| Exam Board | AQA |
|---|---|
| Module | AS Paper 2 (AS Paper 2) |
| Year | 2019 |
| Session | June |
| Marks | 6 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Independent Events |
| Type | Both independence and mutual exclusivity |
| Difficulty | Moderate -0.3 This is a straightforward application of the addition rule for independent events. Part (a)(i) requires solving P(A∪B) = P(A) + P(B) - P(A)P(B) for P(B), which is algebraically simple. Parts (a)(ii) and (b) are direct consequences requiring only basic definitions. Slightly easier than average due to being a standard textbook exercise with no conceptual surprises. |
| Spec | 2.03a Mutually exclusive and independent events |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Uses \(P(A \cup B) = P(A) + P(B) - P(A \cap B)\), PI by \(0.8 = 0.2 + P(B) -\) stated term/value; OR States \(P(A' \cap B') = 0.2\) and \(P(A') = 0.8\) | M1 | AO 3.1a |
| Uses \(P(A \cap B) = P(A) \times P(B)\) OR \(P(A' \cap B') = P(A') \times P(B')\) | M1 | AO 3.1a |
| Obtains correct equation in \(P(B)\) or \(P(B')\) only | A1 | AO 1.1a |
| \(0.8 = 0.2 + P(B) - 0.2P(B)\), \(0.6 = 0.8P(B)\), \(P(B) = 0.75\) | A1 | AO 1.1b |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(P(A \cap B) = 0.15\) (their correct value, provided \(P(B)\) lies between 0 and 0.8) | B1F | AO 1.1b |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| A and B are not mutually exclusive since independent events cannot be mutually exclusive. Other valid reasons: \(P(A \cap B) \neq 0\); or \(P(A \cup B) \neq P(A) + P(B)\); or both can occur at the same time | R1 | AO 2.2a |
## Question 15:
### Part (a)(i):
| Answer/Working | Mark | Guidance |
|---|---|---|
| Uses $P(A \cup B) = P(A) + P(B) - P(A \cap B)$, PI by $0.8 = 0.2 + P(B) -$ stated term/value; OR States $P(A' \cap B') = 0.2$ **and** $P(A') = 0.8$ | M1 | AO 3.1a |
| Uses $P(A \cap B) = P(A) \times P(B)$ OR $P(A' \cap B') = P(A') \times P(B')$ | M1 | AO 3.1a |
| Obtains correct equation in $P(B)$ or $P(B')$ only | A1 | AO 1.1a |
| $0.8 = 0.2 + P(B) - 0.2P(B)$, $0.6 = 0.8P(B)$, $P(B) = 0.75$ | A1 | AO 1.1b |
### Part (a)(ii):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $P(A \cap B) = 0.15$ (their correct value, provided $P(B)$ lies between 0 and 0.8) | B1F | AO 1.1b |
### Part (b):
| Answer/Working | Mark | Guidance |
|---|---|---|
| A and B are not mutually exclusive since independent events cannot be mutually exclusive. Other valid reasons: $P(A \cap B) \neq 0$; or $P(A \cup B) \neq P(A) + P(B)$; or both can occur at the same time | R1 | AO 2.2a |
---15 Two independent events, $A$ and $B$, are such that
$$\begin{aligned}
\mathrm { P } ( A ) & = 0.2 \\
\mathrm { P } ( A \cup B ) & = 0.8
\end{aligned}$$
15
\begin{enumerate}[label=(\alph*)]
\item (i) Find $\mathrm { P } ( B )$\\
15 (a) (ii) Find $\mathrm { P } ( A \cap B )$\\
15
\item State, with a reason, whether or not the events $A$ and $B$ are mutually exclusive.
\end{enumerate}
\hfill \mbox{\textit{AQA AS Paper 2 2019 Q15 [6]}}