| Exam Board | CAIE |
|---|---|
| Module | S1 (Statistics 1) |
| Year | 2018 |
| Session | November |
| Marks | 5 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Permutations & Arrangements |
| Type | Basic arrangements with repeated letters |
| Difficulty | Moderate -0.8 Part (i) is a direct application of the formula for arrangements with repeated items (11!/(4!4!2!)), requiring only recall of a standard technique. Part (ii) involves straightforward counting of favorable outcomes and total outcomes using combinations. Both parts are routine textbook exercises with no problem-solving insight required, making this easier than average. |
| Spec | 2.03e Model with probability: critiquing assumptions5.01a Permutations and combinations: evaluate probabilities |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(P(S,S') = \frac{4}{11}\times\frac{7}{10} = \frac{28}{110}\); \(P(P,P') = \frac{2}{11}\times\frac{9}{10} = \frac{18}{110}\); \(P(I,I') = \frac{4}{11}\times\frac{7}{10} = \frac{28}{110}\); \(P(M,M') = \frac{1}{11}\times\frac{10}{10} = \frac{10}{110}\); Total \(= \frac{84}{110}\) | B1 | One of products correct |
| \(P(\text{Same}) = 1 - \frac{84}{110} = \frac{26}{110}\) | M1 | \(1 -\) sum of probabilities from 4 appropriate scenarios |
| A1 | Correct final answer |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(PP' = \frac{2\times9}{2}=9\); \(SS' = \frac{4\times7}{2}=14\); \(II' = \frac{4\times7}{2}=14\); \(MM' = \frac{1\times10}{2}=5\); Total ways \(= \frac{10\times11}{2}=55\); Number of ways letters repeating \(= 55-(9+14+14+5)=13\) | B1 | \(^{11}C_2\) seen as denominator of fraction (no extra terms), allow unsimplified |
| \(P(\text{Same}) = \frac{13}{55}\) | M1 | \(1 -\) sum of 4 appropriate scenarios |
| A1 | Correct final answer |
## Question 1(ii) — Alternative Methods:
**Method 3:**
| Answer | Mark | Guidance |
|--------|------|----------|
| $P(S,S') = \frac{4}{11}\times\frac{7}{10} = \frac{28}{110}$; $P(P,P') = \frac{2}{11}\times\frac{9}{10} = \frac{18}{110}$; $P(I,I') = \frac{4}{11}\times\frac{7}{10} = \frac{28}{110}$; $P(M,M') = \frac{1}{11}\times\frac{10}{10} = \frac{10}{110}$; Total $= \frac{84}{110}$ | B1 | One of products correct |
| $P(\text{Same}) = 1 - \frac{84}{110} = \frac{26}{110}$ | M1 | $1 -$ sum of probabilities from 4 appropriate scenarios |
| | A1 | Correct final answer |
**Method 4:**
| Answer | Mark | Guidance |
|--------|------|----------|
| $PP' = \frac{2\times9}{2}=9$; $SS' = \frac{4\times7}{2}=14$; $II' = \frac{4\times7}{2}=14$; $MM' = \frac{1\times10}{2}=5$; Total ways $= \frac{10\times11}{2}=55$; Number of ways letters repeating $= 55-(9+14+14+5)=13$ | B1 | $^{11}C_2$ seen as denominator of fraction (no extra terms), allow unsimplified |
| $P(\text{Same}) = \frac{13}{55}$ | M1 | $1 -$ sum of 4 appropriate scenarios |
| | A1 | Correct final answer |1 (i) How many different arrangements are there of the 11 letters in the word MISSISSIPPI?\\
(ii) Two letters are chosen at random from the 11 letters in the word MISSISSIPPI. Find the probability that these two letters are the same.\\
\hfill \mbox{\textit{CAIE S1 2018 Q1 [5]}}