| Exam Board | CAIE |
|---|---|
| Module | S1 (Statistics 1) |
| Year | 2024 |
| Session | June |
| Marks | 7 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Discrete Probability Distributions |
| Type | Construct probability distribution from scenario |
| Difficulty | Standard +0.3 Part (a) requires systematic enumeration of outcomes for three independent coins with different probabilities—straightforward but tedious calculation. Part (b) involves setting up an equation from binomial-type probabilities and solving for p, which is standard S1 material requiring careful algebraic manipulation but no novel insight. |
| Spec | 5.02a Discrete probability distributions: general5.02b Expectation and variance: discrete random variables |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| Table with \(x\): 0, 1, 2, 3 and \(P(X=x)\): \(\frac{24}{60}\), \(\frac{26}{60}\), \(\frac{9}{60}\), \(\frac{1}{60}\) | B1 | Table with correct \(X\) values and at least one probability. Values need not be in order, lines may not be drawn, may be vertical, \(X\) and \(P(X)\) may be omitted. Condone any additional \(X\) values if probability stated as 0. |
| B1 | \(P(X=1)\) or \(P(X=2)\) correct and identified, need not be in table, accept unsimplified. | |
| B1 | Two more correct and identified probabilities, need not be in table, accept unsimplified. | |
| B1 | 4 correct probabilities linked with correct outcomes, may not be in table. Decimals correct to at least 3sf. SC B1 for four probabilities summing to 1 placed in a probability distribution table with the correct \(x\) values. |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \([P(Y=0) =] \frac{2}{3} \times \frac{3}{4} \times \frac{4}{5} \times (1-p)^2\); \([P(Y=5) =] \frac{1}{3} \times \frac{1}{4} \times \frac{1}{5} \times p^2\) | B1 | Either \(\frac{2}{3} \times \frac{3}{4} \times \frac{4}{5} \times (1-p)^2\), not \(\frac{2}{5}(1-p)^2\); or \(\frac{1}{3} \times \frac{1}{4} \times \frac{1}{5} \times p^2\), not \(\frac{1}{60} \times p^2\) |
| \(\frac{2}{3} \times \frac{3}{4} \times \frac{4}{5} \times (1-p)^2 = 6 \times \frac{1}{3} \times \frac{1}{4} \times \frac{1}{5} \times p^2\) | M1 | Equating and forming a 3 term quadratic equation. *Their* \(P(Y=0) = 6 \times\) *their* \(P(Y=5)\). |
| \(\left[24(1-p)^2 = 6 \times p^2\right]\) \(3p^2 - 8p + 4 = 0\) | ||
| \(p = \frac{2}{3}\) | A1 | Not dependent on B1. A0 if \(p = 2\) seen and not clearly rejected. SC B1 if \(p = \frac{2}{3}\) obtained from a correct quadratic with more than three terms. If \(p = 2\) seen and not clearly rejected, SC B0. |
## Question 6(a):
| Answer | Marks | Guidance |
|--------|-------|----------|
| Table with $x$: 0, 1, 2, 3 and $P(X=x)$: $\frac{24}{60}$, $\frac{26}{60}$, $\frac{9}{60}$, $\frac{1}{60}$ | B1 | Table with correct $X$ values and at least one probability. Values need not be in order, lines may not be drawn, may be vertical, $X$ and $P(X)$ may be omitted. Condone any additional $X$ values if probability stated as 0. |
| | B1 | $P(X=1)$ or $P(X=2)$ correct and identified, need not be in table, accept unsimplified. |
| | B1 | Two more correct and identified probabilities, need not be in table, accept unsimplified. |
| | B1 | 4 correct probabilities linked with correct outcomes, may not be in table. Decimals correct to at least 3sf. **SC B1** for four probabilities summing to 1 placed in a probability distribution table with the correct $x$ values. |
---
## Question 6(b):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $[P(Y=0) =] \frac{2}{3} \times \frac{3}{4} \times \frac{4}{5} \times (1-p)^2$; $[P(Y=5) =] \frac{1}{3} \times \frac{1}{4} \times \frac{1}{5} \times p^2$ | B1 | Either $\frac{2}{3} \times \frac{3}{4} \times \frac{4}{5} \times (1-p)^2$, not $\frac{2}{5}(1-p)^2$; or $\frac{1}{3} \times \frac{1}{4} \times \frac{1}{5} \times p^2$, not $\frac{1}{60} \times p^2$ |
| $\frac{2}{3} \times \frac{3}{4} \times \frac{4}{5} \times (1-p)^2 = 6 \times \frac{1}{3} \times \frac{1}{4} \times \frac{1}{5} \times p^2$ | M1 | Equating and forming a 3 term quadratic equation. *Their* $P(Y=0) = 6 \times$ *their* $P(Y=5)$. |
| $\left[24(1-p)^2 = 6 \times p^2\right]$ $3p^2 - 8p + 4 = 0$ | | |
| $p = \frac{2}{3}$ | A1 | Not dependent on B1. A0 if $p = 2$ seen and not clearly rejected. **SC B1** if $p = \frac{2}{3}$ obtained from a correct quadratic with more than three terms. If $p = 2$ seen and not clearly rejected, **SC B0**. |
---6 Harry has three coins:
\begin{itemize}
\item One coin is biased so that the probability of obtaining a head when it is thrown is $\frac { 1 } { 3 }$.
\item The second coin is biased so that the probability of obtaining a head when it is thrown is $\frac { 1 } { 4 }$.
\item The third coin is biased so that the probability of obtaining a head when it is thrown is $\frac { 1 } { 5 }$.
\end{itemize}
Harry throws the three coins. The random variable $X$ is the number of heads that he obtains.
\begin{enumerate}[label=(\alph*)]
\item Draw up the probability distribution table for $X$.\\
Harry has two other coins, each of which is biased so that the probability of obtaining a head when it is thrown is $p$. He throws all five coins at the same time. The random variable $Y$ is the number of heads that he obtains.
\item Given that $\mathrm { P } ( Y = 0 ) = 6 \mathrm { P } ( Y = 5 )$, find the value of $p$.
\end{enumerate}
\hfill \mbox{\textit{CAIE S1 2024 Q6 [7]}}