| Exam Board | OCR MEI |
|---|---|
| Module | Paper 2 (Paper 2) |
| Year | 2023 |
| Session | June |
| Marks | 8 |
| 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 basic set notation and Venn diagram logic. Part (a) is a simple binomial calculation (0.16 × 0.84 × 2). Parts (b)-(d) involve routine manipulation of given percentages to find intersections and apply the definition of conditional probability P(S|A) = P(S∩A)/P(A), then check independence. All steps are standard textbook exercises with no novel insight required, making this easier than average. |
| Spec | 2.03a Mutually exclusive and independent events2.03c Conditional probability: using diagrams/tables2.03d Calculate conditional probability: from first principles |
| Answer | Marks | Guidance |
|---|---|---|
| \(0.16 \times 0.84 \times 2\) or \(B(2, 0.16)\) or \(B(2, 0.84)\) seen or \(1-(0.84^2+0.16^2)\) | M1 | condone omission of 2; allow recovery from bracket error |
| \(\frac{168}{625}\) or \(0.2688\) or \(0.269\) or \(0.27\) | A1 | mark the final answer; allow SC1 for correct answer unsupported |
| Answer | Marks | Guidance |
|---|---|---|
| \(0.75 - 0.66 = 0.09\) | M1 | allow 0.09 embedded in correct place in Venn diagram or contingency table; allow M1 for 9% |
| \([0.16 - 0.09 =]\ 0.07\) | A1 | allow SC1 for correct answer unsupported |
| Answer | Marks | Guidance |
|---|---|---|
| \(\frac{0.09}{0.75}\) | M1 | M0 for 0.12 from wrong working |
| \(0.12\) | A1 | allow SC1 for correct answer unsupported |
| Answer | Marks | Guidance |
|---|---|---|
| \(0.12 \neq 0.16\) | M1 | |
| so not independent | A1 | if M0 allow SCB1 for \(0.16 \times 0.75 \neq 0.09\) so not independent |
## Question 16:
### Part (a):
$0.16 \times 0.84 \times 2$ **or** $B(2, 0.16)$ or $B(2, 0.84)$ seen **or** $1-(0.84^2+0.16^2)$ | M1 | condone omission of 2; allow recovery from bracket error
$\frac{168}{625}$ or $0.2688$ or $0.269$ or $0.27$ | A1 | mark the final answer; allow SC1 for correct answer unsupported
### Part (b):
$0.75 - 0.66 = 0.09$ | M1 | allow 0.09 embedded in correct place in Venn diagram or contingency table; allow M1 for 9%
$[0.16 - 0.09 =]\ 0.07$ | A1 | allow SC1 for correct answer unsupported
### Part (c):
$\frac{0.09}{0.75}$ | M1 | M0 for 0.12 from wrong working
$0.12$ | A1 | allow SC1 for correct answer unsupported
### Part (d):
$0.12 \neq 0.16$ | M1 |
so not independent | A1 | if M0 allow SCB1 for $0.16 \times 0.75 \neq 0.09$ so not independent
---16 Research conducted by social scientists has shown that $16 \%$ of young adults smoke cigarettes.
Two young adults are selected at random.
\begin{enumerate}[label=(\alph*)]
\item Determine the probability that one smokes cigarettes and the other doesn't.
The same research has also shown that
\begin{itemize}
\item 75\% of young adults drink alcohol.
\item $66 \%$ of young adults drink alcohol, but do not smoke cigarettes.
\item Determine the probability that a young adult selected at random does smoke cigarettes, but does not drink alcohol.
\item A young adult who drinks alcohol is selected at random. Determine the probability that this young adult smokes cigarettes.
\item Using your answer to part (c), explain whether the event that a young adult selected at random smokes cigarettes is independent of the event that a young adult selected at random drinks alcohol.
\end{itemize}
\end{enumerate}
\hfill \mbox{\textit{OCR MEI Paper 2 2023 Q16 [8]}}