CAIE S1 2016 June — Question 4 6 marks

Exam BoardCAIE
ModuleS1 (Statistics 1)
Year2016
SessionJune
Marks6
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TopicDiscrete Probability Distributions
TypeSampling without replacement
DifficultyModerate -0.8 This is a straightforward application of calculating expectation and variance for a discrete probability distribution with sampling without replacement. Students need to find P(X=0), P(X=1), P(X=2) using basic combinatorics or tree diagrams, then apply standard formulas E(X) = Σxp(x) and Var(X) = E(X²) - [E(X)]². The calculations are routine with no conceptual challenges beyond understanding the setup.
Spec2.04a Discrete probability distributions2.04b Binomial distribution: as model B(n,p)
4 A box contains 2 green sweets and 5 blue sweets. Two sweets are taken at random from the box, without replacement. The random variable \(X\) is the number of green sweets taken. Find \(\mathrm { E } ( X )\) and \(\operatorname { Var } ( X )\).
Question 4:
AnswerMarks Guidance
AnswerMarks Guidance
\([P(X=0)] = P(B,B) = \frac{5}{7}\times\frac{4}{6} = \frac{10}{21}\)M1 Attempt to find \(P(0)\) or \(P(1)\) or \(P(2)\); can be seen as \(P(BB)\) etc. or table unsimplified
\([P(X=1)] = P(G,B) + P(B,G) = \frac{2}{7}\times\frac{5}{6}\times2 = \frac{10}{21}\)A1 \(P(1)\) or \(P(BG)+P(GB)\) correct
\([P(X=2)] = P(G,G) = \frac{2}{7}\times\frac{1}{6} = \frac{1}{21}\)A1 \(P(0)\) or \(P(2)\) correct; must see \(X\) value
\(E(X) = 0 + \frac{10}{21} + \frac{2}{21} = \frac{4}{7}\ (0.571)\)B1\(\checkmark\) Correct answer ft their probs \(P(1)\) and \(P(2)\)
\(\text{Var}(X) = 0 + \frac{10}{21} + \frac{4}{21} - \left(\frac{4}{7}\right)^2 = \frac{50}{147}\ (0.340)\)M1, A1 [6] Attempt at \(\Sigma x^2p - [E(X)]^2\) 
## Question 4:

| Answer | Marks | Guidance |
|--------|-------|----------|
| $[P(X=0)] = P(B,B) = \frac{5}{7}\times\frac{4}{6} = \frac{10}{21}$ | M1 | Attempt to find $P(0)$ or $P(1)$ or $P(2)$; can be seen as $P(BB)$ etc. or table unsimplified |
| $[P(X=1)] = P(G,B) + P(B,G) = \frac{2}{7}\times\frac{5}{6}\times2 = \frac{10}{21}$ | A1 | $P(1)$ or $P(BG)+P(GB)$ correct |
| $[P(X=2)] = P(G,G) = \frac{2}{7}\times\frac{1}{6} = \frac{1}{21}$ | A1 | $P(0)$ or $P(2)$ correct; must see $X$ value |
| $E(X) = 0 + \frac{10}{21} + \frac{2}{21} = \frac{4}{7}\ (0.571)$ | B1$\checkmark$ | Correct answer ft their probs $P(1)$ and $P(2)$ |
| $\text{Var}(X) = 0 + \frac{10}{21} + \frac{4}{21} - \left(\frac{4}{7}\right)^2 = \frac{50}{147}\ (0.340)$ | M1, A1 [6] | Attempt at $\Sigma x^2p - [E(X)]^2$ |

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4 A box contains 2 green sweets and 5 blue sweets. Two sweets are taken at random from the box, without replacement. The random variable $X$ is the number of green sweets taken. Find $\mathrm { E } ( X )$ and $\operatorname { Var } ( X )$.

\hfill \mbox{\textit{CAIE S1 2016 Q4 [6]}}
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