| Exam Board | Edexcel |
|---|---|
| Module | S1 (Statistics 1) |
| Session | Specimen |
| Marks | 10 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Tree Diagrams |
| Type | Two independent categorical choices |
| Difficulty | Easy -1.2 This is a straightforward tree diagram question testing basic probability concepts: completing a tree diagram with simple fractions, using the multiplication rule (part b), applying Bayes' theorem with given information (part c), and calculating independent event probabilities (part d). All techniques are standard S1 content with no novel problem-solving required. |
| Spec | 2.03b Probability diagrams: tree, Venn, sample space2.03c Conditional probability: using diagrams/tables2.03d Calculate conditional probability: from first principles |
| Answer | Marks | Guidance |
|---|---|---|
| \(P(R)\) and \(P(B)\) correctly shown with 5/12 and 7/12 | B1 | First branches |
| \(2^{nd}\) set of probabilities: 2/3, 1/3, 1/2, 1/2 shown correctly | B1 | Accept 0.41\(\dot{6}\), 0.58\(\dot{3}\), 0.6\(\dot{6}\), 0.\(\dot{3}\), 0.5 and \(\frac{1.5}{3}\) |
| Answer | Marks | Guidance |
|---|---|---|
| \(P(H) = \frac{5}{12} \times \frac{2}{3} + \frac{7}{12} \times \frac{1}{2} = \frac{41}{72}\) or awrt 0.569 | M1 A1 | M1 for sum of two products of probabilities from tree |
| Answer | Marks | Guidance |
|---|---|---|
| \(P(R | H) = \frac{\frac{5}{12} \times \frac{2}{3}}{\frac{41}{72}} = \frac{20}{41}\) or awrt 0.488 | M1 A1ft A1 |
| Answer | Marks | Guidance |
|---|---|---|
| \(\left(\frac{5}{12}\right)^2 + \left(\frac{7}{12}\right)^2 = \frac{25}{144} + \frac{49}{144} = \frac{74}{144}\) or \(\frac{37}{72}\) or awrt 0.514 | M1 A1ft A1 | M1 for \(\left(\frac{5}{12}\right)^2\) or \(\left(\frac{7}{12}\right)^2\). Special case \(\frac{5}{12}\times\frac{4}{11}\) or \(\frac{7}{12}\times\frac{6}{11}\) award M1A0A0 |
# Question 2:
## Part (a)
| $P(R)$ and $P(B)$ correctly shown with 5/12 and 7/12 | B1 | First branches |
|---|---|---|
| $2^{nd}$ set of probabilities: 2/3, 1/3, 1/2, 1/2 shown correctly | B1 | Accept 0.41$\dot{6}$, 0.58$\dot{3}$, 0.6$\dot{6}$, 0.$\dot{3}$, 0.5 and $\frac{1.5}{3}$ |
## Part (b)
| $P(H) = \frac{5}{12} \times \frac{2}{3} + \frac{7}{12} \times \frac{1}{2} = \frac{41}{72}$ or awrt 0.569 | M1 A1 | M1 for sum of two products of probabilities from tree |
## Part (c)
| $P(R|H) = \frac{\frac{5}{12} \times \frac{2}{3}}{\frac{41}{72}} = \frac{20}{41}$ or awrt 0.488 | M1 A1ft A1 | M1 for $\frac{P(R \cap H)}{P(H)}$ with denominator their (b). 2nd A1 for $\frac{20}{41}$ |
## Part (d)
| $\left(\frac{5}{12}\right)^2 + \left(\frac{7}{12}\right)^2 = \frac{25}{144} + \frac{49}{144} = \frac{74}{144}$ or $\frac{37}{72}$ or awrt 0.514 | M1 A1ft A1 | M1 for $\left(\frac{5}{12}\right)^2$ or $\left(\frac{7}{12}\right)^2$. Special case $\frac{5}{12}\times\frac{4}{11}$ or $\frac{7}{12}\times\frac{6}{11}$ award M1A0A0 |
---2. An experiment consists of selecting a ball from a bag and spinning a coin. The bag contains 5 red balls and 7 blue balls. A ball is selected at random from the bag, its colour is noted and then the ball is returned to the bag.
When a red ball is selected, a biased coin with probability $\frac { 2 } { 3 }$ of landing heads is spun.\\
When a blue ball is selected a fair coin is spun.
\begin{enumerate}[label=(\alph*)]
\item Complete the tree diagram below to show the possible outcomes and associated probabilities.\\
\includegraphics[max width=\textwidth, alt={}, center]{61983561-79f7-4883-8ae7-ab1f4955d444-04_785_385_744_568}\\
\includegraphics[max width=\textwidth, alt={}, center]{61983561-79f7-4883-8ae7-ab1f4955d444-04_1054_483_760_954}
Shivani selects a ball and spins the appropriate coin.
\item Find the probability that she obtains a head.
Given that Tom selected a ball at random and obtained a head when he spun the appropriate coin,
\item find the probability that Tom selected a red ball.
Shivani and Tom each repeat this experiment.
\item Find the probability that the colour of the ball Shivani selects is the same as the colour of the ball Tom selects.
\end{enumerate}
\hfill \mbox{\textit{Edexcel S1 Q2 [10]}}