CAIE S1 2010 June — Question 5 9 marks

Exam BoardCAIE
ModuleS1 (Statistics 1)
Year2010
SessionJune
Marks9
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicDiscrete Probability Distributions
TypeConstruct probability distribution from scenario
DifficultyStandard +0.3 This is a straightforward probability distribution question requiring systematic enumeration of outcomes, basic probability calculations, and standard expectation/variance formulas. Part (iv) adds conditional probability but remains routine. The question is slightly easier than average due to small sample spaces and clear structure, though it requires careful organization across multiple parts.
Spec2.03d Calculate conditional probability: from first principles5.02a Discrete probability distributions: general5.02b Expectation and variance: discrete random variables

5 Set \(A\) consists of the ten digits \(0,0,0,0,0,0,2,2,2,4\).
Set \(B\) consists of the seven digits \(0,0,0,0,2,2,2\).
One digit is chosen at random from each set. The random variable \(X\) is defined as the sum of these two digits.
  1. Show that \(\mathrm { P } ( X = 2 ) = \frac { 3 } { 7 }\).
  2. Tabulate the probability distribution of \(X\).
  3. Find \(\mathrm { E } ( X )\) and \(\operatorname { Var } ( X )\).
  4. Given that \(X = 2\), find the probability that the digit chosen from set \(A\) was 2 .

AnswerMarks Guidance
(i) \(P(2) = P(0,2) + P(2,0) = \frac{6}{10} \times \frac{3}{7} + \frac{3}{10} \times \frac{4}{7} = \frac{30}{70} = \frac{3}{7}\) AGM1 Summing two 2-factor probabilities
A1Correct answer legit obtained
(ii)X 0
P(X = y)24/70 30/70
B1Correct probs
(iii) \(E(X) = \frac{13}{7}\) Var\((X) = \frac{120}{70} + \frac{208}{70} + \frac{108}{70} - (13/7)^2 = 2.78\)B1ft Using variance formula correctly with mean² subtracted numerically, no extra division
M1Using variance formula correctly with mean² subtracted numerically, no extra division
A1Correct final answer
(iv) \(P(A2\text{Sum } 2) = \frac{3}{10} \times \frac{4}{7}}{30/70} = 0.4\) M1
A1Correct answer
**(i)** $P(2) = P(0,2) + P(2,0) = \frac{6}{10} \times \frac{3}{7} + \frac{3}{10} \times \frac{4}{7} = \frac{30}{70} = \frac{3}{7}$ AG | M1 | Summing two 2-factor probabilities |
| A1 | Correct answer legit obtained |

**(ii)** | X | 0 | 2 | 4 | 6 |
| P(X = y) | 24/70 | 30/70 | 13/70 | 3/70 | | B1 | Correct values for rv X |
| B1 | Correct probs |

**(iii)** $E(X) = \frac{13}{7}$ Var$(X) = \frac{120}{70} + \frac{208}{70} + \frac{108}{70} - (13/7)^2 = 2.78$ | B1ft | Using variance formula correctly with mean² subtracted numerically, no extra division |
| M1 | Using variance formula correctly with mean² subtracted numerically, no extra division |
| A1 | Correct final answer |

**(iv)** $P(A2|\text{Sum } 2) = \frac{3}{10} \times \frac{4}{7}}{30/70} = 0.4$ | M1 | Correct numerator with a $0 < \text{denom} < 1$ |
| A1 | Correct answer |

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5 Set $A$ consists of the ten digits $0,0,0,0,0,0,2,2,2,4$.\\
Set $B$ consists of the seven digits $0,0,0,0,2,2,2$.\\
One digit is chosen at random from each set. The random variable $X$ is defined as the sum of these two digits.\\
(i) Show that $\mathrm { P } ( X = 2 ) = \frac { 3 } { 7 }$.\\
(ii) Tabulate the probability distribution of $X$.\\
(iii) Find $\mathrm { E } ( X )$ and $\operatorname { Var } ( X )$.\\
(iv) Given that $X = 2$, find the probability that the digit chosen from set $A$ was 2 .

\hfill \mbox{\textit{CAIE S1 2010 Q5 [9]}}