CAIE S1 2004 November — Question 6 9 marks

Exam BoardCAIE
ModuleS1 (Statistics 1)
Year2004
SessionNovember
Marks9
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicCombinations & Selection
TypeProbability distributions from selection
DifficultyStandard +0.3 This is a straightforward combinatorics problem requiring systematic listing of outcomes and basic probability calculations. While part (iv) requires computing expectation and variance from a distribution, all steps are routine applications of standard formulas with no novel problem-solving insight needed. The small sample space (10 outcomes) makes enumeration manageable, placing this slightly above average difficulty due to the multiple parts and careful bookkeeping required.
Spec5.01a Permutations and combinations: evaluate probabilities5.02a Discrete probability distributions: general5.02b Expectation and variance: discrete random variables

6 A box contains five balls numbered \(1,2,3,4,5\). Three balls are drawn randomly at the same time from the box.
  1. By listing all possible outcomes (123, 124, etc.), find the probability that the sum of the three numbers drawn is an odd number. The random variable \(L\) denotes the largest of the three numbers drawn.
  2. Find the probability that \(L\) is 4 .
  3. Draw up a table to show the probability distribution of \(L\).
  4. Calculate the expectation and variance of \(L\).

Question 6:
Part (i)
AnswerMarks Guidance
AnswerMarks Guidance
Options 123, 124, 125, 134, 135, 145, 234, 235, 245, 345M1 For listing options, at least 4 different ones
\(P(\text{odd}) = 0.4\)B1 2 For correct answer, legit obtained
Part (ii)
AnswerMarks Guidance
AnswerMarks Guidance
\(P(\text{largest is } 4) = 0.3\)B1 1 For correct answer
OR \(\frac{1 \times {}_3C_2}{{}_5C_3}\) SR if 9 options in (i) give B1 for 3/9 or 2/9 depending on their missing option
Part (iii)
AnswerMarks Guidance
AnswerMarks Guidance
\(l \quad\quad 3 \quad\quad 4 \quad\quad 5\)M1 For 3, 4, 5 in table or 1, 2 as well; no need for any probs but need to see an (uncompleted) second line
\(P(L=l) \quad 0.1 \quad 0.3 \quad 0.6\)M1 For evaluating another probability based on their list
A1 3For correct answer
Part (iv)
AnswerMarks Guidance
AnswerMarks Guidance
\(E(L) = \sum lp = 3 \times 0.1 + 4 \times 0.4 + 5 \times 0.6 = 4.5\)B1ft For correct answer, ft if their \(\sum p = 1\)
\(\text{Var}(L) = 3^2 \times 0.1 + 4^2 \times 0.3 + 5^2 \times 0.6 - (\text{their } 4.5^2)\)M1 For evaluating their \(\sum l^2 p - (\text{their } 4.5^2)\) (must see \(- \text{their } 4.5^2\)) each \(p < 1\), in first numerical instance, ie can forget the sq rt subsequently
\(= 0.45\)A1 3 For correct answer
# Question 6:

## Part (i)
| Answer | Marks | Guidance |
|--------|-------|----------|
| Options 123, 124, 125, 134, 135, 145, 234, 235, 245, 345 | M1 | For listing options, at least 4 different ones |
| $P(\text{odd}) = 0.4$ | B1 **2** | For correct answer, legit obtained |

## Part (ii)
| Answer | Marks | Guidance |
|--------|-------|----------|
| $P(\text{largest is } 4) = 0.3$ | B1 **1** | For correct answer |
| OR $\frac{1 \times {}_3C_2}{{}_5C_3}$ | | **SR** if 9 options in **(i)** give B1 for 3/9 or 2/9 depending on their missing option |

## Part (iii)
| Answer | Marks | Guidance |
|--------|-------|----------|
| $l \quad\quad 3 \quad\quad 4 \quad\quad 5$ | M1 | For 3, 4, 5 in table or 1, 2 as well; no need for any probs but need to see an (uncompleted) second line |
| $P(L=l) \quad 0.1 \quad 0.3 \quad 0.6$ | M1 | For evaluating another probability based on their list |
| | A1 **3** | For correct answer |

## Part (iv)
| Answer | Marks | Guidance |
|--------|-------|----------|
| $E(L) = \sum lp = 3 \times 0.1 + 4 \times 0.4 + 5 \times 0.6 = 4.5$ | B1ft | For correct answer, ft if their $\sum p = 1$ |
| $\text{Var}(L) = 3^2 \times 0.1 + 4^2 \times 0.3 + 5^2 \times 0.6 - (\text{their } 4.5^2)$ | M1 | For evaluating their $\sum l^2 p - (\text{their } 4.5^2)$ (must see $- \text{their } 4.5^2$) each $p < 1$, in first numerical instance, ie can forget the sq rt subsequently |
| $= 0.45$ | A1 **3** | For correct answer |

---
6 A box contains five balls numbered $1,2,3,4,5$. Three balls are drawn randomly at the same time from the box.\\
(i) By listing all possible outcomes (123, 124, etc.), find the probability that the sum of the three numbers drawn is an odd number.

The random variable $L$ denotes the largest of the three numbers drawn.\\
(ii) Find the probability that $L$ is 4 .\\
(iii) Draw up a table to show the probability distribution of $L$.\\
(iv) Calculate the expectation and variance of $L$.

\hfill \mbox{\textit{CAIE S1 2004 Q6 [9]}}