| Exam Board | CAIE |
|---|---|
| Module | S1 (Statistics 1) |
| Year | 2011 |
| Session | November |
| Marks | 6 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Combinations & Selection |
| Type | Probability distributions from selection |
| Difficulty | Moderate -0.3 This is a straightforward hypergeometric distribution problem requiring systematic calculation of probabilities using combinations, followed by a routine variance calculation using the given expectation. The question is slightly easier than average because it provides E(X) rather than requiring students to calculate it, and the numbers are small enough to make calculations manageable. It tests standard S1 content without requiring novel insight. |
| Spec | 2.04a Discrete probability distributions5.02b Expectation and variance: discrete random variables |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(P(X=1) = P(\text{GBBB}) \times 4C_1\) | M1 | Considering values of \(X\) of 1, 2, 3, 4 |
| \(= 5/8 \times 3/7 \times 2/6 \times 1/5 \times 4 = 1/14\) | M1 | Attempting to find probability of at least 2 values of \(X\) |
| \(P(X=2) = P(\text{GGBB}) \times {}_4C_2 = 3/7\) | ||
| \(P(X=3) = P(\text{GGGB}) \times {}_4C_3 = 3/7\) | A1 | One correct probability |
| \(P(X=4) = P(\text{GGGG}) \times {}_4C_4 = 1/14\) | A1 | All correct |
| OR | ||
| \(P(1) = {}_5C_1 / {}_8C_4 = 1/14\) | M1 | Considering values of \(X\) of 1, 2, 3, 4 |
| \(P(2) = {}_3C_2 \times {}_5C_2 / {}_8C_4 = 3/7\) | M1 | Dividing by \({}_8C_4\) |
| \(P(3) = {}_3C_1 \times {}_5C_3 / {}_8C_4 = 3/7\) | A1 | One correct probability |
| \(P(4) = {}_5C_4 / {}_8C_4 = 1/14\) | A1 [4] | All correct |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(\text{Var}(X) = 1/14 + 12/7 + 27/7 + 16/14 - (5/2)^2\) | M1 | Using variance formula correctly with mean\(^2\) subtracted numerically, no extra division |
| \(= 15/28\ (0.536)\) | A1 [2] | Correct final answer |
## Question 3:
### Part (i):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $P(X=1) = P(\text{GBBB}) \times 4C_1$ | M1 | Considering values of $X$ of 1, 2, 3, 4 |
| $= 5/8 \times 3/7 \times 2/6 \times 1/5 \times 4 = 1/14$ | M1 | Attempting to find probability of at least 2 values of $X$ |
| $P(X=2) = P(\text{GGBB}) \times {}_4C_2 = 3/7$ | | |
| $P(X=3) = P(\text{GGGB}) \times {}_4C_3 = 3/7$ | A1 | One correct probability |
| $P(X=4) = P(\text{GGGG}) \times {}_4C_4 = 1/14$ | A1 | All correct |
| **OR** | | |
| $P(1) = {}_5C_1 / {}_8C_4 = 1/14$ | M1 | Considering values of $X$ of 1, 2, 3, 4 |
| $P(2) = {}_3C_2 \times {}_5C_2 / {}_8C_4 = 3/7$ | M1 | Dividing by ${}_8C_4$ |
| $P(3) = {}_3C_1 \times {}_5C_3 / {}_8C_4 = 3/7$ | A1 | One correct probability |
| $P(4) = {}_5C_4 / {}_8C_4 = 1/14$ | A1 [4] | All correct |
### Part (ii):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $\text{Var}(X) = 1/14 + 12/7 + 27/7 + 16/14 - (5/2)^2$ | M1 | Using variance formula correctly with mean$^2$ subtracted numerically, no extra division |
| $= 15/28\ (0.536)$ | A1 [2] | Correct final answer |
---3 A team of 4 is to be randomly chosen from 3 boys and 5 girls. The random variable $X$ is the number of girls in the team.\\
(i) Draw up a probability distribution table for $X$.\\
(ii) Given that $\mathrm { E } ( X ) = \frac { 5 } { 2 }$, calculate $\operatorname { Var } ( X )$.
\hfill \mbox{\textit{CAIE S1 2011 Q3 [6]}}