CAIE S1 2014 June — Question 1 4 marks

Exam BoardCAIE
ModuleS1 (Statistics 1)
Year2014
SessionJune
Marks4
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicBinomial Distribution
TypeDirect binomial probability calculation
DifficultyModerate -0.5 This is a straightforward binomial probability calculation requiring only the application of P(X < 4) = P(X ≤ 3) using the binomial formula or tables with n=19, p=0.12. It involves routine computation with no conceptual challenges beyond recognizing the binomial setup and summing probabilities for X=0,1,2,3.
Spec2.04b Binomial distribution: as model B(n,p)2.04c Calculate binomial probabilities
1 In a certain country \(12 \%\) of houses have solar heating. 19 houses are chosen at random. Find the probability that fewer than 4 houses have solar heating.
Question 1:
AnswerMarks Guidance
WorkingMark Guidance
\(X \sim B(19, 0.12)\), \(P(X < 4) = P(0,1,2,3)\)M1 Any binomial term \(^{19}C_r p^r(1-p)^{19-r}\), \(0 < p < 1\)
\(= (0.88)^{19} + ^{19}C_1(0.12)^1(0.88)^{18} + ^{19}C_2(0.12)^2(0.88)^{17} + ^{19}C_3(0.12)^3(0.88)^{16}\)M1 Any binomial term \(^nCx(0.12\) or \(0.88)^x(0.88\) or \(0.12)^{n-x}\)
M1\(P(0,1,2,3)\) binomial expression with at least 2 consistent terms
\(= 0.813\)A1 [4] Correct answer 
## Question 1:

| Working | Mark | Guidance |
|---------|------|----------|
| $X \sim B(19, 0.12)$, $P(X < 4) = P(0,1,2,3)$ | M1 | Any binomial term $^{19}C_r p^r(1-p)^{19-r}$, $0 < p < 1$ |
| $= (0.88)^{19} + ^{19}C_1(0.12)^1(0.88)^{18} + ^{19}C_2(0.12)^2(0.88)^{17} + ^{19}C_3(0.12)^3(0.88)^{16}$ | M1 | Any binomial term $^nCx(0.12$ or $0.88)^x(0.88$ or $0.12)^{n-x}$ |
| | M1 | $P(0,1,2,3)$ binomial expression with at least 2 consistent terms |
| $= 0.813$ | A1 **[4]** | Correct answer |

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1 In a certain country $12 \%$ of houses have solar heating. 19 houses are chosen at random. Find the probability that fewer than 4 houses have solar heating.

\hfill \mbox{\textit{CAIE S1 2014 Q1 [4]}}
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