| Exam Board | CAIE |
|---|---|
| Module | S1 (Statistics 1) |
| Year | 2024 |
| Session | November |
| Marks | 6 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Conditional Probability |
| Type | Basic two-way table probability |
| Difficulty | Easy -1.3 This is a straightforward conditional probability question using a two-way table with clear data. Parts (a)-(c) require only basic probability formulas P(A|B) = P(A∩B)/P(B) and checking independence, while part (d) is a standard binomial calculation. All steps are routine applications of formulas with no problem-solving insight required. |
| Spec | 2.03a Mutually exclusive and independent events2.03c Conditional probability: using diagrams/tables2.03d Calculate conditional probability: from first principles2.04c Calculate binomial probabilities |
| Tennis | Hockey | Netball | |
| Year 1 | 16 | 22 | 12 |
| Year 2 | 24 | 18 | 28 |
| Answer | Marks | Guidance |
|---|---|---|
| \([P(X | N) =] \frac{12}{40}\) | B1 |
Total: 1 mark
| Answer | Marks | Guidance |
|---|---|---|
| \([P(N | X) =] \frac{12}{50}\) | B1 |
Total: 1 mark
| Answer | Marks | Guidance |
|---|---|---|
| \(P(N \cap X) = \frac{12}{120}\), \(P(N) = \frac{40}{120}\), \(P(X) = \frac{50}{120}\) | B1 | \(P(N)\), \(P(X)\) and \(P(N \cap X)\) notation seen and equated to the values for \(P(N)\), \(P(X)\) and \(P(N \cap X)\). Calculation stated and evaluated. Not independent clearly stated. \(\frac{5}{36} \neq \frac{12}{120}\) does not need to be stated. All values OE. Condone consistent use of \(A\), \(B\) etc. If values for \(P(N)\), \(P(X)\) stated, accept \(P(N) \times P(X) = \frac{5}{36}\). |
Total: 1 mark
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(P(0,1,2) = (0.85)^8 + {}^8C_1(0.85)^7(0.15) + {}^8C_2(0.85)^6(0.15)^2\) \([= 0.27249 + 0.38469 + 0.23760]\) | M1 | One term of form \({}^8C_x(p)^x(1-p)^{8-x}\), with \(0 < p < 1\), \(x \neq 0\) or 8 |
| Correct unsimplified expression, no terms omitted leading to final answer | A1 | Correct unsimplified expression, no terms omitted leading to final answer |
| \(= 0.895\) | B1 | \(0.8945 \leqslant p \leqslant 0.895\) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(P(0,1,2) = 1 - \{{}^8C_3(0.85)^5(0.15)^3 + {}^8C_4(0.85)^4(0.15)^4 + {}^8C_5(0.85)^3(0.15)^5 + {}^8C_6(0.85)^2(0.15)^6 + {}^8C_7(0.85)(0.15)^7 + (0.15)^8\}\) | M1 | One term of form \({}^8C_x(p)^x(1-p)^{8-x}\), with \(0 < p < 1\), \(x \neq 0\) or 8 |
| Correct unsimplified expression. Condone omission of final bracket. If other brackets omitted, allow recovery if \(1 - 0.1052[\ldots]\) seen | A1 | |
| \(= 0.895\) | B1 | \(0.8945 \leqslant p \leqslant 0.895\) |
**Question 1:**
**Part (a):**
$[P(X|N) =] \frac{12}{40}$ | B1 | $0.3, \frac{3}{10}, 30\%$ OE.
Total: 1 mark
---
**Part (b):**
$[P(N|X) =] \frac{12}{50}$ | B1 | $0.24, \frac{6}{25}$ OE.
Total: 1 mark
---
**Part (c):**
$P(N \cap X) = \frac{12}{120}$, $P(N) = \frac{40}{120}$, $P(X) = \frac{50}{120}$ | B1 | $P(N)$, $P(X)$ and $P(N \cap X)$ notation seen and equated to the values for $P(N)$, $P(X)$ and $P(N \cap X)$. Calculation stated and evaluated. Not independent clearly stated. $\frac{5}{36} \neq \frac{12}{120}$ does not need to be stated. All values OE. Condone consistent use of $A$, $B$ etc. If values for $P(N)$, $P(X)$ stated, accept $P(N) \times P(X) = \frac{5}{36}$.
$\frac{40}{120} \times \frac{50}{120} = \frac{5}{36}, 0.138[8...] \neq \frac{12}{120}, 0.1$ Not independent
Total: 1 mark
## Question 1(d):
**Method 1**
| Answer | Marks | Guidance |
|--------|-------|----------|
| $P(0,1,2) = (0.85)^8 + {}^8C_1(0.85)^7(0.15) + {}^8C_2(0.85)^6(0.15)^2$ $[= 0.27249 + 0.38469 + 0.23760]$ | M1 | One term of form ${}^8C_x(p)^x(1-p)^{8-x}$, with $0 < p < 1$, $x \neq 0$ or 8 |
| Correct unsimplified expression, no terms omitted leading to final answer | A1 | Correct unsimplified expression, no terms omitted leading to final answer |
| $= 0.895$ | B1 | $0.8945 \leqslant p \leqslant 0.895$ |
**Method 2**
| Answer | Marks | Guidance |
|--------|-------|----------|
| $P(0,1,2) = 1 - \{{}^8C_3(0.85)^5(0.15)^3 + {}^8C_4(0.85)^4(0.15)^4 + {}^8C_5(0.85)^3(0.15)^5 + {}^8C_6(0.85)^2(0.15)^6 + {}^8C_7(0.85)(0.15)^7 + (0.15)^8\}$ | M1 | One term of form ${}^8C_x(p)^x(1-p)^{8-x}$, with $0 < p < 1$, $x \neq 0$ or 8 |
| Correct unsimplified expression. Condone omission of final bracket. If other brackets omitted, allow recovery if $1 - 0.1052[\ldots]$ seen | A1 | |
| $= 0.895$ | B1 | $0.8945 \leqslant p \leqslant 0.895$ |
**Total: 3 marks**
---1 At a college, the students choose exactly one of tennis, hockey or netball to play. The table shows the numbers of students in Year 1 and Year 2 at the college playing each of these sports.
\begin{center}
\begin{tabular}{ | c | c | c | c | }
\hline
& Tennis & Hockey & Netball \\
\hline
Year 1 & 16 & 22 & 12 \\
\hline
Year 2 & 24 & 18 & 28 \\
\hline
\end{tabular}
\end{center}
One student is chosen at random from the 120 students. Events $X$ and $N$ are defined as follows:\\
$X$ : the student is in Year 1\\
$N$ : the student plays netball.
\begin{enumerate}[label=(\alph*)]
\item Find $\mathrm { P } ( X \mid N )$.
\item Find $\mathrm { P } ( N \mid X )$.
\item Determine whether or not $X$ and $N$ are independent events.\\
One of the students who plays netball takes 8 shots at goal. On each shot, the probability that she will succeed is 0.15 , independently of all other shots.
\item Find the probability that she succeeds on fewer than 3 of these shots.
\end{enumerate}
\hfill \mbox{\textit{CAIE S1 2024 Q1 [6]}}