CAIE S1 2011 June — Question 1 4 marks

Exam BoardCAIE
ModuleS1 (Statistics 1)
Year2011
SessionJune
Marks4
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicBinomial Distribution
TypeProbability of range of values
DifficultyModerate -0.3 This is a straightforward binomial probability question requiring identification of parameters (n=18, p=2.7/18=0.15), then calculating P(2≤X≤4) = P(X=2) + P(X=3) + P(X=4) using the binomial formula. It's slightly easier than average as it's a direct application with clear setup and routine calculation, though the cumulative probability adds minor complexity.
Spec2.04b Binomial distribution: as model B(n,p)2.04c Calculate binomial probabilities
1 Biscuits are sold in packets of 18. There is a constant probability that any biscuit is broken, independently of other biscuits. The mean number of broken biscuits in a packet has been found to be 2.7 . Find the probability that a packet contains between 2 and 4 (inclusive) broken biscuits.
AnswerMarks Guidance
\(18p = 2.7, p = 0.15\)B1 Correct value for \(p\)
\(P(2, 3, 4) = {}^{18}C_2 \times (0.15)^2(0.85)^{16} + {}^{18}C_3(0.15)^3(0.85)^{15} + {}^{18}C_4(0.15)^4(0.85)^{14} = 0.655\)M1 Summing 3 binomial probs o.e
A1Correct unsimplified answer
A1 [4]Correct answer 
$18p = 2.7, p = 0.15$ | B1 | Correct value for $p$
$P(2, 3, 4) = {}^{18}C_2 \times (0.15)^2(0.85)^{16} + {}^{18}C_3(0.15)^3(0.85)^{15} + {}^{18}C_4(0.15)^4(0.85)^{14} = 0.655$ | M1 | Summing 3 binomial probs o.e
| A1 | Correct unsimplified answer
| A1 [4] | Correct answer
1 Biscuits are sold in packets of 18. There is a constant probability that any biscuit is broken, independently of other biscuits. The mean number of broken biscuits in a packet has been found to be 2.7 . Find the probability that a packet contains between 2 and 4 (inclusive) broken biscuits.

\hfill \mbox{\textit{CAIE S1 2011 Q1 [4]}}
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