=r) = p(1-p)^(r-1) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \([P(0,1,2) =]\ ^8C_2(0.75)^6(0.25)^2 + ^8C_1(0.75)^7(0.25)^1 + (0.75)^8\) | M1 | One term \(^8C_x\). \((p)^x(1-p)^{8-x}\), \(0 < p < 1\), \(0 < x < 8\) |
| \([= 0.31146 + 0.26697 + 0.10011]\) | A1 | Correct expression, accept unsimplified, no terms omitted leading to final answer |
| \(= 0.679\) | B1 | AWRT |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \([P(0,1,2) = 1 - P(3,4,5,6,7,8) =]\ 1 - \{^8C_3(0.75)^5(0.25)^3 + ^8C_4(0.75)^4(0.25)^4 + ^8C_5(0.75)^3(0.25)^5 + ^8C_6(0.75)^2(0.25)^6 + ^8C_7(0.75)(0.25)^7 + (0.25)^8\}\) | M1 | One term \(^8C_x(p)^x(1-p)^{8-x}\), \(0 < p < 1\), \(0 < x < 8\) |
| A1 | Correct expression, accept unsimplified, condone omission of up to 3 'middle' terms, leading to final answer | |
| \(= 0.679\) | B1 | AWRT |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(1 - 0.75^6\) | M1 | \(1 - 0.75^n\), \(n = 6, 7\) |
| \(= 0.822,\ \frac{3367}{4096}\) | A1 | \(0.82202148\ldots\) to at least 3SF |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(0.25 + 0.25 \times 0.75 + 0.25 \times 0.75^2 + 0.25 \times 0.75^3 + 0.25 \times 0.75^4 + 0.25 \times 0.75^5\) | M1 | Summing 6 or 7 terms – condone extra term \(0.25 \times 0.75^6\) |
| \(= 0.822\) | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(1 - 0.75^7 - 0.25 \times 0.75^6\) | M1 | Correct expression |
| \(= 0.822\) | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(P(\text{2nd gold} \cap \text{5th is first unwrapped})\): R G RorG RorG U | ||
| \(0.25 \times 0.3 \times 0.55 \times 0.55 \times 0.45\ [= 0.01021]\) on its own or as a numerator | M1 | \(a \times 0.3 \times b \times c \times 0.45\), \(0 < a, b, c < 1\), \(a \neq 0.3, 0.45\ b\), \(c \neq 0.45\), multiplied in that order, or correct |
| A1 | 5 correct probabilities multiplied |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| RGRGU: \(0.25 \times 0.3 \times 0.25 \times 0.3 \times 0.45\ [= 0.00253125]\) | M1 | \(a \times 0.3 \times b \times c \times 0.45\), \(0 < a, b, c < 1\), \(a \neq 0.3, 0.45\ b\), \(c \neq 0.45\). 4 terms in this form seen added on their own or as a numerator |
| RGRRU: \(0.25 \times 0.3 \times 0.25 \times 0.25 \times 0.45\ [= 0.002109375]\) | ||
| RGGRU: \(0.25 \times 0.3 \times 0.3 \times 0.25 \times 0.45\ [= 0.00253125]\) | A1 | All probabilities correct and attempt to sum the 4 scenarios |
| RGGGU: \(0.25 \times 0.3 \times 0.3 \times 0.3 \times 0.45\ [= 0.0030375]\) | ||
| Total: \(0.010209375\) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \([P(\text{5th is first unwrapped}) =]\ (0.55)^4(0.45)\ [= 0.041178]\) | B1 | |
| \([P(\text{2nd is first gold} \mid \text{5th is first unwrapped}) =]\ \frac{0.25 \times 0.3 \times 0.55 \times 0.55 \times 0.45}{(0.55)^4(0.45)}\) | M1 | \(\frac{\textit{their}\ P(\text{2nd gold} \cap \text{5th is first unwrapped})}{\textit{their}\ P(\text{5th is first unwrapped})}\). Their probabilities must be clearly identified if incorrect |
| \(= 0.248,\ \frac{30}{121}\) | A1 | \(0.24793\ldots\) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(P(\text{red given that it is wrapped}) \times P(\text{gold given that it is wrapped}) = \frac{0.25}{0.55} \times \frac{0.3}{0.55}\) | M1 | Either \(\frac{0.25}{0.55\ \text{or}\ 0.45}\) or \(\frac{0.3}{0.55\ \text{or}\ 0.45}\) |
| A1 | Either \(\frac{0.25}{0.55}\) or \(\frac{0.3}{0.55}\) | |
| B1 | Both probs correct, can be unsimplified | |
| M1 | Multiplying their identified \(P(\text{red given wrapped})\) by their identified \(P(\text{gold given wrapped})\) or correct | |
| \(= 0.248,\ \frac{30}{121}\) | A1 |
## Question 5(a):
**Method 1**
| Answer | Marks | Guidance |
|--------|-------|----------|
| $[P(0,1,2) =]\ ^8C_2(0.75)^6(0.25)^2 + ^8C_1(0.75)^7(0.25)^1 + (0.75)^8$ | M1 | One term $^8C_x$. $(p)^x(1-p)^{8-x}$, $0 < p < 1$, $0 < x < 8$ |
| $[= 0.31146 + 0.26697 + 0.10011]$ | A1 | Correct expression, accept unsimplified, no terms omitted leading to final answer |
| $= 0.679$ | B1 | AWRT |
**Method 2**
| Answer | Marks | Guidance |
|--------|-------|----------|
| $[P(0,1,2) = 1 - P(3,4,5,6,7,8) =]\ 1 - \{^8C_3(0.75)^5(0.25)^3 + ^8C_4(0.75)^4(0.25)^4 + ^8C_5(0.75)^3(0.25)^5 + ^8C_6(0.75)^2(0.25)^6 + ^8C_7(0.75)(0.25)^7 + (0.25)^8\}$ | M1 | One term $^8C_x(p)^x(1-p)^{8-x}$, $0 < p < 1$, $0 < x < 8$ |
| | A1 | Correct expression, accept unsimplified, condone omission of up to 3 'middle' terms, leading to final answer |
| $= 0.679$ | B1 | AWRT |
---
## Question 5(b):
**Method 1**
| Answer | Marks | Guidance |
|--------|-------|----------|
| $1 - 0.75^6$ | M1 | $1 - 0.75^n$, $n = 6, 7$ |
| $= 0.822,\ \frac{3367}{4096}$ | A1 | $0.82202148\ldots$ to at least 3SF |
**Method 2**
| Answer | Marks | Guidance |
|--------|-------|----------|
| $0.25 + 0.25 \times 0.75 + 0.25 \times 0.75^2 + 0.25 \times 0.75^3 + 0.25 \times 0.75^4 + 0.25 \times 0.75^5$ | M1 | Summing 6 or 7 terms – condone extra term $0.25 \times 0.75^6$ |
| $= 0.822$ | A1 | |
**Method 3**
| Answer | Marks | Guidance |
|--------|-------|----------|
| $1 - 0.75^7 - 0.25 \times 0.75^6$ | M1 | Correct expression |
| $= 0.822$ | A1 | |
---
## Question 5(c):
**Method 1**
| Answer | Marks | Guidance |
|--------|-------|----------|
| $P(\text{2nd gold} \cap \text{5th is first unwrapped})$: R G RorG RorG U | | |
| $0.25 \times 0.3 \times 0.55 \times 0.55 \times 0.45\ [= 0.01021]$ on its own or as a numerator | M1 | $a \times 0.3 \times b \times c \times 0.45$, $0 < a, b, c < 1$, $a \neq 0.3, 0.45\ b$, $c \neq 0.45$, multiplied in that order, or correct |
| | A1 | 5 correct probabilities multiplied |
**Method 2** (4 possible scenarios)
| Answer | Marks | Guidance |
|--------|-------|----------|
| RGRGU: $0.25 \times 0.3 \times 0.25 \times 0.3 \times 0.45\ [= 0.00253125]$ | M1 | $a \times 0.3 \times b \times c \times 0.45$, $0 < a, b, c < 1$, $a \neq 0.3, 0.45\ b$, $c \neq 0.45$. 4 terms in this form seen added on their own or as a numerator |
| RGRRU: $0.25 \times 0.3 \times 0.25 \times 0.25 \times 0.45\ [= 0.002109375]$ | | |
| RGGRU: $0.25 \times 0.3 \times 0.3 \times 0.25 \times 0.45\ [= 0.00253125]$ | A1 | All probabilities correct and attempt to sum the 4 scenarios |
| RGGGU: $0.25 \times 0.3 \times 0.3 \times 0.3 \times 0.45\ [= 0.0030375]$ | | |
| Total: $0.010209375$ | | |
**For either approach:**
| Answer | Marks | Guidance |
|--------|-------|----------|
| $[P(\text{5th is first unwrapped}) =]\ (0.55)^4(0.45)\ [= 0.041178]$ | B1 | |
| $[P(\text{2nd is first gold} \mid \text{5th is first unwrapped}) =]\ \frac{0.25 \times 0.3 \times 0.55 \times 0.55 \times 0.45}{(0.55)^4(0.45)}$ | M1 | $\frac{\textit{their}\ P(\text{2nd gold} \cap \text{5th is first unwrapped})}{\textit{their}\ P(\text{5th is first unwrapped})}$. Their probabilities must be clearly identified if incorrect |
| $= 0.248,\ \frac{30}{121}$ | A1 | $0.24793\ldots$ |
**Method 3**
| Answer | Marks | Guidance |
|--------|-------|----------|
| $P(\text{red given that it is wrapped}) \times P(\text{gold given that it is wrapped}) = \frac{0.25}{0.55} \times \frac{0.3}{0.55}$ | M1 | Either $\frac{0.25}{0.55\ \text{or}\ 0.45}$ or $\frac{0.3}{0.55\ \text{or}\ 0.45}$ |
| | A1 | Either $\frac{0.25}{0.55}$ or $\frac{0.3}{0.55}$ |
| | B1 | Both probs correct, can be unsimplified |
| | M1 | Multiplying their identified $P(\text{red given wrapped})$ by their identified $P(\text{gold given wrapped})$ or correct |
| $= 0.248,\ \frac{30}{121}$ | A1 | |
---5 A factory produces chocolates. 30\% of the chocolates are wrapped in gold foil, 25\% are wrapped in red foil and the remainder are unwrapped.
Indigo chooses 8 chocolates at random from the production line.
\begin{enumerate}[label=(\alph*)]
\item Find the probability that she obtains no more than 2 chocolates that are wrapped in red foil.\\
Jake chooses chocolates one at a time at random from the production line.
\item Find the probability that the first time he obtains a chocolate that is wrapped in red foil is before the 7th choice.\\
\includegraphics[max width=\textwidth, alt={}, center]{915661eb-2544-4293-af72-608fedb43d70-08_2720_35_106_2015}\\
\includegraphics[max width=\textwidth, alt={}, center]{915661eb-2544-4293-af72-608fedb43d70-09_2717_29_105_22}
Keifa chooses chocolates one at a time at random from the production line.
\item Find the probability that the second chocolate chosen is the first one wrapped in gold foil given that the fifth chocolate chosen is the first unwrapped chocolate.
\end{enumerate}
\hfill \mbox{\textit{CAIE S1 2024 Q5 [10]}}