CAIE S1 2016 November — Question 3 6 marks

Exam BoardCAIE
ModuleS1 (Statistics 1)
Year2016
SessionNovember
Marks6
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicBinomial Distribution
TypeMultiple independent binomial calculations
DifficultyModerate -0.8 This question involves straightforward application of basic probability rules and binomial distribution formulas with no problem-solving required. Part (i) is simple multiplication of independent probabilities, part (ii) is routine binomial probability calculation using P(X<3), and part (iii) requires only direct recall of mean=np and variance=np(1-p). All three parts are standard textbook exercises with clear setup and no conceptual challenges.
Spec2.03a Mutually exclusive and independent events2.04b Binomial distribution: as model B(n,p)2.04c Calculate binomial probabilities5.02b Expectation and variance: discrete random variables

3 Visitors to a Wildlife Park in Africa have independent probabilities of 0.9 of seeing giraffes, 0.95 of seeing elephants, 0.85 of seeing zebras and 0.1 of seeing lions.
  1. Find the probability that a visitor to the Wildlife Park sees all these animals.
  2. Find the probability that, out of 12 randomly chosen visitors, fewer than 3 see lions.
  3. 50 people independently visit the Wildlife Park. Find the mean and variance of the number of these people who see zebras.

(i)
AnswerMarks Guidance
B1[1] Bin term \(\binom{12}{x}p^x(1-p)^{12-x}\) where \(p < 1\), \(x \neq 0\)
(ii)
\(0.9 \times 0.95 \times 0.85 \times 0.1 = 0.0727\)
AnswerMarks
M1Bin expression with \(p = 0.1\) or \(0.9\), \(n = 12\), 2 or 3 terms
\(P(0, 1, 2) = (0.9)^{12} + \binom{12}{1}(0.1)(0.9)^{11} + \binom{12}{2}(0.1)^2(0.9)^{10} = 0.889\)
AnswerMarks
M1[3]
A1
(iii)
\(X \sim B(50, 0.85)\)
AnswerMarks
M1\(50 \times 0.85\) seen or implied
Expectation \(= 50 \times 0.85 = 42.5\)
Var \(= 50 \times 0.85 \times 0.15 = 6.375\)
AnswerMarks
A1Correct unsimplified mean and var
[2]
**(i)**

B1 | [1] | Bin term $\binom{12}{x}p^x(1-p)^{12-x}$ where $p < 1$, $x \neq 0$

**(ii)**

$0.9 \times 0.95 \times 0.85 \times 0.1 = 0.0727$

M1 | Bin expression with $p = 0.1$ or $0.9$, $n = 12$, 2 or 3 terms

$P(0, 1, 2) = (0.9)^{12} + \binom{12}{1}(0.1)(0.9)^{11} + \binom{12}{2}(0.1)^2(0.9)^{10} = 0.889$

M1 | [3]

A1

**(iii)**

$X \sim B(50, 0.85)$

M1 | $50 \times 0.85$ seen or implied

Expectation $= 50 \times 0.85 = 42.5$

Var $= 50 \times 0.85 \times 0.15 = 6.375$

A1 | Correct unsimplified mean and var

[2]

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3 Visitors to a Wildlife Park in Africa have independent probabilities of 0.9 of seeing giraffes, 0.95 of seeing elephants, 0.85 of seeing zebras and 0.1 of seeing lions.\\
(i) Find the probability that a visitor to the Wildlife Park sees all these animals.\\
(ii) Find the probability that, out of 12 randomly chosen visitors, fewer than 3 see lions.\\
(iii) 50 people independently visit the Wildlife Park. Find the mean and variance of the number of these people who see zebras.

\hfill \mbox{\textit{CAIE S1 2016 Q3 [6]}}