| Exam Board | AQA |
|---|---|
| Module | S1 (Statistics 1) |
| Year | 2013 |
| Session | June |
| Marks | 11 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Conditional Probability |
| Type | Tree diagram with two-stage events |
| Difficulty | Moderate -0.8 This is a straightforward conditional probability question using basic probability rules (multiplication and addition) with clearly stated probabilities. Part (a) involves simple independent events, while part (b) applies the definition of conditional probability in a structured way. All required probabilities are given explicitly, requiring only careful application of P(A∩B) = P(A)×P(B|A) with no complex reasoning or insight needed. |
| Spec | 2.03a Mutually exclusive and independent events2.03b Probability diagrams: tree, Venn, sample space2.03c Conditional probability: using diagrams/tables |
| Answer | Marks |
|---|---|
| \(P(\text{both}) = 0.90 \times 0.95\) | M1 |
| \(= 0.855\) | A1 |
| Answer | Marks |
|---|---|
| \(P(\text{exactly one}) = 0.90 \times 0.05 + 0.10 \times 0.95\) | M1 |
| \(= 0.045 + 0.095 = 0.140\) | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| \(P = P(D | A) \times P(A) = 0.80 \times 0.90\) | M1 |
| \(= 0.720\) | A1 |
| Answer | Marks |
|---|---|
| \(P(\text{both Wed}) \times P(\text{both Thu}) = 0.720 \times (1 \times 0.95)\) | M1 |
| \(= 0.720 \times 0.95 = 0.684\) | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| \(P(A_T) \times P(D' | A_T) = 0.95 \times (1-1)= 0.95 \times 0 = 0\) | B1 |
| Answer | Marks |
|---|---|
| \(P(\text{neither Wed}) = 1 - P(\text{at least one Wed})\) | M1 |
| \(P(\text{Alison not Wed}) = 0.10\); \(P(\text{David not Wed} | \text{Alison not}) = 1 - 0.15 = 0.85\) |
| \(P(\text{neither Wed}) = 0.10 \times 0.85 + ... \) working through tree | M1 |
| \(P(\text{neither Thu}) = 0.05 \times 1 = 0.05\) (since if Alison doesn't bowl Thu, David doesn't) | |
| \(P(\text{neither either}) = (0.10 \times 0.85) \times 0.05\) | |
| \(= 0.085 \times 0.05 = 0.00425\) | A1 |
# Question 5:
## Part (a)(i) - Two evenings
$P(\text{both}) = 0.90 \times 0.95$ | M1 | |
$= 0.855$ | A1 | |
## Part (a)(ii) - Exactly one evening
$P(\text{exactly one}) = 0.90 \times 0.05 + 0.10 \times 0.95$ | M1 | |
$= 0.045 + 0.095 = 0.140$ | A1 | |
## Part (b)(i) - Alison and David bowl Wednesday
$P = P(D|A) \times P(A) = 0.80 \times 0.90$ | M1 | |
$= 0.720$ | A1 | |
## Part (b)(ii) - Both bowl on both evenings
$P(\text{both Wed}) \times P(\text{both Thu}) = 0.720 \times (1 \times 0.95)$ | M1 | |
$= 0.720 \times 0.95 = 0.684$ | A1 | |
## Part (b)(iii) - Alison but not David, Thursday
$P(A_T) \times P(D'|A_T) = 0.95 \times (1-1)= 0.95 \times 0 = 0$ | B1 | |
## Part (b)(iv) - Neither bowls either evening
$P(\text{neither Wed}) = 1 - P(\text{at least one Wed})$ | M1 | |
$P(\text{Alison not Wed}) = 0.10$; $P(\text{David not Wed}| \text{Alison not}) = 1 - 0.15 = 0.85$ | | |
$P(\text{neither Wed}) = 0.10 \times 0.85 + ... $ working through tree | M1 | |
$P(\text{neither Thu}) = 0.05 \times 1 = 0.05$ (since if Alison doesn't bowl Thu, David doesn't) | | |
$P(\text{neither either}) = (0.10 \times 0.85) \times 0.05$ | | |
$= 0.085 \times 0.05 = 0.00425$ | A1 | |5 Alison is a member of a tenpin bowling club which meets at a bowling alley on Wednesday and Thursday evenings.
The probability that she bowls on a Wednesday evening is 0.90 . Independently, the probability that she bowls on a Thursday evening is 0.95 .
\begin{enumerate}[label=(\alph*)]
\item Calculate the probability that, during a particular week, Alison bowls on:
\begin{enumerate}[label=(\roman*)]
\item two evenings;
\item exactly one evening.
\end{enumerate}\item David, a friend of Alison, is a member of the same club.
The probability that he bowls on a Wednesday evening, given that Alison bowls on that evening, is 0.80 . The probability that he bowls on a Wednesday evening, given that Alison does not bowl on that evening, is 0.15 .
The probability that he bowls on a Thursday evening, given that Alison bowls on that evening, is 1 . The probability that he bowls on a Thursday evening, given that Alison does not bowl on that evening, is 0 .
Calculate the probability that, during a particular week:
\begin{enumerate}[label=(\roman*)]
\item Alison and David bowl on a Wednesday evening;
\item Alison and David bowl on both evenings;
\item Alison, but not David, bowls on a Thursday evening;
\item neither bowls on either evening.
\end{enumerate}\end{enumerate}
\hfill \mbox{\textit{AQA S1 2013 Q5 [11]}}