| Exam Board | Edexcel |
|---|---|
| Module | S1 (Statistics 1) |
| Year | 2012 |
| Session | June |
| Marks | 9 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Probability Definitions |
| Type | Mutually exclusive events |
| Difficulty | Easy -1.3 This is a straightforward Venn diagram question testing basic probability definitions (mutually exclusive, independence) and simple probability calculations. All parts require only reading values from the diagram and applying standard formulas with no problem-solving or insight needed. This is easier than average A-level content. |
| Spec | 2.03a Mutually exclusive and independent events2.03c Conditional probability: using diagrams/tables |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(B, W\) or \(T, W\) [accept \(B \cup T, W\) or \(B \cap T, W\)] [condone P(\(B\)), P(\(W\)) etc] | B1 | 1st B1 for a suitable pair; do not accept universally exclusive such as \(B\) and \(B'\) etc |
| Since there is no overlap between the events, or cannot happen together (mutually exclusive); e.g. "no one walks and takes the train" | B1 | 2nd B1 for any correct statement; accept use of symbols e.g. \(B \cap W = \varnothing\) or \(P(T \cap W) = 0\); but \(T \cap W = 0\) is B0 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| e.g. \(P(B) = \frac{9}{25}\), \(P(T) = \frac{8}{25}\), \(P(B \cap T) = \frac{5}{25}\) | M1 | 1st M1 for attempt at all required probabilities with labels; allow one error; must be probabilities not integers |
| \(P(B \cap T) \neq P(B) \times P(T)\) \([0.2 \neq 0.36 \times 0.32 = 0.1152]\) | M1 | 2nd M1 for \(P(B) \times P(T)\) evaluated; or \(P(B \cap T) \neq P(B) \times P(T)\) stated or implied |
| So \(B\) and \(T\) are not independent | A1cso | Requires all probabilities correct and seen; A mark dependent on both M marks. NB: \(P(B |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(P(W) = \frac{7}{25}\) or \(0.28\) | B1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(P(B \cap T) = \frac{5}{25}\) or \(\frac{1}{5}\) or \(0.2\) | B1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(P(T | B) = \frac{P(T \cap B)}{P(B)} = \frac{\text{"(d)"}}{(5+4)/25}\) | M1 |
| \(= \frac{5}{9}\) or \(0.\overline{5}\) | A1 | \(\frac{5}{9}\) with no incorrect working; \(\frac{5}{9}\) following from \(P(B |
# Question 4:
## Part (a):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $B, W$ or $T, W$ [accept $B \cup T, W$ or $B \cap T, W$] [condone P($B$), P($W$) etc] | B1 | 1st B1 for a suitable pair; do not accept universally exclusive such as $B$ and $B'$ etc |
| Since there is no overlap between the events, or cannot happen together (mutually exclusive); e.g. "no one walks and takes the train" | B1 | 2nd B1 for any correct statement; accept use of symbols e.g. $B \cap W = \varnothing$ or $P(T \cap W) = 0$; but $T \cap W = 0$ is B0 |
## Part (b):
| Answer/Working | Marks | Guidance |
|---|---|---|
| e.g. $P(B) = \frac{9}{25}$, $P(T) = \frac{8}{25}$, $P(B \cap T) = \frac{5}{25}$ | M1 | 1st M1 for attempt at all required probabilities with labels; allow one error; must be probabilities not integers |
| $P(B \cap T) \neq P(B) \times P(T)$ $[0.2 \neq 0.36 \times 0.32 = 0.1152]$ | M1 | 2nd M1 for $P(B) \times P(T)$ evaluated; or $P(B \cap T) \neq P(B) \times P(T)$ stated or implied |
| So $B$ and $T$ are not independent | A1cso | Requires all probabilities correct and seen; A mark dependent on both M marks. NB: $P(B|T) = \frac{5}{8}$ & $P(B) = \frac{9}{25}$ or $P(T|B) = \frac{5}{9}$ & $P(T) = \frac{8}{25}$ seen with correct conclusion scores 3/3 |
## Part (c):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $P(W) = \frac{7}{25}$ or $0.28$ | B1 | |
## Part (d):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $P(B \cap T) = \frac{5}{25}$ or $\frac{1}{5}$ or $0.2$ | B1 | |
## Part (e):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $P(T|B) = \frac{P(T \cap B)}{P(B)} = \frac{\text{"(d)"}}{(5+4)/25}$ | M1 | For a correct ratio of probabilities e.g. $\frac{5/25}{(5+4)/25}$ or $\frac{5}{5+4}$; correct ratio expression with at least one correct probability substituted |
| $= \frac{5}{9}$ or $0.\overline{5}$ | A1 | $\frac{5}{9}$ with no incorrect working; $\frac{5}{9}$ following from $P(B|T)$ is 0/2; $\frac{5}{9}$ alone is 2/2 |
---4.
\begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{0593544d-392d-465b-b922-c9cb1435abb5-06_611_1127_237_447}
\captionsetup{labelformat=empty}
\caption{Figure 1}
\end{center}
\end{figure}
Figure 1 shows how 25 people travelled to work.\\
Their travel to work is represented by the events
$$\begin{array} { l l }
B & \text { bicycle } \\
T & \text { train } \\
W & \text { walk }
\end{array}$$
\begin{enumerate}[label=(\alph*)]
\item Write down 2 of these events that are mutually exclusive. Give a reason for your answer.
\item Determine whether or not $B$ and $T$ are independent events.
One person is chosen at random.\\
Find the probability that this person
\item walks to work,
\item travels to work by bicycle and train.
\item Given that this person travels to work by bicycle, find the probability that they will also take the train.
\end{enumerate}
\hfill \mbox{\textit{Edexcel S1 2012 Q4 [9]}}