| Exam Board | Edexcel |
|---|---|
| Module | S1 (Statistics 1) |
| Session | Specimen |
| Marks | 18 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Conditional Probability |
| Type | Venn diagram with two events |
| Difficulty | Moderate -0.8 This is a straightforward S1 conditional probability question requiring basic application of P(A|B) = P(A∩B)/P(B), Venn diagram construction, and simple probability calculations. Part (a) is given as 'show that', parts (b)-(d) follow directly from the Venn diagram, and part (e) requires combining independent events but with clear structure. All techniques are standard textbook exercises with no novel insight required. |
| 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 |
|---|---|---|
| \(P(H \cap W) = P(H \mid W)P(W) = \frac{11}{12} \times \frac{1}{2} = \frac{11}{24}\) | M1 A1 | (2 marks) |
| Answer | Marks | Guidance |
|---|---|---|
| [Venn diagram with \(H\) and \(W\) circles showing: \(\frac{17}{120}\), \(\frac{11}{24}\), \(\frac{1}{24}\), \(\frac{43}{120}\)] | Diagram | M1 |
| \(H \cap W'\) | M1 A1 | |
| \(H' \cap W\) | A1 | |
| \(H \cap W\) | B1 | (5 marks) |
| Answer | Marks | Guidance |
|---|---|---|
| \(P(\text{only one has a degree}) = \frac{17}{120} + \frac{1}{24} = \frac{11}{60}\) | M1 A1 | (2 marks) |
| Answer | Marks |
|---|---|
| \(P(\text{neither has a degree}) = 1 - \left\{\frac{17}{120} + \frac{11}{24} + \frac{1}{24}\right\} = \frac{43}{120}\) | M1 A1 |
| A1 | (3 marks) |
| Answer | Marks |
|---|---|
| Any one | B1 |
| All correct | B1 |
| Answer | Marks | Guidance |
|---|---|---|
| \(2 \times \frac{17}{120} \times \frac{1}{24}\) | B1 ft | |
| \(2 \times \frac{11}{24} \times \frac{43}{120}\) | B1 ft | |
| Adding their probabilities | M1 | |
| \(= \frac{49}{144}\) | A1 | (6 marks) |
## (a)
$P(H \cap W) = P(H \mid W)P(W) = \frac{11}{12} \times \frac{1}{2} = \frac{11}{24}$ | M1 A1 | (2 marks)
## (b)
[Venn diagram with $H$ and $W$ circles showing: $\frac{17}{120}$, $\frac{11}{24}$, $\frac{1}{24}$, $\frac{43}{120}$] | Diagram | M1 |
$H \cap W'$ | M1 A1 |
$H' \cap W$ | A1 |
$H \cap W$ | B1 | (5 marks)
## (c)
$P(\text{only one has a degree}) = \frac{17}{120} + \frac{1}{24} = \frac{11}{60}$ | M1 A1 | (2 marks)
## (d)
$P(\text{neither has a degree}) = 1 - \left\{\frac{17}{120} + \frac{11}{24} + \frac{1}{24}\right\} = \frac{43}{120}$ | M1 A1 |
| A1 | (3 marks)
## (e)
**Possibilities:**
$(HW')(H'W); (H'W)(HW'); (HW)(H'W'); (H'W')(HW)$
Any one | B1 |
All correct | B1 |
$P(\text{only 1 H or 1 W}) = \left(2 \times \frac{17}{120} \times \frac{1}{24}\right) + \left(2 \times \frac{11}{24} \times \frac{43}{120}\right)$
$2 \times \frac{17}{120} \times \frac{1}{24}$ | B1 ft |
$2 \times \frac{11}{24} \times \frac{43}{120}$ | B1 ft |
Adding their probabilities | M1 |
$= \frac{49}{144}$ | A1 | (6 marks)
**(18 marks)**For any married couple who are members of a tennis club, the probability that the husband has a degree is $\frac{3}{5}$ and the probability that the wife has a degree is $\frac{1}{2}$. The probability that the husband has a degree, given that the wife has a degree, is $\frac{11}{12}$.
A married couple is chosen at random.
\begin{enumerate}[label=(\alph*)]
\item Show that the probability that both of them have degrees is $\frac{11}{24}$. [2]
\item Draw a Venn diagram to represent these data. [5]
\end{enumerate}
Find the probability that
\begin{enumerate}[label=(\alph*)]
\setcounter{enumi}{2}
\item only one of them has a degree, [2]
\item neither of them has a degree. [3]
\end{enumerate}
Two married couples are chosen at random.
\begin{enumerate}[label=(\alph*)]
\setcounter{enumi}{4}
\item Find the probability that only one of the two husbands and only one of the two wives have degrees. [6]
\end{enumerate}
\hfill \mbox{\textit{Edexcel S1 Q6 [18]}}