S1 June 2017 Q3

EdexcelOld spec11 marksProbability

3. The Venn diagram shows three events \(A\), \(B\) and \(C\), where \(p\), \(q\), \(r\), \(s\) and \(t\) are probabilities.

Venn diagram in a box: A and B overlap, B and C overlap, A and C do not overlap. A only r, A and B s, B only t, B and C p, C only q, outside 0.08

\(\mathrm{P}(A) = 0.5\), \(\mathrm{P}(B) = 0.6\) and \(\mathrm{P}(C) = 0.25\) and the events \(B\) and \(C\) are independent.

(a) Find the value of \(p\) and the value of \(q\). (2)

(b) Find the value of \(r\). (2)

(d) State, giving a reason, whether or not the events \(A\) and \(B\) are independent. (2)

(e) Find \(\mathrm{P}(B \mid A \cup C)\). (3)

Completed Venn diagram: A only 0.22, A and B 0.28, B only 0.17, B and C 0.15, C only 0.10, outside 0.08 — Mark scheme (a)

Scheme	Marks
\(p = \mathrm{P}(B \cap C) = \ \mathrm{P}(B) \times \mathrm{P}(C) = 0.6 \times 0.25 = \)0.15	M1
\(q = [\mathrm{P}(C) - p] = \)0.10	A1
	(2)

Mark scheme (b)

Scheme	Marks
\(r = 1 - 0.08 - [\mathrm{P}(B) + q] = 1 - 0.08 - 0.6 - 0.1\) (o.e.) or 1 – 0.08 – (0.6+0.25 – \(p\))	M1
\(= \)0.22	A1cao
	(2)

Notes

Mark (b) & (c) together

M1 for a correct expression for \(r\) using \(\mathrm{P}(B \cup C)\). Can ft their \(q \in [0, 0.32]\)
A1cao for \(r = 0.22\) (correct ans only 2/2)

ALT Find \(t\) then \(s\) then \(r\)
(c) 2nd B1 for \(t = 0.17\) [ from 1 – 0.08 – P(\(A\)) – P(\(C\))]
1st B1ft for \(s = 0.28\) or P(\(B\)) – “0.17” – “0.15”
(b) M1 for \(r = \mathrm{P}(A) - s\) and the A1 for 0.22

\(s = 0.3\) They assume \(A\) and \(B\) are independent and get \(s = 0.3\) [from P(\(A\))\(\times\)P(\(B\))]
(c) 1st B0 for \(s = 0.3\) BUT can get 2nd B1ft for either case in the scheme
(b) M1 for \(r = \mathrm{P}(A) - s\) BUT then A0cao for \(r = 0.2\)

Mark scheme (c)

Scheme	Marks
\(s = [\mathrm{P}(A) - r\,] = \)0.28	B1ft
\(t = [\ \mathrm{P}(B) - p - s\) or use \(\mathrm{P}(B \cap C') - s = 0.6 \times 0.75 - \text{"}0.28\text{"}\ ] = \)0.17	B1ft
	(2)

Notes

1st B1ft for \(s = 0.28\) or 0.5 – their “0.22”

2nd B1ft for \(t = 0.17\) or 0.6 – their “0.15” – their “0.28”

Mark scheme (d)

Scheme	Marks
\(\mathrm{P}(A) \times \mathrm{P}(B) = 0.5 \times 0.6 = 0.3\) which is not equal to \(s\) (= 0.28)	M1
So \(A\) and \(B\) are not independent	A1
	(2)

Notes

M1 for a correct \(\mathrm{P}(A) \times \mathrm{P}(B) = 0.5 \times 0.6\) or 0.3 and a clear comparison with their \(s\,(\neq 0.3)\)
Or calculation of \(\mathrm{P}(A \mid B) = \frac{7}{15}\) or 0.467 or \(\dfrac{\text{their } s}{0.6}\) and comparison with P(\(A\)) = 0.5 (o.e.)

A1 dep. on M1 being earned and clear statement that \(A\) and \(B\) are not independent

SC \(s = 0.3\) dep on 1st B1ft for \(s = 0.5 - 0.2\) in (c); for correct calc. and conclusion seen (B1). On epen M0A1

Mark scheme (e)

Scheme	Marks
\(\dfrac{(s + p) \text{ or } (0.6 - t)}{\mathrm{P}(A \cup C) \text{ or } [\mathrm{P}(A) + \mathrm{P}(C)] \text{ or } (r + s + p + q)},\ = \dfrac{(\text{"}0.28\text{"} + \text{"}0.15\text{"}) \text{ or } (0.6 - \text{"}0.17\text{"})}{0.5 + 0.25}\)	M1, A1ft
\(= \dfrac{43}{75}\)	A1
	(3)
	(11 marks)

Notes

M1 for a correct ratio expression of probs: num. < den. Allow 1 – (0.08+their “\(t\)”) on den.
Any sight of multiplication on the numerator e.g. \(0.6 \times 0.75\) is M0

1st A1ft for correct ratio or ft using their values in numerator but correct denominator.

2nd A1 for \(\frac{43}{75}\) or accept awrt 0.573