October 2021 Paper 2 Q12

OCR ACurrent spec13 marksProbability

12 Anika and Beth are playing a game which consists of several points.

The probability that Anika will win any point is 0.7.
The probability that Beth will win any point is 0.3.
The outcome of each point is independent of the outcome of every other point.

The first player to win two points wins the game.

(a) Write down the probability that the game consists of more than three points. [1]

(b) Complete the probability tree diagram in the Printed Answer Booklet showing all the possibilities for the game. [3]

Start of a tree diagram from the Printed Answer Booklet: one branch with probability 0.7 to Anika wins, and one branch with probability 0.3 to Beth wins

(d) Determine the probability that the game consists of exactly three points. [2]

(e) Given that Beth wins the game, determine the probability that the game consists of exactly three points. [4]

Scheme	Marks
0	B1
	[1]

Completed tree: first point (A) 0.7 or (B) 0.3. After A: A 0.7 (game ends) or B 0.3, then A 0.7 or B 0.3. After B: A 0.7, then A 0.7 or B 0.3; or B 0.3 (game ends) — Mark scheme (b)

Scheme	Marks
	B1 B1 B1
	[3]

Mark scheme (c)

Scheme	Marks
\(0.3 \times 0.3 + 0.7 \times 0.3 \times 0.3 + 0.3 \times 0.7 \times 0.3\) or \(0.09 + 0.063 + 0.063\) oe	M1 M1
\(= \dfrac{27}{125}\) or 0.216	A1
	[3]

Notes

All correct M2 ft their diagram
Two products correct M1 ft their diagram

SC Correct answer with no working: B2

Mark scheme (d)

Scheme	Marks
\(0.7 \times 0.3 + 0.3 \times 0.7\) or \(0.21 + 0.21\)	M1
\(= \dfrac{21}{50}\) or 0.42	A1
	[2]

Notes

M1: or \(1 - (0.7^2 + 0.3^2)\) or \(1 - (0.49 + 0.09)\) Condone missing brackets
or \(0.7 \times 0.3 \times 0.7 + 0.7 \times 0.3 \times 0.3 + 0.3 \times 0.7 \times 0.7 + 0.3 \times 0.7 \times 0.3\) oe
or \(2 \times 0.147 + 2 \times 0.063\) oe Wholly correct method ft their diagram

A1: SC Correct answer with no working: B1

Mark scheme (e)

Scheme	Marks
P(B wins and 3 points) \(= 0.7 \times 0.3 \times 0.3 + 0.3 \times 0.7 \times 0.3\) or \(2 \times 0.063\) \((= 0.126\ \text{oe})\)	M2
\(\dfrac{\text{P(B wins \& 3 points)}}{\text{P(B wins)}} = \dfrac{\text{'}0.126\text{'}}{\text{'}0.216\text{'}}\)	M1
\(= \dfrac{7}{12}\)	A1
	[4]

Notes

ft their diagram for M-marks

M2: soi M1 for one correct product or 0.063

M1: Must attempt \(\dfrac{\text{P(B wins \& 3 points)}}{\text{Their (c) NOT (d) or } 0.3 \times 0.3 + 0.7 \times 0.3 \times 0.3 + 0.3 \times 0.7 \times 0.3}\)

A1: Allow 0.583 (3 sf) SC no working \(\frac{7}{12}\) B2
But minimal working is OK, eg \(\dfrac{2 \times 0.063}{0.216} = \dfrac{7}{12}\) M2M1A1