| Exam Board | OCR |
|---|---|
| Module | S1 (Statistics 1) |
| Session | Specimen |
| Marks | 10 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Discrete Probability Distributions |
| Type | Sampling without replacement |
| Difficulty | Moderate -0.8 This is a straightforward hypergeometric distribution question requiring only standard combinatorial calculations (choosing 3 from 12 students) and basic expectation/variance formulas. The 'show that' parts guide students through the method, and all calculations involve simple arithmetic with small numbers. This is easier than average for A-level, being a routine textbook exercise with no problem-solving insight required. |
| Spec | 5.01a Permutations and combinations: evaluate probabilities5.02a Discrete probability distributions: general5.02b Expectation and variance: discrete random variables |
| \(x\) | 0 | 1 | 2 | 3 |
| \(\mathrm { P } ( X = x )\) | \(\frac { 7 } { 44 }\) | \(\frac { 21 } { 44 }\) | \(\frac { 7 } { 22 }\) | \(\frac { 1 } { 22 }\) |
| Answer | Marks | Guidance |
|---|---|---|
| EITHER: \(P(X = 0) = \frac{\binom{5}{0}}{\binom{13}{3}} = \frac{33}{220} = \frac{3}{44}\) | M1 | For ratio of relevant \(\binom{\ }{\ }\) terms |
| A1 | For showing the given answer correctly | |
| OR: \(P(X = 0) = \frac{7}{13} \times \frac{6}{12} \times \frac{5}{10} = \frac{7}{44}\) | M1 | For multiplication of relevant 'girl' probs |
| A1 | For showing the given answer correctly |
| Answer | Marks | Guidance |
|---|---|---|
| \(= \frac{2 \times 10}{220} = \frac{7}{22}\) | M1 | For use of three \(\binom{\ }{\ }\) terms relevant to the 2B, 1G case |
| B1 | For both \(\binom{5}{2}\) and \(\binom{13}{5}\) correct | |
| A1 | For showing the given answer correctly |
| Answer | Marks | Guidance |
|---|---|---|
| \(= \frac{5}{13} \times \frac{4}{12} \times \frac{8}{10} \times 3 = \frac{2}{11}\) | M1 | For three probabilities multiplied relevant to the 2B, 1G case |
| B1 | For inclusion of factor 3 | |
| A1 | For showing the given answer correctly |
| Answer | Marks | Guidance |
|---|---|---|
| \(E(X) = 0 \times \frac{7}{44} + 1 \times \frac{21}{44} + 2 \times \frac{28}{52} + 3 \times \frac{7}{44} = \frac{3}{2}\) | M1 | For correct calculation process |
| A1 | For correct answer | |
| \(E(X^2) = 0 \times \frac{7}{44} + 1 \times \frac{21}{44} + 4 \times \frac{7}{52} + 9 \times \frac{1}{22} = \frac{95}{44}\) | B1 | For correct numerical expression for \(E(X^2) p\) |
| M1 | For correct overall method for variance | |
| \(\text{Var}(X) = \frac{95}{44} - \left(\frac{3}{2}\right)^2 = \frac{105}{176}\) or 0.597 (to 3dp) | A1√ | For correct answer |
**Part (i)**
EITHER: $P(X = 0) = \frac{\binom{5}{0}}{\binom{13}{3}} = \frac{33}{220} = \frac{3}{44}$ | M1 | For ratio of relevant $\binom{\ }{\ }$ terms
| A1 | For showing the given answer correctly
OR: $P(X = 0) = \frac{7}{13} \times \frac{6}{12} \times \frac{5}{10} = \frac{7}{44}$ | M1 | For multiplication of relevant 'girl' probs
| A1 | For showing the given answer correctly
**Part (ii)**
EITHER: $P(X = 2) = P(2 \text{ boys and 1 girl})$
$= \frac{\binom{5}{2} \times \binom{8}{1}}{\binom{13}{3}}$
$= \frac{2 \times 10}{220} = \frac{7}{22}$ | M1 | For use of three $\binom{\ }{\ }$ terms relevant to the 2B, 1G case
| B1 | For both $\binom{5}{2}$ and $\binom{13}{5}$ correct
| A1 | For showing the given answer correctly
OR: $P(X = 2) = P(2 \text{ boys and 1 girl})$
$= \frac{5}{13} \times \frac{4}{12} \times \frac{8}{10} \times 3 = \frac{2}{11}$ | M1 | For three probabilities multiplied relevant to the 2B, 1G case
| B1 | For inclusion of factor 3
| A1 | For showing the given answer correctly
**Part (iii)**
$E(X) = 0 \times \frac{7}{44} + 1 \times \frac{21}{44} + 2 \times \frac{28}{52} + 3 \times \frac{7}{44} = \frac{3}{2}$ | M1 | For correct calculation process
| A1 | For correct answer
$E(X^2) = 0 \times \frac{7}{44} + 1 \times \frac{21}{44} + 4 \times \frac{7}{52} + 9 \times \frac{1}{22} = \frac{95}{44}$ | B1 | For correct numerical expression for $E(X^2) p$
| M1 | For correct overall method for variance
$\text{Var}(X) = \frac{95}{44} - \left(\frac{3}{2}\right)^2 = \frac{105}{176}$ or 0.597 (to 3dp) | A1√ | For correct answer
---5 A sixth-form class consists of 7 girls and 5 boys. Three students from the class are chosen at random. The number of boys chosen is denoted by the random variable $X$. Show that\\
(i) $\quad \mathrm { P } ( X = 0 ) = \frac { 7 } { 44 }$,\\
(ii) $\mathrm { P } ( X = 2 ) = \frac { 7 } { 22 }$.
The complete probability distribution of $X$ is shown in the following table.
\begin{center}
\begin{tabular}{ | c | c | c | c | c | }
\hline
$x$ & 0 & 1 & 2 & 3 \\
\hline
$\mathrm { P } ( X = x )$ & $\frac { 7 } { 44 }$ & $\frac { 21 } { 44 }$ & $\frac { 7 } { 22 }$ & $\frac { 1 } { 22 }$ \\
\hline
\end{tabular}
\end{center}
(iii) Calculate $\mathrm { E } ( X )$ and $\operatorname { Var } ( X )$.
\hfill \mbox{\textit{OCR S1 Q5 [10]}}