| Exam Board | Edexcel |
|---|---|
| Module | S1 (Statistics 1) |
| Marks | 13 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Probability Definitions |
| Type | Multiple independent trials |
| Difficulty | Moderate -0.8 This is a straightforward S1 probability question testing basic probability rules (complementary events, independent events, binomial probability) and conditional probability/Bayes' theorem. All parts use standard techniques with clear setups: (a) is simple subtraction, (b-c) are routine probability calculations, and (d) is a textbook application of conditional probability with given proportions. No novel insight or complex problem-solving required—purely mechanical application of formulas. |
| Spec | 2.03a Mutually exclusive and independent events2.03c Conditional probability: using diagrams/tables2.03d Calculate conditional probability: from first principles2.04b Binomial distribution: as model B(n,p)2.04c Calculate binomial probabilities5.01a Permutations and combinations: evaluate probabilities |
| Answer | Marks | Guidance |
|---|---|---|
| (a) \(P(\text{C.S.}) = 1 - \frac{3}{8} - \frac{1}{5} = \frac{17}{40}\) | M1 A1 | |
| (b) \(P(\text{both do the same}) = \frac{3}{8}^2 + \frac{1}{5}^2 + \frac{17}{40}^2 = 0.361\) | M1 A1 A1 | |
| (c) \(P(\text{1 games, other 2 not}) = \frac{3}{8} \times \frac{4}{5} \times \frac{5}{8} \times 3 = 0.439\) | M1 A1 A1 | |
| (d) Let \(P(\text{Boy does P.S.}) = p\). By tree diagram or otherwise, \(\frac{2}{3}p + \frac{1}{10} = \frac{1}{3}\) so \(p = \frac{1}{6}\) | M1 A1 | |
| \(P(\text{Boy} \mid \text{P.S.}) = \left(\frac{2}{3} \times \frac{1}{6}\right) \div \frac{1}{3} = \frac{1}{2}\) | M1 M1 A1 | Total: 13 marks |
(a) $P(\text{C.S.}) = 1 - \frac{3}{8} - \frac{1}{5} = \frac{17}{40}$ | M1 A1 |
(b) $P(\text{both do the same}) = \frac{3}{8}^2 + \frac{1}{5}^2 + \frac{17}{40}^2 = 0.361$ | M1 A1 A1 |
(c) $P(\text{1 games, other 2 not}) = \frac{3}{8} \times \frac{4}{5} \times \frac{5}{8} \times 3 = 0.439$ | M1 A1 A1 |
(d) Let $P(\text{Boy does P.S.}) = p$. By tree diagram or otherwise, $\frac{2}{3}p + \frac{1}{10} = \frac{1}{3}$ so $p = \frac{1}{6}$ | M1 A1 |
$P(\text{Boy} \mid \text{P.S.}) = \left(\frac{2}{3} \times \frac{1}{6}\right) \div \frac{1}{3} = \frac{1}{2}$ | M1 M1 A1 | **Total: 13 marks**The students in a large Sixth Form can choose to do exactly one of Community Service, Games or Private Study on Wednesday afternoons. The probabilities that a randomly chosen student does Games and Private Study are $\frac{3}{8}$ and $\frac{1}{5}$ respectively. It may be assumed that the number of students is large enough for these probabilities to be treated as constant.
\begin{enumerate}[label=(\alph*)]
\item Find the probability that a randomly chosen student does Community Service. [2 marks]
\item If two students are chosen at random, find the probability that they both do the same activity. [3 marks]
\item If three students are chosen at random, find the probability that exactly one of them does Games. [3 marks]
\end{enumerate}
Two-fifths of the students are girls, and a quarter of these girls do Private Study.
\begin{enumerate}[label=(\alph*)]
\setcounter{enumi}{3}
\item Find the probability that a randomly chosen student who does Private Study is a boy. [5 marks]
\end{enumerate}
\hfill \mbox{\textit{Edexcel S1 Q5 [13]}}