| Exam Board | Edexcel |
|---|---|
| Module | S1 (Statistics 1) |
| Year | 2013 |
| Session | January |
| Marks | 12 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Principle of Inclusion/Exclusion |
| Type | Three Events with Independence Constraints |
| Difficulty | Moderate -0.3 This is a standard S1 probability question testing inclusion-exclusion principle, conditional probability, independence, and Venn diagrams. Parts (a)-(c) are direct formula applications with given values. Part (d) requires careful calculation of regions but follows a routine method. Part (e) is straightforward once the Venn diagram is complete. While multi-part with several concepts, each step is textbook-standard with no novel problem-solving required, 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/tables2.03d Calculate conditional probability: from first principles |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(P(A \cup B) = 0.35 + 0.45 - 0.13\) or \(0.22 + 0.13 + 0.32\) | M1 | For attempt to use the addition rule. Correct substitution i.e. correct expression seen |
| \(= \mathbf{0.67}\) | A1 | For 0.67 only. Correct answer only scores 2/2 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(P(A' \mid B') = \dfrac{P(A' \cap B')}{P(B')}\) or \(\dfrac{0.33}{0.55}\) | M1 | For a correct ratio of probabilities or correct formula with at least one correct prob. Do not award for assuming independence i.e. \(\frac{P(A' \cap B')}{P(B')} = \frac{0.65 \times 0.55}{0.55}\) is M0 |
| \(= \dfrac{3}{5}\) or \(0.6\) | A1 | For 3/5 or any exact equivalent |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(P(B \cap C) = 0.45 \times 0.2\) | M1 | For correct expression. Need correct values for \(P(B)\) and \(P(C)\) seen |
| \(= \mathbf{0.09}\) | A1 | For 0.09 or any exact equivalent. Correct answer only scores 2/2 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| 3 intersecting circles in a box with zeros in regions for \(A \cap C\) | B1 | Allow 1st B1 for 3 intersecting circles in box with zeros in regions for \(A \cap C\). No box is B0 |
| 0.13 and 0.09 in correct places | B1ft | ft \(P(B \cap C)\) from (c). Do not accept "blank" for zero |
| Any 2 of 0.22, 0.22, 0.11 and 0.23 correct | B1 | |
| All 4 values correct | B1 | No labels \(A\), \(B\), \(C\) in (d) loses 1st B1 but can score other 3 by implication |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(P(B \cup C)' = 0.22 + 0.22\) or \(1 - [0.56]\) or \(1 - [0.13 + 0.23 + 0.09 + 0.11]\) | M1 | For a correct expression or follow through from Venn diagram. NB \(P(B') \times P(C') = 0.55 \times 0.8\) is OK. Do not ft "blank" for zero and M0 for negative probs |
| \(= \mathbf{0.44}\) | A1 | For 0.44 only. Correct answer only scores 2/2 |
## Question 7:
### Part (a):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $P(A \cup B) = 0.35 + 0.45 - 0.13$ or $0.22 + 0.13 + 0.32$ | M1 | For attempt to use the addition rule. Correct substitution i.e. correct expression seen |
| $= \mathbf{0.67}$ | A1 | For 0.67 only. Correct answer only scores 2/2 |
### Part (b):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $P(A' \mid B') = \dfrac{P(A' \cap B')}{P(B')}$ or $\dfrac{0.33}{0.55}$ | M1 | For a correct ratio of probabilities or correct formula with at least one correct prob. Do not award for assuming independence i.e. $\frac{P(A' \cap B')}{P(B')} = \frac{0.65 \times 0.55}{0.55}$ is M0 |
| $= \dfrac{3}{5}$ or $0.6$ | A1 | For 3/5 or any exact equivalent |
### Part (c):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $P(B \cap C) = 0.45 \times 0.2$ | M1 | For correct expression. Need correct values for $P(B)$ and $P(C)$ seen |
| $= \mathbf{0.09}$ | A1 | For 0.09 or any exact equivalent. Correct answer only scores 2/2 |
### Part (d):
| Answer/Working | Marks | Guidance |
|---|---|---|
| 3 intersecting circles in a box with zeros in regions for $A \cap C$ | B1 | Allow 1st B1 for 3 intersecting circles in box with zeros in regions for $A \cap C$. No box is B0 |
| 0.13 and 0.09 in correct places | B1ft | ft $P(B \cap C)$ from (c). Do not accept "blank" for zero |
| Any 2 of 0.22, 0.22, 0.11 and 0.23 correct | B1 | |
| All 4 values correct | B1 | No labels $A$, $B$, $C$ in (d) loses 1st B1 but can score other 3 by implication |
### Part (e):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $P(B \cup C)' = 0.22 + 0.22$ or $1 - [0.56]$ or $1 - [0.13 + 0.23 + 0.09 + 0.11]$ | M1 | For a correct expression or follow through from Venn diagram. NB $P(B') \times P(C') = 0.55 \times 0.8$ is OK. Do not ft "blank" for zero and M0 for negative probs |
| $= \mathbf{0.44}$ | A1 | For 0.44 only. Correct answer only scores 2/2 |
**Total: 12 marks**\begin{enumerate}
\item Given that
\end{enumerate}
$$\mathrm { P } ( A ) = 0.35 , \quad \mathrm { P } ( B ) = 0.45 \quad \text { and } \quad \mathrm { P } ( A \cap B ) = 0.13$$
find\\
(a) $\mathrm { P } ( A \cup B )$\\
(b) $\mathrm { P } \left( A ^ { \prime } \mid B ^ { \prime } \right)$
The event $C$ has $\mathrm { P } ( C ) = 0.20$\\
The events $A$ and $C$ are mutually exclusive and the events $B$ and $C$ are independent.\\
(c) Find $\mathrm { P } ( B \cap C )$\\
(d) Draw a Venn diagram to illustrate the events $A , B$ and $C$ and the probabilities for each region.\\
(e) Find $\mathrm { P } \left( [ B \cup C ] ^ { \prime } \right)$\\
\hfill \mbox{\textit{Edexcel S1 2013 Q7 [12]}}