CAIE S1 2009 November — Question 6 14 marks

Exam BoardCAIE
ModuleS1 (Statistics 1)
Year2009
SessionNovember
Marks14
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicApproximating Binomial to Normal Distribution
TypeProbability between two values
DifficultyModerate -0.3 Parts (i)-(iii) are standard S1 probability questions involving combinations and conditional probability with straightforward calculations. Part (iv) requires a routine binomial-to-normal approximation with continuity correction—a standard textbook application. The multi-part structure adds length but not conceptual difficulty, making this slightly easier than average overall.
Spec2.03a Mutually exclusive and independent events2.03c Conditional probability: using diagrams/tables2.04d Normal approximation to binomial
6 A box contains 4 pears and 7 oranges. Three fruits are taken out at random and eaten. Find the probability that
  1. 2 pears and 1 orange are eaten, in any order,
  2. the third fruit eaten is an orange,
  3. the first fruit eaten was a pear, given that the third fruit eaten is an orange. There are 121 similar boxes in a warehouse. One fruit is taken at random from each box.
  4. Using a suitable approximation, find the probability that fewer than 39 are pears.
Question 6:
Part (i):
AnswerMarks Guidance
Answer/WorkingMark Guidance
\(\frac{{}^4C_2\times{}^7C_1}{{}^{11}C_3} = 0.255\)M1 Using 2 combs mult for numerator and 1 comb for denom
M1Correct denom or num unsimplified
A1Correct answer
OR \(\frac{4}{11}\times\frac{3}{10}\times\frac{7}{9}\times3\)M1 Multiplying 3 correct probs
\(= 0.255\ (14/55)\ (42/165)\)M1, A1 [3] Mult by 3 or \(\Sigma\) their 3 options; Correct answer
Part (ii):
AnswerMarks Guidance
Answer/WorkingMark Guidance
\(P(3^{\text{rd}}\text{ is orange}) = P(P,P,O)+P(P,O,P)+P(O,P,O)+P(O,O,O)\)M1 Summing four 3-factor options with or without replacement
\(=\frac{4}{11}\times\frac{3}{10}\times\frac{7}{9}+\frac{4}{11}\times\frac{7}{10}\times\frac{6}{9}+\frac{7}{11}\times\frac{4}{10}\times\frac{6}{9}+\frac{7}{11}\times\frac{6}{10}\times\frac{5}{9}\)A1 At least 3 correct unsimplified options
\(=\left[\frac{14}{165}+\frac{28}{165}+\frac{28}{165}+\frac{7}{33}\right]\)
\(= 7/11\ (0.636)\)A1 [3] Correct answer. Award B3 if correct answer stated with no working
Part (iii):
AnswerMarks Guidance
Answer/WorkingMark Guidance
\(P(P\O) = \frac{P(P\cap O)}{P(O)}\) M1
\(= \frac{P(P,P,O)+P(P,O,O)}{P(O)}\)M1 Summing exactly 2 three-factor products in num
\(= \frac{28/110}{7/11} = \frac{28}{70} = \frac{4}{10} = 0.4\)A1 [3] Correct answer
Part (iv):
AnswerMarks Guidance
Answer/WorkingMark Guidance
\(\mu = 121\times\frac{4}{11} = 44\)B1 44 and 28 or 5.29 seen
\(\sigma^2 = 121\times\frac{4}{11}\times\frac{7}{11} = 28\)M1 Standardising, with or without cc, must have sq rt on denom
\(P(X<39) = \Phi\left(\frac{38.5-44}{\sqrt{28}}\right)\)M1 cc either 39.5 or 38.5
\(= \Phi(-1.039) = 1-0.8506\)M1 Correct area "\(1-\Phi\)" seen
\(= 0.149\)A1 [5] Correct answer 
# Question 6:

## Part (i):

| Answer/Working | Mark | Guidance |
|---|---|---|
| $\frac{{}^4C_2\times{}^7C_1}{{}^{11}C_3} = 0.255$ | M1 | Using 2 combs mult for numerator and 1 comb for denom |
| | M1 | Correct denom or num unsimplified |
| | A1 | Correct answer |
| OR $\frac{4}{11}\times\frac{3}{10}\times\frac{7}{9}\times3$ | M1 | Multiplying 3 correct probs |
| $= 0.255\ (14/55)\ (42/165)$ | M1, A1 **[3]** | Mult by 3 or $\Sigma$ their 3 options; Correct answer |

## Part (ii):

| Answer/Working | Mark | Guidance |
|---|---|---|
| $P(3^{\text{rd}}\text{ is orange}) = P(P,P,O)+P(P,O,P)+P(O,P,O)+P(O,O,O)$ | M1 | Summing four 3-factor options with or without replacement |
| $=\frac{4}{11}\times\frac{3}{10}\times\frac{7}{9}+\frac{4}{11}\times\frac{7}{10}\times\frac{6}{9}+\frac{7}{11}\times\frac{4}{10}\times\frac{6}{9}+\frac{7}{11}\times\frac{6}{10}\times\frac{5}{9}$ | A1 | At least 3 correct unsimplified options |
| $=\left[\frac{14}{165}+\frac{28}{165}+\frac{28}{165}+\frac{7}{33}\right]$ | | |
| $= 7/11\ (0.636)$ | A1 **[3]** | Correct answer. Award B3 if correct answer stated with no working |

## Part (iii):

| Answer/Working | Mark | Guidance |
|---|---|---|
| $P(P\|O) = \frac{P(P\cap O)}{P(O)}$ | M1 | Substituting in cond prob formula with at least one 3-factor product in num, and denom their (ii) or $7/11$ |
| $= \frac{P(P,P,O)+P(P,O,O)}{P(O)}$ | M1 | Summing exactly 2 three-factor products in num |
| $= \frac{28/110}{7/11} = \frac{28}{70} = \frac{4}{10} = 0.4$ | A1 **[3]** | Correct answer |

## Part (iv):

| Answer/Working | Mark | Guidance |
|---|---|---|
| $\mu = 121\times\frac{4}{11} = 44$ | B1 | 44 and 28 or 5.29 seen |
| $\sigma^2 = 121\times\frac{4}{11}\times\frac{7}{11} = 28$ | M1 | Standardising, with or without cc, must have sq rt on denom |
| $P(X<39) = \Phi\left(\frac{38.5-44}{\sqrt{28}}\right)$ | M1 | cc either 39.5 or 38.5 |
| $= \Phi(-1.039) = 1-0.8506$ | M1 | Correct area "$1-\Phi$" seen |
| $= 0.149$ | A1 **[5]** | Correct answer |
6 A box contains 4 pears and 7 oranges. Three fruits are taken out at random and eaten. Find the probability that\\
(i) 2 pears and 1 orange are eaten, in any order,\\
(ii) the third fruit eaten is an orange,\\
(iii) the first fruit eaten was a pear, given that the third fruit eaten is an orange.

There are 121 similar boxes in a warehouse. One fruit is taken at random from each box.\\
(iv) Using a suitable approximation, find the probability that fewer than 39 are pears.

\hfill \mbox{\textit{CAIE S1 2009 Q6 [14]}}
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