| Exam Board | Edexcel |
|---|---|
| Module | S1 (Statistics 1) |
| Marks | 15 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Conditional Probability |
| Type | Tree diagram with two-stage events |
| Difficulty | Moderate -0.3 This is a standard S1 conditional probability question using tree diagrams. While it requires careful bookkeeping of probabilities and understanding of conditional probability notation, the methods are routine for this module. The multi-part structure guides students through the problem systematically, and part (d) is a straightforward binomial probability application. Slightly easier than average due to the scaffolding and standard techniques required. |
| Spec | 2.03b Probability diagrams: tree, Venn, sample space2.03c 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) Let \(P(\text{younger child is a boy, given elder is a boy}) = p\). \(\frac{5}{12}p + \frac{7}{12} \times \frac{3}{4} = \frac{9}{16}\) giving \(\frac{5}{12}p = \frac{1}{8}\) and \(p = \frac{3}{10}\) | M1 A1 A1 | |
| \(P(B, B) = \frac{5}{12} \times \frac{3}{10} = \frac{1}{8}\) | M1 A1 | |
| (b) \(P(B, G \text{ or } G, B) = \frac{5}{12} \times \frac{7}{10} + \frac{7}{12} \times \frac{3}{4} = \frac{35}{48}\) or \(0.729\) | M1 A1 A1 | |
| (c) \(P(B, G \text{ or } G, B | \text{ at least 1 girl}) = \frac{35}{48} \div \frac{7}{8} = \frac{5}{6}\) | M1 M1 A1 |
| (d) \(\frac{1}{8} \times \frac{1}{3} \times \frac{7}{8} \times 3 = \frac{21}{512}\) or \(0.0410\) | M1 A1 A1 | |
| (e) Assumed independence | B1 | Total: 15 marks |
(a) Let $P(\text{younger child is a boy, given elder is a boy}) = p$. $\frac{5}{12}p + \frac{7}{12} \times \frac{3}{4} = \frac{9}{16}$ giving $\frac{5}{12}p = \frac{1}{8}$ and $p = \frac{3}{10}$ | M1 A1 A1 |
$P(B, B) = \frac{5}{12} \times \frac{3}{10} = \frac{1}{8}$ | M1 A1 |
(b) $P(B, G \text{ or } G, B) = \frac{5}{12} \times \frac{7}{10} + \frac{7}{12} \times \frac{3}{4} = \frac{35}{48}$ or $0.729$ | M1 A1 A1 |
(c) $P(B, G \text{ or } G, B | \text{ at least 1 girl}) = \frac{35}{48} \div \frac{7}{8} = \frac{5}{6}$ | M1 M1 A1 |
(d) $\frac{1}{8} \times \frac{1}{3} \times \frac{7}{8} \times 3 = \frac{21}{512}$ or $0.0410$ | M1 A1 A1 |
(e) Assumed independence | B1 | Total: 15 marksAmong the families with two children in a large city, the probability that the elder child is a boy is $\frac{5}{12}$ and the probability that the younger child is a boy is $\frac{9}{16}$. The probability that the younger child is a girl, given that the elder child is a girl, is $\frac{1}{4}$.
One of the families is chosen at random. Using a tree diagram, or otherwise,
\begin{enumerate}[label=(\alph*)]
\item show that the probability that both children are boys is $\frac{1}{8}$. [5 marks]
\end{enumerate}
Find the probability that
\begin{enumerate}[label=(\alph*)]
\setcounter{enumi}{1}
\item one child is a boy and the other is a girl, [3 marks]
\item one child is a boy given that the other is a girl. [3 marks]
\end{enumerate}
If three of the families are chosen at random,
\begin{enumerate}[label=(\alph*)]
\setcounter{enumi}{3}
\item find the probability that exactly two of the families have two boys. [3 marks]
\item State an assumption that you have made in answering part (d). [1 mark]
\end{enumerate}
\hfill \mbox{\textit{Edexcel S1 Q7 [15]}}