AQA S1 2014 June — Question 3 12 marks

Exam BoardAQA
ModuleS1 (Statistics 1)
Year2014
SessionJune
Marks12
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicConditional Probability
TypeBasic two-way table probability
DifficultyEasy -1.2 This is a straightforward conditional probability question using a two-way table with clearly presented data. Students need only to identify the relevant cells and apply the basic conditional probability formula P(A|B) = P(A∩B)/P(B), which is a standard S1 technique requiring minimal problem-solving beyond careful reading of the table.
Spec5.06a Chi-squared: contingency tables

3 The table shows the colour of hair and the colour of eyes of a sample of 750 people from a particular population.

Question 3:
(a)(i)
AnswerMarks Guidance
\(\frac{220}{750} = 0.293\)B1 Accept \(\frac{22}{75}\)
(a)(ii)
AnswerMarks Guidance
\(\frac{24}{750} = 0.032\)B1 Accept \(\frac{4}{125}\)
(a)(iii)
AnswerMarks Guidance
\(\frac{110 + 215 - 24}{750} = \frac{301}{750}\)M1 For adding totals and subtracting overlap
\(= 0.401\)A1 Accept \(\frac{301}{750}\)
(a)(iv)
AnswerMarks Guidance
\(\frac{64}{220}\)M1 Correct numerator and denominator
\(= 0.291\)A1 Accept \(\frac{16}{55}\)
(a)(v)
AnswerMarks Guidance
\(\frac{64}{195}\)M1 Correct numerator and denominator
\(= 0.328\)A1 Accept \(\frac{64}{195}\)
(b)
AnswerMarks Guidance
Number with dark hair and brown eyes \(= 92\)B1 Seen or implied
Number with medium hair and green eyes \(= 55\)B1 Seen or implied
\(3 \times \frac{92}{750} \times \frac{91}{749} \times \frac{55}{748}\)M1 Three fractions multiplied with factor of 3 (or equivalent arrangements)
\(= 0.00603\)A1 Answer to 3 s.f.
## Question 3:

**(a)(i)**

$\frac{220}{750} = 0.293$ | B1 | Accept $\frac{22}{75}$

**(a)(ii)**

$\frac{24}{750} = 0.032$ | B1 | Accept $\frac{4}{125}$

**(a)(iii)**

$\frac{110 + 215 - 24}{750} = \frac{301}{750}$ | M1 | For adding totals and subtracting overlap |
$= 0.401$ | A1 | Accept $\frac{301}{750}$

**(a)(iv)**

$\frac{64}{220}$ | M1 | Correct numerator and denominator |
$= 0.291$ | A1 | Accept $\frac{16}{55}$

**(a)(v)**

$\frac{64}{195}$ | M1 | Correct numerator and denominator |
$= 0.328$ | A1 | Accept $\frac{64}{195}$

**(b)**

Number with dark hair and brown eyes $= 92$ | B1 | Seen or implied

Number with medium hair and green eyes $= 55$ | B1 | Seen or implied

$3 \times \frac{92}{750} \times \frac{91}{749} \times \frac{55}{748}$ | M1 | Three fractions multiplied with factor of 3 (or equivalent arrangements)

$= 0.00603$ | A1 | Answer to 3 s.f.
3 The table shows the colour of hair and the colour of eyes of a sample of 750 people from a particular population.

\hfill \mbox{\textit{AQA S1 2014 Q3 [12]}}