| Exam Board | OCR MEI |
|---|---|
| Module | Paper 2 (Paper 2) |
| Year | 2020 |
| Session | November |
| Marks | 5 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Probability Definitions |
| Type | Independent events test |
| Difficulty | Moderate -0.5 This is a straightforward application of basic probability laws (addition rule, conditional probability, independence test) with standard notation. Part (a) uses P(A∪B)' = 0.2 to find P(A∪B) = 0.8, then applies the addition rule. Parts (b) and (c) are direct calculations requiring no problem-solving insight. Slightly easier than average due to being a routine multi-part question with clear signposting. |
| Spec | 2.03a Mutually exclusive and independent events2.03c Conditional probability: using diagrams/tables |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(0.6 + 0.5 - P(A \cap B) = 1 - 0.2\) oe soi | M1 | or M1 for probabilities in bold correct in table or marked correctly on Venn diagram |
| \(= 0.3\) | A1 | NB \(0.3\) from \(0.6 \times 0.5\) does not score |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(\frac{0.3}{0.5} = 0.6\) | M1 | \(\frac{\text{their } P(A \cap B)}{0.5}\) |
| A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| independent since \(p(A) = p(A/B)\) oe | B1 | or \(0.6 \times 0.5 = 0.3\) or \(P(A \cap B) = P(A) \times P(B)\); FT their values with correct argument |
## Question 7:
### Part (a):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $0.6 + 0.5 - P(A \cap B) = 1 - 0.2$ oe soi | M1 | or M1 for probabilities in bold correct in table or marked correctly on Venn diagram |
| $= 0.3$ | A1 | NB $0.3$ from $0.6 \times 0.5$ does not score |
### Part (b):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $\frac{0.3}{0.5} = 0.6$ | M1 | $\frac{\text{their } P(A \cap B)}{0.5}$ |
| | A1 | |
### Part (c):
| Answer | Marks | Guidance |
|--------|-------|----------|
| independent since $p(A) = p(A/B)$ oe | B1 | or $0.6 \times 0.5 = 0.3$ or $P(A \cap B) = P(A) \times P(B)$; FT their values with correct argument |
---7 You are given that $P ( A ) = 0.6 , P ( B ) = 0.5$ and $P ( A \cup B ) ^ { \prime } = 0.2$.
\begin{enumerate}[label=(\alph*)]
\item Find $\mathrm { P } ( \mathrm { A } \cap \mathrm { B } )$.
\item Find $\mathrm { P } ( \mathrm { A } \mid \mathrm { B } )$.
\item State, with a reason, whether $A$ and $B$ are independent.
\end{enumerate}
\hfill \mbox{\textit{OCR MEI Paper 2 2020 Q7 [5]}}