June 2025 Paper 2 Q7

OCR MEICurrent spec7 marksProbability

7 Some people have blood type A and some people have brown hair. A person may have both, just one, or neither of these characteristics.

A person is selected at random.

A is the event ‘the person has blood type A’.
B is the event ‘the person has brown hair’.

You are given the following probabilities.

\(\mathrm{P}(\mathrm{A} \cap \mathrm{B}^{\prime}) = 0.3382,\ \mathrm{P}(\mathrm{A}^{\prime} \cap \mathrm{B}) = 0.0682\) and \(\mathrm{P}((\mathrm{A} \cup \mathrm{B})^{\prime}) = 0.5518\).

(a) Show this information on the Venn diagram in the Printed Answer Booklet. [2]

(b) Determine whether A and B are independent events. [4]

Scheme	Marks	AO
	B1 B1	1.1 1.1
	[2]

Mark scheme (b)

Scheme	Marks	AO
\([\mathrm{P}(\mathrm{A} \cap \mathrm{B}) =]\ 1 - 0.5518 - 0.3382 - 0.0682\)	M1	3.1a
P(A) = 0.3382 + their 0.0418 or P(B) = 0.0682 + their 0.0418	M1	2.1
[P(A) =] 0.38 and [P(B) =] 0.11	A1	1.1
\(0.38\times 0.11 = 0.0418\) so [A and B are] independent	A1	3.2a
	[4]

Notes

M1: value for \(\mathrm{P}(\mathrm{A} \cap \mathrm{B})\) found; allow M1 for 0.0418 unsupported;
allow if seen in part (a) on diagram and nowhere else

M1: may be implied by 0.38 or 0.11 even if wrongly attributed

A1: both correct; allow unsimplified

A1: allow \((0.3382 + 0.0418)\times(0.0682 + 0.0418)\); allow P(A) \(\times\) P(B) = 0.0418 if P(A) = 0.38 and P(B) = 0.11 seen elsewhere

Alternative method

Scheme	Marks
\([\mathrm{P}(\mathrm{A} \cap \mathrm{B}) =]\ 1 - 0.5518 - 0.3382 - 0.0682\)	M1
P(B) = 0.0682 + their 0.0418 or P(A) = 0.3382 + their 0.0418	M1
\(\mathrm{P}(A\|B) = \dfrac{0.0418}{(0.0682 + 0.0418)} = 0.38\)	A1
\(\mathrm{P}(A\|B) = 0.38 = \mathrm{P}(A)\) so independent	A1

M1: value for \(\mathrm{P}(\mathrm{A} \cap \mathrm{B})\) found; allow M1 for 0.0418 unsupported
allow if seen in part (a) on diagram

M1: may be implied by 0.11
or may be implied by 0.38

A1: or \(\mathrm{P}(B|A) = \dfrac{0.0418}{(0.3382 + 0.0418)} = 0.11\); allow eg \(\mathrm{P}(A|B) = \dfrac{\mathrm{P}(\mathrm{A}\cap\mathrm{B})}{\mathrm{P}(\mathrm{B})} = 0.38\) if P(B) = 0.11 and \(\mathrm{P}(\mathrm{A}\cap\mathrm{B}) = 0.0418\) seen elsewhere

A1: or \(\mathrm{P}(B|A) = 0.11 = \mathrm{P}(B)\) so independent

Mark scheme (c)

Scheme	Marks	AO
\(\mathrm{P}(\mathrm{A} \cap \mathrm{B}) \neq 0\) so not mutually exclusive or \(1 - 0.5518 \neq\) their 0.11 + their 0.38 so \(\mathrm{P}(A \cup B) \neq \mathrm{P}(A) + \mathrm{P}(B)\) so not mutually exclusive	B1	2.4
	[1]

Notes

B1: allow \(\mathrm{P}(\mathrm{A} \cap \mathrm{B}) =\) their 0.0418 so not mutually exclusive; must refer to probability;
do not allow
eg they have an intersection;
eg the intersection is 0.0418