| Exam Board | Edexcel |
|---|---|
| Module | S1 (Statistics 1) |
| Year | 2010 |
| Session | June |
| 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: calculating branch probabilities (simple fractions), using the addition rule for part (b), applying Bayes' theorem for part (c), and considering independent events for part (d). All techniques are standard S1 content with no novel problem-solving required—purely methodical application of learned procedures. |
| 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 |
|---|---|---|
| Answer | Marks | Guidance |
| P(R) and P(B) | B1 | 1st B1 for the probabilities on the first 2 branches. Accept 0.416 and 0.583 |
| 2nd set of probabilities | B1 | 2nd B1 for probabilities on the second set of branches. Accept 0.6, 0.3, 0.5 and \(\frac{1.5}{3}\) |
| Allow exact decimal equivalents using clear recurring notation if required. | ||
| \(P(H) = \frac{5}{12} \times \frac{2}{3} + \frac{7}{12} \times \frac{1}{2} = \frac{41}{72}\) or awrt 0.569 | M1 A1 | M1 for an expression for P(H) that follows through their sum of two products of probabilities from their tree diagram |
| \(P(R \mid H) = \frac{\frac{5}{12} \times \frac{2}{3}}{\frac{41}{72}} = \frac{20}{41}\) or awrt 0.488 | M1 A1ft A1 | Formula seen: M1 for \(\frac{P(R \cap H)}{P(H)}\) with denominator their (b) substituted e.g. \(P(R \cap H) = \frac{5}{12}\) award M1. Formula not seen: M1 for \(\frac{\text{probability} \times \text{probability}}{\text{their b}}\) but M0 if fraction repeated e.g. \(\frac{5}{12} \times \frac{2}{3} \div \frac{2}{3}\). 1st A1ft for a fully correct expression or correct follow through. 2nd A1 for \(\frac{20}{41}\) o.e. |
| \(\left(\frac{5}{12}\right)^2 + \left(\frac{7}{12}\right)^2 = \frac{25}{144} + \frac{49}{144} = \frac{74}{144}\) or \(\frac{37}{72}\) or awrt 0.514 | M1 A1ft A1 | M1 for \(\left(\frac{5}{12}\right)^2\) or \(\left(\frac{7}{12}\right)^2\) can follow through their equivalent values from tree diagram. 1st A1 for both values correct or follow through from their original tree and +. 2nd A1 for a correct answer. Special Case: \(\frac{5}{12} \times \frac{4}{11}\) or \(\frac{7}{12} \times \frac{6}{11}\) seen award M1A0A0 |
| Total 10 |
| Answer | Marks | Guidance |
|--------|-------|----------|
| P(R) and P(B) | B1 | 1st B1 for the probabilities on the first 2 branches. Accept 0.416 and 0.583 |
| 2nd set of probabilities | B1 | 2nd B1 for probabilities on the second set of branches. Accept 0.6, 0.3, 0.5 and $\frac{1.5}{3}$ |
| | | Allow exact decimal equivalents using clear recurring notation if required. |
| $P(H) = \frac{5}{12} \times \frac{2}{3} + \frac{7}{12} \times \frac{1}{2} = \frac{41}{72}$ or awrt 0.569 | M1 A1 | M1 for an expression for P(H) that follows through their sum of two products of probabilities from their tree diagram |
| $P(R \mid H) = \frac{\frac{5}{12} \times \frac{2}{3}}{\frac{41}{72}} = \frac{20}{41}$ or awrt 0.488 | M1 A1ft A1 | **Formula seen:** M1 for $\frac{P(R \cap H)}{P(H)}$ with denominator their (b) substituted e.g. $P(R \cap H) = \frac{5}{12}$ award M1. **Formula not seen:** M1 for $\frac{\text{probability} \times \text{probability}}{\text{their b}}$ but M0 if fraction repeated e.g. $\frac{5}{12} \times \frac{2}{3} \div \frac{2}{3}$. 1st A1ft for a fully correct expression or correct follow through. 2nd A1 for $\frac{20}{41}$ o.e. |
| $\left(\frac{5}{12}\right)^2 + \left(\frac{7}{12}\right)^2 = \frac{25}{144} + \frac{49}{144} = \frac{74}{144}$ or $\frac{37}{72}$ or awrt 0.514 | M1 A1ft A1 | M1 for $\left(\frac{5}{12}\right)^2$ or $\left(\frac{7}{12}\right)^2$ can follow through their equivalent values from tree diagram. 1st A1 for both values correct or follow through from their original tree and +. 2nd A1 for a correct answer. **Special Case:** $\frac{5}{12} \times \frac{4}{11}$ or $\frac{7}{12} \times \frac{6}{11}$ seen award M1A0A0 |
| | | **Total 10** |
---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]{039e6fcf-3222-40cc-95ea-37b8dc4a4ddb-03_787_395_734_548}
\section*{Coin}
\begin{center}
\includegraphics[max width=\textwidth, alt={}]{039e6fcf-3222-40cc-95ea-37b8dc4a4ddb-03_1007_488_808_950}
\end{center}
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 2010 Q2 [10]}}