S1 June 2015 Q3

EdexcelOld spec13 marksProbability

3. A college has 80 students in Year 12.

20 students study Biology
28 students study Chemistry
30 students study Physics
7 students study both Biology and Chemistry
11 students study both Chemistry and Physics
5 students study both Physics and Biology
3 students study all 3 of these subjects

(a) Draw a Venn diagram to represent this information. (5)

A Year 12 student at the college is selected at random.

(b) Find the probability that the student studies Chemistry but not Biology or Physics. (1)

Given that the student studies Chemistry or Physics or both,

(d) find the probability that the student does not study Biology. (2)

(e) Determine whether studying Biology and studying Chemistry are statistically independent. (3)

Venn diagram in a box: Biology only 11, Chemistry only 13, Physics only 17, Biology and Chemistry only 4, Biology and Physics only 2, Chemistry and Physics only 8, all three 3, outside 22 — Mark scheme (a)

Scheme	Marks
	B1 M1 A1 A1 B1
	(5)

Scheme	Marks
\(\dfrac{\text{'}13\text{'}}{80}\) or 0.1625	B1ft
	(1)

Mark scheme (c)

Scheme	Marks
\(\dfrac{28 + 30 - 11}{80}\) or \(\dfrac{2 + 3 + 4 + 8 + 13 + 17}{80}\) or \(1 - \dfrac{(11 + 22)}{80} = \dfrac{47}{80}\) or 0.5875	M1 A1
	(2)

Notes

M1 for a correct expression seen in (c) ( or ft their diagram). Correct ans M1A1

Mark scheme (d)

Scheme	Marks
\(\dfrac{\text{"}17 + 8 + 13\text{"}}{\text{"}47\text{"}}\) or \(\dfrac{\frac{\text{"}38\text{"}}{80}}{\frac{\text{"}47\text{"}}{80}}\) or \(1 - \dfrac{\text{"}2 + 3 + 4\text{"}}{\text{"}47\text{"}} = \dfrac{38}{47}\) (condone awrt 0.809)	M1 A1cao
	(2)

Notes

M1 for denominator of 47 or ft their numerator from part (c) and numerator of 38 or their (17 + 8 + 13) or (their 47) – their (2 + 3 + 4). Correct ans M1A1

Mark scheme (e)

Scheme	Marks
\(\mathrm{P}(B \mid C) = \dfrac{7}{28},\ \mathrm{P}(B) = \dfrac{20}{80}\) \(\mathrm{P}(C \mid B) = \dfrac{7}{20},\ \mathrm{P}(C) = \dfrac{28}{80}\) \(\mathrm{P}(B \cap C) = \dfrac{7}{80},\ \mathrm{P}(B) = \dfrac{20}{80}\ \mathrm{P}(C) = \dfrac{28}{80}\)	M1
\(\mathrm{P}(B \mid C) = \mathrm{P}(B),\ \mathrm{P}(C \mid B) = \mathrm{P}(C)\) these may be implied by correct conclusion \(\mathrm{P}(B \cap C) = \mathrm{P}(B) \times \mathrm{P}(C)\) this approach requires the product to be seen	M1
So, they are independent.	A1
	(3)
	(13 marks)

Notes

M1 for stating at least the required probs.& labelled for a correct test (can ft their diagram)

M1 for use of a correct test with \(B\) and \(C\) Must see product attempted for \(\mathrm{P}(B \cap C)\) test.

A1 for a correct test with all probabilities correct and a correct concluding statement.
NB M0M1A0 should be possible but A1 requires both Ms