| Exam Board | CAIE |
|---|---|
| Module | S1 (Statistics 1) |
| Year | 2010 |
| Session | November |
| Marks | 11 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Combinations & Selection |
| Type | Probability of specific committee composition |
| Difficulty | Standard +0.3 This is a multi-part combinatorics question requiring systematic application of selection formulas and basic probability. Parts (i)-(iii) involve standard committee selection with constraints, while part (iv) adds arrangement probability. The techniques are straightforward (combinations, complementary counting, arrangements) with no novel insights required, making it slightly easier than average but still requiring careful case-by-case analysis. |
| Spec | 2.03d Calculate conditional probability: from first principles5.01a Permutations and combinations: evaluate probabilities |
| Answer | Marks | Guidance |
|---|---|---|
| 4M 2W or 5M 1W | M1 | At least 1 of \(^{10}C_4 \times ^8C_2\) and \(^{10}C_5 \times ^8C_1\) seen |
| chosen in \(^{10}C_4 \times ^8C_2 + ^{10}C_5 \times ^8C_1 = 9828\) | A1 | Correct unsimplified expression |
| A1 [3] | Correct answer |
| Answer | Marks | Guidance |
|---|---|---|
| \(^7C_1 \times ^8C_1 + ^4C_4 = 798\) | M1 | One of \(^7C_1 \times ^8C_1\) and \(^4C_4 \times (^8C_0)\) seen |
| \(\text{Prob} = 798/9828 = 0.0812\) | A1 [2] | Correct answer |
| Answer | Marks | Guidance |
|---|---|---|
| Albert + not T... \(^9C_3 \times ^8C_2 + ^8C_4 \times ^8C_1 = 3360\) | M1 | One of \(^7C_3 \times ^8C_2\) or \(^8C_4 \times ^8C_1\) or \(^9C_3 \times (^8C_0)\) seen |
| Tracey + not A... \(^8C_4 \times ^8C_1 + ^7C_5 = 1134\) | ||
| Number of ways \(= 4494\) | A1 | Unsimplified 3360 or 1134 seen |
| A1 [3] | Correct final answer |
| Answer | Marks | Guidance |
|---|---|---|
| \(6! - 4! \times 5 \times 2\) or \(6! - 5! \times 2\) \((= 480)\) OR \(4! \times 5 \times 4\) or \(4! \times ^5P_2\) \((= 480)\) | B1 | \(6! - 4! \times 5 \times 2\) or \(6! - 5! \times 2\) or \(4! \times 5 \times 4\) or \(4! \times ^5P_2\) dividing by 6! |
| prob \(= 480/6! = 2/3\) \((0.667)\) | M1 | dividing by 6! |
| A1 [3] | correct answer |
| Answer | Marks | Guidance |
|---|---|---|
| OR Women together \(5!/4! (= 5)\) Women not together \(= 15 - 5 = 10\) total ways \(MMMWMW = 6!/4!2! = 15\) prob \(= 2/3\) | B1 | 5 or 10 seen |
| M1 | Dividing by 15 | |
| A1 | Correct answer |
**(i)**
4M 2W or 5M 1W | M1 | At least 1 of $^{10}C_4 \times ^8C_2$ and $^{10}C_5 \times ^8C_1$ seen
chosen in $^{10}C_4 \times ^8C_2 + ^{10}C_5 \times ^8C_1 = 9828$ | A1 | Correct unsimplified expression
| A1 [3] | Correct answer
**(ii)**
$^7C_1 \times ^8C_1 + ^4C_4 = 798$ | M1 | One of $^7C_1 \times ^8C_1$ and $^4C_4 \times (^8C_0)$ seen
$\text{Prob} = 798/9828 = 0.0812$ | A1 [2] | Correct answer
**(iii)**
Albert + not T... $^9C_3 \times ^8C_2 + ^8C_4 \times ^8C_1 = 3360$ | M1 | One of $^7C_3 \times ^8C_2$ or $^8C_4 \times ^8C_1$ or $^9C_3 \times (^8C_0)$ seen
Tracey + not A... $^8C_4 \times ^8C_1 + ^7C_5 = 1134$ | |
Number of ways $= 4494$ | A1 | Unsimplified 3360 or 1134 seen
| A1 [3] | Correct final answer
**(iv)**
$6! - 4! \times 5 \times 2$ or $6! - 5! \times 2$ $(= 480)$ OR $4! \times 5 \times 4$ or $4! \times ^5P_2$ $(= 480)$ | B1 | $6! - 4! \times 5 \times 2$ or $6! - 5! \times 2$ or $4! \times 5 \times 4$ or $4! \times ^5P_2$ dividing by 6!
prob $= 480/6! = 2/3$ $(0.667)$ | M1 | dividing by 6!
| A1 [3] | correct answer
OR using probabilities...as above
OR Women together $5!/4! (= 5)$ Women not together $= 15 - 5 = 10$ total ways $MMMWMW = 6!/4!2! = 15$ prob $= 2/3$ | B1 | 5 or 10 seen
| M1 | Dividing by 15
| A1 | Correct answer7 A committee of 6 people, which must contain at least 4 men and at least 1 woman, is to be chosen from 10 men and 9 women.\\
(i) Find the number of possible committees that can be chosen.\\
(ii) Find the probability that one particular man, Albert, and one particular woman, Tracey, are both on the committee.\\
(iii) Find the number of possible committees that include either Albert or Tracey but not both.\\
(iv) The committee that is chosen consists of 4 men and 2 women. They queue up randomly in a line for refreshments. Find the probability that the women are not next to each other in the queue.
\hfill \mbox{\textit{CAIE S1 2010 Q7 [11]}}