| Exam Board | OCR MEI |
|---|---|
| Module | S1 (Statistics 1) |
| Year | 2014 |
| Session | June |
| Marks | 6 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Combinations & Selection |
| Type | Probability of specific committee composition |
| Difficulty | Moderate -0.8 This is a straightforward combinations probability question requiring only basic counting principles. Part (i) is a direct application of C(16,4)/C(30,4), and part (ii) requires the complement rule (1 - P(all girls) - P(all boys)). Both parts are routine calculations with no conceptual challenges beyond standard S1 material. |
| Spec | 5.01a Permutations and combinations: evaluate probabilities |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(P(\text{All four are girls}) = \frac{16}{30} \times \frac{15}{29} \times \frac{14}{28} \times \frac{13}{27}\) | M1 | For \(\frac{16}{30} \times\) product of other three correct fractions without extra terms |
| \(= 0.0664\) | A1 | CAO. Allow 0.066 with working but not 0.07. Allow full marks for unsimplified fractional answers. |
| OR \(\binom{16}{4} / \binom{30}{4} = \frac{1820}{27405} = \frac{52}{783} = 0.0664\). M1 for either term in correct position in a fraction. M1 for correct fraction. A1 CAO | ||
| Total | [3] | SC2 for \(\frac{14}{30} \times \frac{13}{29} \times \frac{12}{28} \times \frac{11}{27} = 0.0365\). SC2 for \(\binom{14}{4} / \binom{30}{4} = \frac{143}{3915} = 0.0365\) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(P(\text{All four are boys}) = \frac{14}{30} \times \frac{13}{29} \times \frac{12}{28} \times \frac{11}{27} = 0.0365\) | M1 | For P(All four are boys) without extra terms. M1 for this then as per scheme. |
| \(P(\text{At least one girl and at least one boy}) = 1 - (0.0664 + 0.0365)\) | M1 | FT their 'sensible' probabilities. CAO. Allow answer 0.8975 from use of 0.066 |
| \(= 0.897\) | A1 | Allow 0.90 work working, but not 0.9. NB Watch for \((1 - 0.0365) \times (1 - 0.0664) = 0.9635 \times 0.9336 = 0.8995\) Gets just M1 for 0.0365. Accept 0.90 work working, but not 0.9 |
| Total | [3] |
## (i)
| **Answer** | **Marks** | **Guidance** |
|---|---|---|
| $P(\text{All four are girls}) = \frac{16}{30} \times \frac{15}{29} \times \frac{14}{28} \times \frac{13}{27}$ | M1 | For $\frac{16}{30} \times$ product of other three correct fractions without extra terms |
| $= 0.0664$ | A1 | CAO. Allow 0.066 with working but not 0.07. Allow full marks for unsimplified fractional answers. |
| | | OR $\binom{16}{4} / \binom{30}{4} = \frac{1820}{27405} = \frac{52}{783} = 0.0664$. M1 for either term in correct position in a fraction. M1 for correct fraction. A1 CAO |
| **Total** | **[3]** | SC2 for $\frac{14}{30} \times \frac{13}{29} \times \frac{12}{28} \times \frac{11}{27} = 0.0365$. SC2 for $\binom{14}{4} / \binom{30}{4} = \frac{143}{3915} = 0.0365$ |
## (ii)
| **Answer** | **Marks** | **Guidance** |
|---|---|---|
| $P(\text{All four are boys}) = \frac{14}{30} \times \frac{13}{29} \times \frac{12}{28} \times \frac{11}{27} = 0.0365$ | M1 | For P(All four are boys) without extra terms. M1 for this then as per scheme. |
| $P(\text{At least one girl and at least one boy}) = 1 - (0.0664 + 0.0365)$ | M1 | FT their 'sensible' probabilities. CAO. Allow answer 0.8975 from use of 0.066 |
| $= 0.897$ | A1 | Allow 0.90 work working, but not 0.9. NB Watch for $(1 - 0.0365) \times (1 - 0.0664) = 0.9635 \times 0.9336 = 0.8995$ Gets just M1 for 0.0365. Accept 0.90 work working, but not 0.9 |
| **Total** | **[3]** | |
---There are 16 girls and 14 boys in a class. Four of them are to be selected to form a quiz team. The team is to be selected at random.
\begin{enumerate}[label=(\roman*)]
\item Find the probability that all 4 members of the team will be girls. [3]
\item Find the probability that the team will contain at least one girl and at least one boy. [3]
\end{enumerate}
\hfill \mbox{\textit{OCR MEI S1 2014 Q4 [6]}}