| Exam Board | Edexcel |
|---|---|
| Module | S1 (Statistics 1) |
| Year | 2015 |
| Session | January |
| Marks | 9 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Conditional Probability |
| Type | Conditional probability with algebraic expressions |
| Difficulty | Standard +0.3 This is a standard S1 Venn diagram question requiring basic probability rules (addition, conditional probability formula) and solving simultaneous equations. While it involves algebraic manipulation, the concepts are straightforward and the multi-step nature is typical for this level, making it slightly easier than average. |
| Spec | 2.03a Mutually exclusive and independent events2.03b Probability diagrams: tree, Venn, sample space2.03c Conditional probability: using diagrams/tables |
| Answer | Marks | Guidance |
|---|---|---|
| \(x + 0.1\) | B1 | \(P(x+0.1)\) is B0 |
| Answer | Marks | Guidance |
|---|---|---|
| \(P(B\mid A) = \frac{P(B\cap A)}{P(A)} = \frac{0.1}{x+0.1}\) | M1 A1 | M1 for correct ratio formula with at least one correct probability; A1 for \(\frac{0.1}{x+0.1}\) as final answer |
| Answer | Marks | Guidance |
|---|---|---|
| \(x + y + 0.1\) | B1 | \(P(x+y+0.1)\) is B0; condone \(x+y+0.1 = 1-0.32\) or \(0.68\) |
| Answer | Marks | Guidance |
|---|---|---|
| \(x+y+0.1+0.32=1\) or \(x+y+0.1=0.68\); \(x+0.1=2(y+0.1)\); eliminating \(x\) gives \(3y=0.48\); \(x=\mathbf{0.42}\), \(y=\mathbf{0.16}\) | M1, M1, M1, A1 A1 | 1st M1 sum of probs \(=1\); 2nd M1 using \(P(A)=2P(B)\); 3rd M1 solving two linear equations; A1 each for \(x=0.42\), \(y=0.16\) |
# Question 4:
## Part (a)(i)
| $x + 0.1$ | B1 | $P(x+0.1)$ is B0 |
## Part (a)(ii)
| $P(B\mid A) = \frac{P(B\cap A)}{P(A)} = \frac{0.1}{x+0.1}$ | M1 A1 | M1 for correct ratio formula with at least one correct probability; A1 for $\frac{0.1}{x+0.1}$ as final answer |
## Part (b)
| $x + y + 0.1$ | B1 | $P(x+y+0.1)$ is B0; condone $x+y+0.1 = 1-0.32$ or $0.68$ |
## Part (c)
| $x+y+0.1+0.32=1$ or $x+y+0.1=0.68$; $x+0.1=2(y+0.1)$; eliminating $x$ gives $3y=0.48$; $x=\mathbf{0.42}$, $y=\mathbf{0.16}$ | M1, M1, M1, A1 A1 | 1st M1 sum of probs $=1$; 2nd M1 using $P(A)=2P(B)$; 3rd M1 solving two linear equations; A1 each for $x=0.42$, $y=0.16$ |
---4. Events $A$ and $B$ are shown in the Venn diagram below\\
where $x , y , 0.10$ and 0.32 are probabilities.\\
\includegraphics[max width=\textwidth, alt={}, center]{c58f3e88-2dbc-40d6-a966-a5765a7c67ba-08_467_798_408_575}
\begin{enumerate}[label=(\alph*)]
\item Find an expression in terms of $x$ for
\begin{enumerate}[label=(\roman*)]
\item $\mathrm { P } ( A )$
\item $\mathrm { P } ( B \mid A )$
\end{enumerate}\item Find an expression in terms of $x$ and $y$ for $\mathrm { P } ( A \cup B )$
Given that $\mathrm { P } ( A ) = 2 \mathrm { P } ( B )$
\item find the value of $x$ and the value of $y$
\end{enumerate}
\hfill \mbox{\textit{Edexcel S1 2015 Q4 [9]}}