S1 June 2016 Q4

EdexcelOld spec13 marksProbability

4. The Venn diagram shows the probabilities of customer bookings at Harry’s hotel.

\(R\) is the event that a customer books a room
\(B\) is the event that a customer books breakfast
\(D\) is the event that a customer books dinner

\(u\) and \(t\) are probabilities.

Venn diagram in a box: B inside R; D overlaps B and R and extends outside R. B only 0.33, B and D 0.27, R and D but not B 0.15, R only u, D only t, outside 0

(a) Write down the probability that a customer books breakfast but does not book a room. (1)

Given that the events \(B\) and \(D\) are independent

(b) find the value of \(t\) (4)

(d) Find

(i) \(\mathrm{P}(D \mid R \cap B)\)

(ii) \(\mathrm{P}(D \mid R \cap B')\) (4)

A coach load of 77 customers arrive at Harry’s hotel.
Of these 77 customers
40 have booked a room and breakfast
37 have booked a room without breakfast

(e) Estimate how many of these 77 customers will book dinner. (2)

Scheme	Marks
\([\mathrm{P}(B \cap R') =]\) 0	B1
	(1)

Mark scheme (b)

Scheme	Marks
P(\(B\)) = 0.27+0.33 = 0.6, P(\(D\)) = 0.27 + 0.15 + \(t\) , \(\mathrm{P}(B \cap D) = 0.27\)	M1
\([\mathrm{P}(B) \times \mathrm{P}(D) = \mathrm{P}(B \cap D)\) gives] \(0.6 \times (0.42 + t) = 0.27\)	M1
\(0.42 + t = \dfrac{0.27}{0.6}\) or \(0.6t = 0.018\)	A1
\(t = \)0.03	A1
	(4)

Notes

1st M1 for attempting 3 suitable probabilities, one involving \(t\) (at least 2 correct)
e.g. sight of 0.6, 0.27, 0.42 + \(t\) correctly labelled in terms of \(B, D, R\) or in a correct equation.
May see e.g. \(\mathrm{P}(B \mid D) = \dfrac{0.27}{0.42 + t}\)

2nd M1 for using the independence to form a linear equation in \(t\). ft their probs if stated.

1st A1 for solving leading to a correct equation as far as \(p + t = q\) or \(pt = q\)

2nd A1 for 0.03 or exact equivalent

Mark scheme (c)

Scheme	Marks
\([\,u =]\ \ 1 - (0.6 + 0.15 + t)\)	M1
\(u = \)0.22	A1ft
	(2)

Notes

M1 for a correct expression for \(u\) . Allow their \(t\) or just letter \(t\) in a correct expression

A1ft for 0.22 (or exact equivalent) or ft their \(t\). i.e. \(u = 0.25 - t\) provided \(u\) & \(t\) are probs
Can score M1A1ft provided their \(u\) + their \(t\) = 0.25 where \(u\) and \(t\) are both in [0, 1]

Mark scheme (d)

Scheme	Marks
(i) \(\left[\dfrac{\mathrm{P}(D \cap R \cap B)}{\mathrm{P}(R \cap B)} =\right] = \dfrac{0.27}{0.27 + 0.33}\) or \(\mathrm{P}(D \mid R \cap B) = \mathrm{P}(D \mid B) = \mathrm{P}(D)\)	M1
\(= \)0.45	A1
(ii) \(\left[\dfrac{\mathrm{P}(D \cap [R \cap B'])}{\mathrm{P}(R \cap B')} =\right] = \dfrac{0.15}{0.15 + u}\)	M1
\(= \dfrac{15}{37}\)	A1
	(4)

Notes

(i) M1 for a correct numerical ratio of probabilities
A1 for 0.45 or exact equivalent (Answer only 2/2)

(ii) M1 for a correct numerical ratio of probabilities, ft their \(u\), provided \(u\) is a probability
A1 for \(\frac{15}{37}\) or \(0.\dot{4}0\dot{5}\) or allow awrt 0.41 following a correct expression (Ans only 2/2)

Mark scheme (e)

Scheme	Marks
\(40 \times \text{"}0.45\text{"}\) and \(37 \times \text{"}\dfrac{15}{37}\text{"}\)	M1
\(= \)33	A1
	(2)
	(13 marks)

Notes

M1 for a correct method for both 18 and 15 ft their 0.45 and their \(\frac{15}{37}\) provided both in [0,1]
NB \(\mathrm{P}(D) \times 77\) is M0

A1 for 33 only
NB \(\frac{27}{33} \times 40 = 32.7\ldots\) which rounds to 33 but scores M0A0. (Ans only send to review)