| Exam Board | OCR |
|---|---|
| Module | S2 (Statistics 2) |
| Year | 2012 |
| Session | January |
| Marks | 10 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Hypergeometric Distribution |
| Type | Combined hypergeometric and binomial |
| Difficulty | Standard +0.3 This is a straightforward S2 probability question testing binomial distribution and normal approximation. Part (i) requires basic binomial probability calculations with p=12/20=0.6, distinguishing between sampling with and without replacement. Part (ii) applies the standard normal approximation to binomial (n=60, p=0.6) with continuity correction, plus a brief justification that the approximation works equally well in both sampling scenarios due to large population size. All techniques are routine for S2 with no novel problem-solving required, making it slightly easier than average. |
| Spec | 2.04d Normal approximation to binomial5.01a Permutations and combinations: evaluate probabilities |
\begin{enumerate}[label=(\roman*)]
\item Six prizes are allocated, using random numbers, to a group of 12 girls and 8 boys. Calculate the probability that exactly 4 of the prizes are allocated to girls if
\begin{enumerate}[label=(\alph*)]
\item the same child may win more than one prize, [2]
\item no child may win more than one prize. [2]
\end{enumerate}
\item Sixty prizes are allocated, using random numbers, to a group of 1200 girls and 800 boys. Use a suitable approximation to calculate the probability that at least 30 of the prizes are allocated to girls. Does it affect your calculation whether or not the same child may win more than one prize? Justify your answer. [6]
\end{enumerate}
\hfill \mbox{\textit{OCR S2 2012 Q5 [10]}}