| Exam Board | CAIE |
|---|---|
| Module | S1 (Statistics 1) |
| Year | 2020 |
| Session | November |
| Marks | 7 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Permutations & Arrangements |
| Type | Conditional probability in arrangements |
| Difficulty | Standard +0.3 Part (a) is straightforward counting with fixed positions. Part (b) requires conditional probability with arrangements, but the calculation is mechanical: count arrangements with 3Es together and both Rs together, divided by arrangements with 3Es together. This is a standard S1 exercise requiring careful counting but no novel insight, making it slightly easier than average. |
| Spec | 2.03c Conditional probability: using diagrams/tables5.01a Permutations and combinations: evaluate probabilities |
| Answer | Marks | Guidance |
|---|---|---|
| Total number of ways \(= \dfrac{8!}{3!\,2!} = 3360\) | B1 | Correct unsimplified expression for total number of ways |
| Number of ways with V and E in correct positions \(= \dfrac{6!}{2 \times 2!} = 180\) | B1 | \(\dfrac{6!}{2\times2!}\) alone or as numerator in attempt to find number of ways with V and E in correct positions. No \(\times\), \(\pm\) |
| Probability \(= \dfrac{180}{3360} = \dfrac{3}{56}\) or \(0.0536\) | B1 FT | Final answer from their \(\dfrac{6!}{2\times2!}\) divided by their total number of ways |
| Answer | Marks | Guidance |
|---|---|---|
| \(\dfrac{1}{8} \times \dfrac{3}{7}\) | M1 | \(\dfrac{a}{8} \times \dfrac{b}{7}\) seen, no other terms (correct denominators) |
| M1 | \(\dfrac{1}{c} \times \dfrac{3}{d}\) seen, no other terms (correct numerators) | |
| \(\dfrac{3}{56}\) or \(0.0536\) | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Rs together and Es together: \(5!\) (120) | B1 | Alone or as numerator of probability to represent number of ways with Rs and Es together, no \(\times\), \(+\), \(-\) |
| Es together: \(\dfrac{6!}{2!} = 360\) | B1 | Alone or as denominator of probability to represent number of ways with Es together, no \(\times\), \(+\) or \(-\) |
| Probability \(= \dfrac{5!}{\dfrac{6!}{2!}}\) | M1 | \(\dfrac{\text{their}\; 5!}{\text{their}\; \frac{6!}{2!}}\) seen |
| \(\dfrac{1}{3}\) | A1 | OE |
| Answer | Marks | Guidance |
|---|---|---|
| \(P(\text{Rs together and Es together}) = \dfrac{5!}{\text{their total}} = \dfrac{1}{28}\) | B1 | |
| \(P(\text{Es together}) = \dfrac{\frac{6!}{2!}}{\text{their total}} = \dfrac{3}{28}\) | B1 | Alone or as numerator of probability representing P(Rs and Es together), no \(\times\), \(+\), \(-\) |
| Probability \(= \dfrac{\frac{1}{28}}{\frac{3}{28}}\) | M1 | Alone or as denominator of probability representing P(Es together), no \(\times\), \(+\) or \(-\) |
| \(\dfrac{1}{3}\) | A1 | OE, \(\dfrac{\text{their}\;\frac{1}{28}}{\text{their}\;\frac{3}{28}}\) seen |
## Question 5:
### Part 5(a):
Total number of ways $= \dfrac{8!}{3!\,2!} = 3360$ | B1 | Correct unsimplified expression for total number of ways
Number of ways with V and E in correct positions $= \dfrac{6!}{2 \times 2!} = 180$ | B1 | $\dfrac{6!}{2\times2!}$ alone or as numerator in attempt to find number of ways with V and E in correct positions. No $\times$, $\pm$
Probability $= \dfrac{180}{3360} = \dfrac{3}{56}$ or $0.0536$ | B1 FT | Final answer from their $\dfrac{6!}{2\times2!}$ divided by their total number of ways
**Alternative method for 5(a):**
$\dfrac{1}{8} \times \dfrac{3}{7}$ | M1 | $\dfrac{a}{8} \times \dfrac{b}{7}$ seen, no other terms (correct denominators)
| M1 | $\dfrac{1}{c} \times \dfrac{3}{d}$ seen, no other terms (correct numerators)
$\dfrac{3}{56}$ or $0.0536$ | A1 |
### Part 5(b):
Rs together and Es together: $5!$ (120) | B1 | Alone or as numerator of probability to represent number of ways with Rs and Es together, no $\times$, $+$, $-$
Es together: $\dfrac{6!}{2!} = 360$ | B1 | Alone or as denominator of probability to represent number of ways with Es together, no $\times$, $+$ or $-$
Probability $= \dfrac{5!}{\dfrac{6!}{2!}}$ | M1 | $\dfrac{\text{their}\; 5!}{\text{their}\; \frac{6!}{2!}}$ seen
$\dfrac{1}{3}$ | A1 | OE
**Alternative method for 5(b):**
$P(\text{Rs together and Es together}) = \dfrac{5!}{\text{their total}} = \dfrac{1}{28}$ | B1 |
$P(\text{Es together}) = \dfrac{\frac{6!}{2!}}{\text{their total}} = \dfrac{3}{28}$ | B1 | Alone or as numerator of probability representing P(Rs and Es together), no $\times$, $+$, $-$
Probability $= \dfrac{\frac{1}{28}}{\frac{3}{28}}$ | M1 | Alone or as denominator of probability representing P(Es together), no $\times$, $+$ or $-$
$\dfrac{1}{3}$ | A1 | OE, $\dfrac{\text{their}\;\frac{1}{28}}{\text{their}\;\frac{3}{28}}$ seen5 The 8 letters in the word RESERVED are arranged in a random order.
\begin{enumerate}[label=(\alph*)]
\item Find the probability that the arrangement has V as the first letter and E as the last letter.
\item Find the probability that the arrangement has both Rs together given that all three Es are together.
\end{enumerate}
\hfill \mbox{\textit{CAIE S1 2020 Q5 [7]}}