| Exam Board | CAIE |
|---|---|
| Module | S1 (Statistics 1) |
| Year | 2024 |
| Session | June |
| Marks | 6 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Binomial Distribution |
| Type | Binomial with complementary events |
| Difficulty | Standard +0.3 This is a straightforward binomial distribution problem with complementary events. Part (a) requires calculating P(at least one 4 in up to 3 throws) = 1 - (3/4)³, which is routine. Parts (b) and (c) apply basic binomial probability formulas with the calculated probability. While multi-step, it requires only standard techniques without novel insight, making it slightly easier than average. |
| Spec | 2.03b Probability diagrams: tree, Venn, sample space2.04c Calculate binomial probabilities |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| Method 1: \(1 - \left(\frac{3}{4}\right)^3 = \frac{37}{64}\) | M1 | \(1-(s)^3\), \(s = \frac{3}{4}\) or \(\frac{1}{4}\) |
| AG | A1 | |
| Method 2: \(\frac{1}{4} + \frac{1}{4} \times \frac{3}{4} + \frac{1}{4} \times \left(\frac{3}{4}\right)^2 = \frac{37}{64}\) | (M1) | \(t + t(1-t) + t(1-t)^2\), \(t = \frac{1}{4}\) or \(\frac{3}{4}\) |
| AG | (A1) | |
| Method 3: \(^3C_1 \times \frac{1}{4} \times \left(\frac{3}{4}\right)^2 + {^3C_2} \times \left(\frac{1}{4}\right)^2 \times \frac{3}{4} + {^3C_3} \times \left(\frac{1}{4}\right)^3 = \frac{37}{64}\) | (M1) | |
| AG | (A1) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| Method 1: \(\left(1 - \frac{37}{64}\right)^4 = 0.0317\) | B1FT | \(\left(1 - their(\textbf{a})\right)^4\), accept unsimplified. |
| Method 2: \(\left(\frac{3}{4}\right)^6 \times \left(\frac{3}{4}\right)^6 = 0.0317\) | (B1FT) | Accept unsimplified. |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| X3, Y1: \(\left(\frac{37}{64}\right)^3 \times \frac{37}{64} \times \left(\frac{27}{64}\right)^2 \times 3\ \ [= 0.059645]\) | B1 | Correct probability for 1 identified scenario. Accept unsimplified. |
| X2, Y0: \(\left(\frac{37}{64}\right)^2 \times \frac{27}{64} \times \left(\frac{27}{64}\right)^3 \times 3\ \ [= 0.03176]\) | M1 | Add values of 2 correct scenarios. Identification may be implied by correct unsimplified expressions (condone omission of \(\times 3\)). Values may not be probabilities. |
| Probability \(= 0.0914\) | A1 | If A0 scored, SC B1 for 0.0914 WWW. |
## Question 4(a):
| Answer | Marks | Guidance |
|--------|-------|----------|
| **Method 1:** $1 - \left(\frac{3}{4}\right)^3 = \frac{37}{64}$ | M1 | $1-(s)^3$, $s = \frac{3}{4}$ or $\frac{1}{4}$ |
| AG | A1 | |
| **Method 2:** $\frac{1}{4} + \frac{1}{4} \times \frac{3}{4} + \frac{1}{4} \times \left(\frac{3}{4}\right)^2 = \frac{37}{64}$ | (M1) | $t + t(1-t) + t(1-t)^2$, $t = \frac{1}{4}$ or $\frac{3}{4}$ |
| AG | (A1) | |
| **Method 3:** $^3C_1 \times \frac{1}{4} \times \left(\frac{3}{4}\right)^2 + {^3C_2} \times \left(\frac{1}{4}\right)^2 \times \frac{3}{4} + {^3C_3} \times \left(\frac{1}{4}\right)^3 = \frac{37}{64}$ | (M1) | |
| AG | (A1) | |
## Question 4(b):
| Answer | Marks | Guidance |
|--------|-------|----------|
| **Method 1:** $\left(1 - \frac{37}{64}\right)^4 = 0.0317$ | B1FT | $\left(1 - their(\textbf{a})\right)^4$, accept unsimplified. |
| **Method 2:** $\left(\frac{3}{4}\right)^6 \times \left(\frac{3}{4}\right)^6 = 0.0317$ | (B1FT) | Accept unsimplified. |
## Question 4(c):
| Answer | Marks | Guidance |
|--------|-------|----------|
| X3, Y1: $\left(\frac{37}{64}\right)^3 \times \frac{37}{64} \times \left(\frac{27}{64}\right)^2 \times 3\ \ [= 0.059645]$ | B1 | Correct probability for 1 identified scenario. Accept unsimplified. |
| X2, Y0: $\left(\frac{37}{64}\right)^2 \times \frac{27}{64} \times \left(\frac{27}{64}\right)^3 \times 3\ \ [= 0.03176]$ | M1 | Add values of 2 correct scenarios. Identification may be implied by correct unsimplified expressions (condone omission of $\times 3$). Values may not be probabilities. |
| Probability $= 0.0914$ | A1 | If A0 scored, **SC B1** for 0.0914 WWW. |4 A game for two players is played using a fair 4-sided dice with sides numbered 1, 2, 3 and 4. One turn consists of throwing the dice repeatedly up to a maximum of three times. When a 4 is obtained, no further throws are made during that turn. A player who obtains a 4 in their turn scores 1 point.
\begin{enumerate}[label=(\alph*)]
\item Show that the probability that a player obtains a 4 in one turn is $\frac { 37 } { 64 }$.\\
Xeno and Yao play this game.
\item Find the probability that neither Xeno nor Yao score any points in their first two turns.
\item Xeno and Yao each have three turns.
Find the probability that Xeno scores 2 more points than Yao.\\
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\includegraphics[max width=\textwidth, alt={}, center]{a909cef1-8a22-4cef-b0b7-c48316304c0c-07_74_1570_1994_324}\\
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\includegraphics[max width=\textwidth, alt={}, center]{a909cef1-8a22-4cef-b0b7-c48316304c0c-07_70_1570_2631_324}
\end{enumerate}
\hfill \mbox{\textit{CAIE S1 2024 Q4 [6]}}