| Exam Board | Edexcel |
|---|---|
| Module | S1 (Statistics 1) |
| Year | 2004 |
| Session | January |
| Marks | 16 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Tree Diagrams |
| Type | Independent stages tree diagram |
| Difficulty | Moderate -0.3 This is a straightforward S1 tree diagram question with independent events and simple fractions. Parts (a)-(c) are routine applications of probability rules, while part (d) requires identifying complementary outcomes but remains accessible. Slightly easier than average due to clear structure and standard techniques. |
| Spec | 2.03b Probability diagrams: tree, Venn, sample space |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| Tree diagram with correct number of branches | M1 | |
| \(\frac{2}{3}, \frac{1}{3}\) | A1 | |
| \(\frac{1}{2}, \frac{1}{2}, \frac{1}{2}, \frac{1}{2}\) | A1 | |
| \(\frac{1}{4}, \frac{3}{4} \cdots \frac{3}{4}\) | A1 | (4) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(P(\text{All 3 Keys}) = \frac{2}{3} \times \frac{1}{2} \times \frac{1}{4} = \frac{2}{24} = \frac{1}{12}\) | M1 A1 | \(\frac{1}{12}\); \(0.083\); \(0.0833\) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(P(\text{exactly } 1 \text{ key}) = \left(\frac{2}{3} \times \frac{1}{2} \times \frac{3}{4}\right) + \left(\frac{1}{3} \times \frac{1}{2} \times \frac{3}{4}\right) + \left(\frac{1}{3} \times \frac{1}{2} \times \frac{1}{4}\right)\) | M1 | 3 triples added |
| \(= \frac{10}{24} = \frac{5}{12}\) | A1 A1 A1 A1 | Each correct; \(\frac{10}{24}\); \(\frac{5}{12}\); \(0.416\); \(0.417\) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(P(\text{Keys not collected on at least 2 successive stages})\) | ||
| \(= \left(\frac{2}{3} \times \frac{1}{2} \times \frac{3}{4}\right) + \left(\frac{1}{3} \times \frac{1}{2} \times \frac{1}{4}\right) + \left(\frac{1}{3} \times \frac{1}{2} \times \frac{3}{4}\right)\) | M1 A1 A1 A1 | 3 triples added; Each correct |
| \(= \frac{10}{24} = \frac{5}{12}\) | A1 | \(\frac{10}{24}\); \(\frac{5}{12}\); \(0.416\); \(0.417\) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(1 - P(\text{Keys collected on at least 2 successive stages})\) | M1 | |
| \(= 1 - \left\{\left(\frac{2}{3} \times \frac{1}{2} \times \frac{1}{4}\right) + \left(\frac{2}{3} \times \frac{1}{2} \times \frac{3}{4}\right) + \left(\frac{1}{3} \times \frac{1}{2} \times \frac{1}{4}\right)\right\}\) | A1 A1 A1 | |
| \(= \frac{5}{8}\) | A1 | (5) |
# Question 6:
## Part (a):
| Answer | Marks | Guidance |
|--------|-------|----------|
| Tree diagram with correct number of branches | M1 | |
| $\frac{2}{3}, \frac{1}{3}$ | A1 | |
| $\frac{1}{2}, \frac{1}{2}, \frac{1}{2}, \frac{1}{2}$ | A1 | |
| $\frac{1}{4}, \frac{3}{4} \cdots \frac{3}{4}$ | A1 | (4) |
## Part (b):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $P(\text{All 3 Keys}) = \frac{2}{3} \times \frac{1}{2} \times \frac{1}{4} = \frac{2}{24} = \frac{1}{12}$ | M1 A1 | $\frac{1}{12}$; $0.083$; $0.0833$ | (2) |
## Part (c):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $P(\text{exactly } 1 \text{ key}) = \left(\frac{2}{3} \times \frac{1}{2} \times \frac{3}{4}\right) + \left(\frac{1}{3} \times \frac{1}{2} \times \frac{3}{4}\right) + \left(\frac{1}{3} \times \frac{1}{2} \times \frac{1}{4}\right)$ | M1 | 3 triples added |
| $= \frac{10}{24} = \frac{5}{12}$ | A1 A1 A1 A1 | Each correct; $\frac{10}{24}$; $\frac{5}{12}$; $0.416$; $0.417$ | (5) |
## Part (d):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $P(\text{Keys not collected on at least 2 successive stages})$ | | |
| $= \left(\frac{2}{3} \times \frac{1}{2} \times \frac{3}{4}\right) + \left(\frac{1}{3} \times \frac{1}{2} \times \frac{1}{4}\right) + \left(\frac{1}{3} \times \frac{1}{2} \times \frac{3}{4}\right)$ | M1 A1 A1 A1 | 3 triples added; Each correct |
| $= \frac{10}{24} = \frac{5}{12}$ | A1 | $\frac{10}{24}$; $\frac{5}{12}$; $0.416$; $0.417$ | (5) |
**Alternative:**
| Answer | Marks | Guidance |
|--------|-------|----------|
| $1 - P(\text{Keys collected on at least 2 successive stages})$ | M1 | |
| $= 1 - \left\{\left(\frac{2}{3} \times \frac{1}{2} \times \frac{1}{4}\right) + \left(\frac{2}{3} \times \frac{1}{2} \times \frac{3}{4}\right) + \left(\frac{1}{3} \times \frac{1}{2} \times \frac{1}{4}\right)\right\}$ | A1 A1 A1 | |
| $= \frac{5}{8}$ | A1 | (5) |6. One of the objectives of a computer game is to collect keys. There are three stages to the game. The probability of collecting a key at the first stage is $\frac { 2 } { 3 }$, at the second stage is $\frac { 1 } { 2 }$, and at the third stage is $\frac { 1 } { 4 }$.
\begin{enumerate}[label=(\alph*)]
\item Draw a tree diagram to represent the 3 stages of the game.
\item Find the probability of collecting all 3 keys.
\item Find the probability of collecting exactly one key in a game.
\item Calculate the probability that keys are not collected on at least 2 successive stages in a game.
\end{enumerate}
\hfill \mbox{\textit{Edexcel S1 2004 Q6 [16]}}