| Exam Board | AQA |
|---|---|
| Module | S1 (Statistics 1) |
| Year | 2012 |
| Session | January |
| Marks | 11 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Conditional Probability |
| Type | Venn diagram with two events |
| Difficulty | Moderate -0.8 This is a straightforward conditional probability question using standard formulas. Part (a) involves basic set operations with given probabilities (P(A∪B) = P(A) + P(B) - P(A∩B)), and part (b) applies the definition of conditional probability with clearly stated conditions. All steps are routine applications of formulas with no conceptual challenges or novel problem-solving 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 |
|---|---|---|
| \(A\) | \(A'\) | Total |
| \(E\) | 0.55 | 0.05 |
| \(E'\) | 0.30 | 0.10 |
| Total | 0.85 | 0.15 |
| 0.15 or 0.4 | B1 | |
| 0.05 and 0.3 | B1 | |
| \(\boxed{0.1}\) | Bdep1 | 3 |
| Answer | Marks | Guidance |
|---|---|---|
| 0.55; CAO | B1 | |
| 0.3 and 0.05; CAO | B1 | |
| \(\boxed{0.1}\); AG so dependent on B1 B1 | Bdep1 | 3 |
| Answer | Marks | Guidance |
|---|---|---|
| \(P(\geq 1) = 0.85 + 0.60 - 0.55\) OR \(0.85 + 0.60 - 0.55 + p = 1\) OR \(0.15 + 0.40 - 0.45\) | M2 (M1) | |
| \(P(0) = 1 - P(\geq 1)\) OR \(= 1 - 0.9 = \boxed{0.1}\) OR \(0.9 + p = 1\) OR \(= \boxed{0.1}\) | A1 | 3 |
| Answer | Marks | Guidance |
|---|---|---|
| \(P(\geq 1) = 0.9\) or \(\frac{9}{10}\) | B1 | 1 |
| Answer | Marks | Guidance |
|---|---|---|
| \(P(1) = 0.3 + 0.05 = 1 - (0.55 + 0.10) = 0.35\) or \(\frac{35}{100}\) or \(\frac{7}{20}\) | B1 | 1 |
| Answer | Marks | Guidance |
|---|---|---|
| \(P(3) = 0.55 \times 0.30\) | B1 | |
| \(= 0.165\) or \(\frac{165}{1000}\) or \(\frac{33}{200}\) | B1 | 2 |
| Answer | Marks | Guidance |
|---|---|---|
| \(0.55 \times (1 - 0.3)\) or \(0.385\) | M1 | |
| \((0.3 \times 0.75)\) or \(0.225\); \((0.05 \times 0.75)\) or \(0.0375\); \((0.35 \times 0.75)\) or \(0.2625\) | M1 | |
| \((0.385 + 0.2625) + 0.165\) | B1 | |
| \(= 0.812\) to \(0.813\) | A1 | 4 |
| or \(\frac{8125}{10000}\) or \(\frac{1625}{2000}\) or \(\frac{325}{400}\) or \(\frac{65}{80}\) or \(\frac{13}{16}\) |
| Answer | Marks | Guidance |
|---|---|---|
| \(0.1 \times (1 - 0.4)\) or \(0.06\) | M1 | |
| \((0.3 \times 0.25)\) or \(0.075\); \((0.05 \times 0.25)\) or \(0.0125\); \((0.35 \times 0.25)\) or \(0.0875\); \((0.1 \times 0.4)\) or \(0.04\) | M1 | |
| \(1 - (0.1875)\) | B1 | |
| \(= 0.812\) to \(0.813\) | A1 | 4 |
| Answer | Marks | Guidance |
|---|---|---|
| \((0.55 + p)\) where \(0 < p < 0.45\) | M1 | |
| \((0.3 \times 0.75)\) or \(0.225\); \((0.05 \times 0.75)\) or \(0.0375\); \((0.35 \times 0.75)\) or \(0.2625\) | M1 | |
| \(0.55 + 0.2625\) | B1 | |
| \(= 0.812\) to \(0.813\) | A1 | 4 |
# Question 6:
## Part (a)(i):
**Table Method** (2-way with either R or C totals)
| | $A$ | $A'$ | Total |
|---|---|---|---|
| $E$ | 0.55 | **0.05** | 0.60 |
| $E'$ | **0.30** | **0.10** | 0.40 |
| Total | 0.85 | 0.15 | 1.00 |
0.15 or 0.4 | B1 | | CAO; allow fractions
0.05 and 0.3 | B1 | | CAO; allow fractions
$\boxed{0.1}$ | Bdep1 | 3 | AG so dependent on B1 B1
**Venn Diagram Method:**
0.55; CAO | B1 | |
0.3 and 0.05; CAO | B1 | |
$\boxed{0.1}$; AG so dependent on B1 B1 | Bdep1 | 3 |
**Formula Method:**
$P(\geq 1) = 0.85 + 0.60 - 0.55$ OR $0.85 + 0.60 - 0.55 + p = 1$ OR $0.15 + 0.40 - 0.45$ | M2 (M1) | | Full justification for numerical expression; Insufficient justification or numerical expression only
$P(0) = 1 - P(\geq 1)$ OR $= 1 - 0.9 = \boxed{0.1}$ OR $0.9 + p = 1$ OR $= \boxed{0.1}$ | A1 | 3 | AG; gained from M2 or M1
## Part (a)(ii):
$P(\geq 1) = 0.9$ or $\frac{9}{10}$ | B1 | 1 | CAO
## Part (a)(iii):
$P(1) = 0.3 + 0.05 = 1 - (0.55 + 0.10) = 0.35$ or $\frac{35}{100}$ or $\frac{7}{20}$ | B1 | 1 | CAO
## Part (b)(i):
$P(3) = 0.55 \times 0.30$ | B1 | | OE; implied by correct answer
$= 0.165$ or $\frac{165}{1000}$ or $\frac{33}{200}$ | B1 | 2 | CAO
## Part (b)(ii):
$0.55 \times (1 - 0.3)$ or $0.385$ | M1 | |
$(0.3 \times 0.75)$ or $0.225$; $(0.05 \times 0.75)$ or $0.0375$; $(0.35 \times 0.75)$ or $0.2625$ | M1 | | At least one of these expressions or values
$(0.385 + 0.2625) + 0.165$ | B1 | | OE; implied by correct answer
$= 0.812$ to $0.813$ | A1 | 4 | AWFW (0.8125)
or $\frac{8125}{10000}$ or $\frac{1625}{2000}$ or $\frac{325}{400}$ or $\frac{65}{80}$ or $\frac{13}{16}$ | | | CAO
**Alternative (b)(ii):**
$0.1 \times (1 - 0.4)$ or $0.06$ | M1 | |
$(0.3 \times 0.25)$ or $0.075$; $(0.05 \times 0.25)$ or $0.0125$; $(0.35 \times 0.25)$ or $0.0875$; $(0.1 \times 0.4)$ or $0.04$ | M1 | | At least one of these expressions or values
$1 - (0.1875)$ | B1 | | OE; implied by correct answer
$= 0.812$ to $0.813$ | A1 | 4 | AWFW (0.8125); CAO for equivalent fraction
**Alternative (b)(ii) using $p$:**
$(0.55 + p)$ where $0 < p < 0.45$ | M1 | |
$(0.3 \times 0.75)$ or $0.225$; $(0.05 \times 0.75)$ or $0.0375$; $(0.35 \times 0.75)$ or $0.2625$ | M1 | | At least one of these expressions or values
$0.55 + 0.2625$ | B1 | | OE; implied by correct answer
$= 0.812$ to $0.813$ | A1 | 4 | AWFW (0.8125); CAO for equivalent fraction
**Total: 11 marks**
---6 Twins Alec and Eric are members of the same local cricket club and play for the club's under 18 team.
The probability that Alec is selected to play in any particular game is 0.85 .\\
The probability that Eric is selected to play in any particular game is 0.60 .\\
The probability that both Alec and Eric are selected to play in any particular game is 0.55 .
\begin{enumerate}[label=(\alph*)]
\item By using a table, or otherwise:
\begin{enumerate}[label=(\roman*)]
\item show that the probability that neither twin is selected for a particular game is 0.10 ;
\item find the probability that at least one of the twins is selected for a particular game;
\item find the probability that exactly one of the twins is selected for a particular game.
\end{enumerate}\item The probability that the twins' younger brother, Cedric, is selected for a particular game is:\\
0.30 given that both of the twins have been selected;\\
0.75 given that exactly one of the twins has been selected;\\
0.40 given that neither of the twins has been selected.
Calculate the probability that, for a particular game:
\begin{enumerate}[label=(\roman*)]
\item all three brothers are selected;
\item at least two of the three brothers are selected.\\
(6 marks)
\end{enumerate}\end{enumerate}
\hfill \mbox{\textit{AQA S1 2012 Q6 [11]}}