| Exam Board | Edexcel |
|---|---|
| Module | S1 (Statistics 1) |
| Year | 2014 |
| Session | June |
| Marks | 9 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Tree Diagrams |
| Type | Population partition tree diagram |
| Difficulty | Easy -1.3 This is a standard S1 tree diagram question requiring straightforward application of probability rules (multiplication along branches, addition across branches, and Bayes' theorem). All probabilities are given explicitly, calculations are routine, and the question follows a textbook template 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 |
|---|---|---|
| Answer/Working | Marks | Guidance |
| (a) Tree diagram with correct probabilities: \(P(J\cap B) = 0.005\), \(P(J\cap B') = 0.245\), \(P(K\cap B) = 0.0135\), \(P(K\cap B') = 0.4365\), \(P(L\cap B) = 0.015\), \(P(L\cap B') = 0.285\) | M1 A1 (2) | M1 for (3+6) tree drawn with 0.25, 0.45, 0.02, 0.03, 0.05 on correct branches. A1 for 0.3, 0.98, 0.97, 0.95 on correct branches and labels, condone missing \(B'\)s. Correct answer only scores full marks for (b), (c), (d). When using "their probability \(p\)" for M1 and A1ft they must have \(0 < p < 1\) |
| (b) \(0.25 \times 0.98 = \mathbf{0.245}\) (or exact equiv. e.g. \(\frac{49}{200}\)) | M1 A1 (2) | M1 for \(0.25 \times\) 'their 0.98' o.e. |
| (c) \(0.25\times0.02 + 0.45\times0.03 + 0.3\times0.05 = \mathbf{0.0335}\) (or exact equiv. e.g. \(\frac{67}{2000}\)) | M1 A1 (2) | M1 for \(0.25\times\text{their } 0.02 + 0.45\times\text{their } 0.03 + 0.3\times\text{their } 0.05\). Condone 1 transcription error. Or \(1-(0.25\times\text{their }0.98+0.45\times\text{their }0.97+0.3\times\text{their }0.95)\) |
| (d) \([P(J\cup L\ | B)] = \frac{0.25\times0.02 + 0.3\times0.05}{0.0335}\) or \(\frac{0.0335 - 0.45\times0.03}{0.0335}\) \(= 0.5970\ldots\) awrt 0.597 (or \(\frac{40}{67}\) or exact equiv.) | M1 A1ft A1 (3) |
# Question 4:
| Answer/Working | Marks | Guidance |
|---|---|---|
| **(a)** Tree diagram with correct probabilities: $P(J\cap B) = 0.005$, $P(J\cap B') = 0.245$, $P(K\cap B) = 0.0135$, $P(K\cap B') = 0.4365$, $P(L\cap B) = 0.015$, $P(L\cap B') = 0.285$ | M1 A1 **(2)** | M1 for (3+6) tree drawn with 0.25, 0.45, 0.02, 0.03, 0.05 on correct branches. A1 for 0.3, 0.98, 0.97, 0.95 on correct branches and labels, condone missing $B'$s. Correct answer only scores full marks for (b), (c), (d). When using "their probability $p$" for M1 and A1ft they must have $0 < p < 1$ |
| **(b)** $0.25 \times 0.98 = \mathbf{0.245}$ (or exact equiv. e.g. $\frac{49}{200}$) | M1 A1 **(2)** | M1 for $0.25 \times$ 'their 0.98' o.e. |
| **(c)** $0.25\times0.02 + 0.45\times0.03 + 0.3\times0.05 = \mathbf{0.0335}$ (or exact equiv. e.g. $\frac{67}{2000}$) | M1 A1 **(2)** | M1 for $0.25\times\text{their } 0.02 + 0.45\times\text{their } 0.03 + 0.3\times\text{their } 0.05$. Condone 1 transcription error. Or $1-(0.25\times\text{their }0.98+0.45\times\text{their }0.97+0.3\times\text{their }0.95)$ |
| **(d)** $[P(J\cup L\|B)] = \frac{0.25\times0.02 + 0.3\times0.05}{0.0335}$ or $\frac{0.0335 - 0.45\times0.03}{0.0335}$ $= 0.5970\ldots$ awrt **0.597** (or $\frac{40}{67}$ or exact equiv.) | M1 A1ft A1 **(3)** | M1 for use of conditional probability with their (c) as denominator, and exactly 2 products on numerator with at least one correct (or correct ft), or their (c) minus one product from their (c). A1ft for $\frac{0.25\times\text{their }0.02 + 0.3\times\text{their }0.05}{\text{their}(c)}$ or $\frac{\text{their}(c)-0.45\times\text{their }0.03}{\text{their}(c)}$ or $\frac{0.02}{\text{their}(c)}$. A1 awrt 0.597 or exact fraction $\frac{40}{67}$ |
---\begin{enumerate}
\item In a factory, three machines, $J , K$ and $L$, are used to make biscuits.
\end{enumerate}
Machine $J$ makes $25 \%$ of the biscuits.
Machine $K$ makes $45 \%$ of the biscuits.
The rest of the biscuits are made by machine $L$.\\
It is known that $2 \%$ of the biscuits made by machine $J$ are broken, $3 \%$ of the biscuits made by machine $K$ are broken and 5\% of the biscuits made by machine $L$ are broken.\\
(a) Draw a tree diagram to illustrate all the possible outcomes and associated probabilities.
A biscuit is selected at random.\\
(b) Calculate the probability that the biscuit is made by machine $J$ and is not broken.\\
(c) Calculate the probability that the biscuit is broken.\\
(d) Given that the biscuit is broken, find the probability that it was not made by machine $K$.\\
\hfill \mbox{\textit{Edexcel S1 2014 Q4 [9]}}