| Exam Board | OCR MEI |
|---|---|
| Module | Paper 2 (Paper 2) |
| Year | 2021 |
| Session | November |
| Marks | 10 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Hypergeometric Distribution |
| Type | Probability with multiple categories |
| Difficulty | Standard +0.8 This multi-part hypergeometric problem requires careful counting across overlapping categories (gender and color), conditional probability calculation, and independence testing. Part (c) demands finding P(A|B) using intersection probabilities, which is non-routine. While the individual calculations are A-level standard, coordinating multiple categories and the conditional/independence analysis elevates this above typical textbook exercises. |
| Spec | 2.03a Mutually exclusive and independent events2.03b Probability diagrams: tree, Venn, sample space2.03d Calculate conditional probability: from first principles |
| male | female | |
| black | 1 | 3 |
| yellow | 2 | 1 |
| chocolate | 1 | 1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(\frac{5}{9} \times \frac{4}{8} \times \frac{3}{7} \times \frac{4}{6}\) oe | M1 | M0 if binomial distribution or sampling with replacement used |
| \(\times 4\) | M1 | dependent on award of first M1 |
| *Alternative:* \(_{9}C_{4}\) \([= 126]\) soi | M1 | |
| \(_{5}C_{3} \times {_{4}C_{1}}\) \([= 10 \times 4]\) soi | M1 | |
| \(\frac{40}{126}\) or \(\frac{20}{63}\) or \(0.317460...\) isw or rounded to 2 sf or better | A1 | |
| [3] |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(\frac{4}{9} \times \frac{3}{8} \times \frac{2}{7} \times \frac{5}{6} \times 4\) oe or \(\frac{4}{9} \times \frac{3}{8} \times \frac{2}{7} \times \frac{1}{6}\) oe | M1 | condone omission of \(\times 4\) in first term; M0 if binomial distribution or sampling with replacement used; NB \(1 + 4\times5 = 21\) |
| for addition of their terms | M1 | dependent on award of first M1 |
| *Alternative:* \(_{4}C_{4} + {_{4}C_{3}} \times {_{5}C_{1}}\) | M1 | for \(_{4}C_{3} \times {_{5}C_{1}}\) |
| for addition of their terms | M1 | |
| *Alternative:* \(\frac{5}{9}\times\frac{4}{8}\times\frac{3}{7}\times\frac{2}{6} + 4\times\frac{4}{9}\times\frac{5}{8}\times\frac{4}{7}\times\frac{3}{6} + 6\times\frac{4}{9}\times\frac{5}{8}\times\frac{4}{7}\times\frac{3}{6}\) oe | M1 | for two of these terms; NB \(\frac{2520}{3024} = \frac{5}{6}\) |
| \(1 -\) their \(\frac{2520}{3024}\) | M1 | for \(1 -\) the sum of their 3 terms |
| \(\frac{1}{6}\) or \(0.166666...\) to 2 sf or better | A1 | |
| [3] |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \((3\ \text{BF},\ 1\text{M}) + (2\text{BF},\ 1\text{NBF},\ 1\text{BM})\) attempted | M1 | M0 if binomial distribution or sampling with replacement used |
| \(\frac{3}{9}\times\frac{2}{8}\times\frac{1}{7}\times\frac{4}{6}\times 4 + \frac{3}{9}\times\frac{2}{8}\times\frac{2}{7}\times\frac{1}{6}\times 12\left[=\frac{5}{63}\right]\) oe | A1 | |
| *Alternative:* \(_{3}C_{3}\times{_{4}C_{1}} + {_{3}C_{2}}\times{_{2}C_{1}}\times{_{1}C_{1}}\) | M1 | for either term |
| \(10\) | A1 | |
| \(\frac{10}{21}\) or \(0.476190476...\) to 2 sf or better | A1 | |
| [3] |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| Not independent since \(\frac{10}{21} \neq \frac{20}{63}\) oe | B1 | FT their probabilities; no FT if binomial distribution or sampling with replacement used |
| *Alternatively:* not independent since \(\frac{20}{63} \times \frac{1}{6} \neq \frac{5}{63}\) | FT their calculated probabilities from first three parts | |
| [1] |
## Question 9(a):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $\frac{5}{9} \times \frac{4}{8} \times \frac{3}{7} \times \frac{4}{6}$ **oe** | M1 | **M0** if binomial distribution or sampling with replacement used |
| $\times 4$ | M1 | dependent on award of first **M1** |
| *Alternative:* $_{9}C_{4}$ $[= 126]$ **soi** | M1 | |
| $_{5}C_{3} \times {_{4}C_{1}}$ $[= 10 \times 4]$ **soi** | M1 | |
| $\frac{40}{126}$ or $\frac{20}{63}$ or $0.317460...$ isw or rounded to 2 sf or better | A1 | |
| | **[3]** | |
---
## Question 9(b):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $\frac{4}{9} \times \frac{3}{8} \times \frac{2}{7} \times \frac{5}{6} \times 4$ **oe** or $\frac{4}{9} \times \frac{3}{8} \times \frac{2}{7} \times \frac{1}{6}$ **oe** | M1 | condone omission of $\times 4$ in first term; **M0** if binomial distribution or sampling with replacement used; **NB** $1 + 4\times5 = 21$ |
| for addition of their terms | M1 | dependent on award of first **M1** |
| *Alternative:* $_{4}C_{4} + {_{4}C_{3}} \times {_{5}C_{1}}$ | M1 | for $_{4}C_{3} \times {_{5}C_{1}}$ |
| for addition of their terms | M1 | |
| *Alternative:* $\frac{5}{9}\times\frac{4}{8}\times\frac{3}{7}\times\frac{2}{6} + 4\times\frac{4}{9}\times\frac{5}{8}\times\frac{4}{7}\times\frac{3}{6} + 6\times\frac{4}{9}\times\frac{5}{8}\times\frac{4}{7}\times\frac{3}{6}$ **oe** | M1 | for two of these terms; **NB** $\frac{2520}{3024} = \frac{5}{6}$ |
| $1 -$ their $\frac{2520}{3024}$ | M1 | for $1 -$ the sum of their 3 terms |
| $\frac{1}{6}$ or $0.166666...$ to 2 sf or better | A1 | |
| | **[3]** | |
---
## Question 9(c):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $(3\ \text{BF},\ 1\text{M}) + (2\text{BF},\ 1\text{NBF},\ 1\text{BM})$ attempted | M1 | **M0** if binomial distribution or sampling with replacement used |
| $\frac{3}{9}\times\frac{2}{8}\times\frac{1}{7}\times\frac{4}{6}\times 4 + \frac{3}{9}\times\frac{2}{8}\times\frac{2}{7}\times\frac{1}{6}\times 12\left[=\frac{5}{63}\right]$ **oe** | A1 | |
| *Alternative:* $_{3}C_{3}\times{_{4}C_{1}} + {_{3}C_{2}}\times{_{2}C_{1}}\times{_{1}C_{1}}$ | M1 | for either term |
| $10$ | A1 | |
| $\frac{10}{21}$ or $0.476190476...$ to 2 sf or better | A1 | |
| | **[3]** | |
---
## Question 9(d):
| Answer | Marks | Guidance |
|--------|-------|----------|
| Not independent since $\frac{10}{21} \neq \frac{20}{63}$ **oe** | B1 | FT their probabilities; no FT if binomial distribution or sampling with replacement used |
| *Alternatively:* not independent since $\frac{20}{63} \times \frac{1}{6} \neq \frac{5}{63}$ | | FT their calculated probabilities from first three parts |
| | **[1]** | |
---9 Labrador puppies may be black, yellow or chocolate in colour. Some information about a litter of 9 puppies is given in the table.
\begin{center}
\begin{tabular}{ | l | c | c | }
\hline
& male & female \\
\hline
black & 1 & 3 \\
\hline
yellow & 2 & 1 \\
\hline
chocolate & 1 & 1 \\
\hline
\end{tabular}
\end{center}
Four puppies are chosen at random to train as guide dogs.
\begin{enumerate}[label=(\alph*)]
\item Determine the probability that exactly 3 females are chosen.
\item Determine the probability that at least 3 black puppies are chosen.
\item Determine the probability that exactly 3 females are chosen given that at least 3 black puppies are chosen.
\item Explain whether the 2 events 'choosing exactly 3 females' and 'choosing at least 3 black puppies' are independent events.
\end{enumerate}
\hfill \mbox{\textit{OCR MEI Paper 2 2021 Q9 [10]}}