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Date May 2014 Marks available 5 Reference code 14M.1.sl.TZ1.9
Level SL only Paper 1 Time zone TZ1
Command term Find Question number 9 Adapted from N/A

Question

Bill and Andrea play two games of tennis. The probability that Bill wins the first game is \(\frac{4}{5}\).

If Bill wins the first game, the probability that he wins the second game is \(\frac{5}{6}\).

If Bill loses the first game, the probability that he wins the second game is \(\frac{2}{3}\).

Copy and complete the following tree diagram. (Do not write on this page.)

 


[3]
a.

Find the probability that Bill wins the first game and Andrea wins the second game.

[2]
b.

Find the probability that Bill wins at least one game.

[4]
c.

Given that Bill wins at least one game, find the probability that he wins both games.

[5]
d.

Markscheme


     A1A1A1     N3

 

Note:   Award A1 for each correct bold probability.

 

[3 marks]

a.

multiplying along the branches (may be seen on diagram)     (M1)

eg   \(\frac{4}{5} \times \frac{1}{6}\)

\(\frac{4}{{30}}\left( {\frac{2}{{15}}} \right)\)     A1     N2

[2 marks]

b.

METHOD 1

multiplying along the branches (may be seen on diagram)     (M1)

eg   \(\frac{4}{5} \times \frac{5}{6},{\text{ }}\frac{4}{5} \times \frac{1}{6},{\text{ }}\frac{1}{5} \times \frac{2}{3}\)

adding their probabilities of three mutually exclusive paths     (M1)

eg   \(\frac{4}{5} \times \frac{5}{6} + \frac{4}{5} \times \frac{1}{6} + \frac{1}{5} \times \frac{2}{3},{\text{ }}\frac{4}{5} + \frac{1}{5} \times \frac{2}{3}\)

correct simplification     (A1)

eg   \(\frac{{20}}{{30}} + \frac{4}{{30}} + \frac{2}{{15}},{\text{ }}\frac{2}{3} + \frac{2}{{15}} + \frac{2}{{15}}\)

\(\frac{{28}}{{30}}{\text{ }}\left( { = \frac{{14}}{{15}}} \right)\)     A1     N3

METHOD 2

recognizing “Bill wins at least one” is complement of “Andrea wins 2”     (R1)

eg   finding P (Andrea wins 2)

P (Andrea wins both) \( = \frac{1}{5} \times \frac{1}{3}\)     (A1)

evidence of complement     (M1)

eg   \(1 - p,{\text{ }}1 - \frac{1}{{15}}\)

\(\frac{{14}}{{15}}\)     A1     N3

[4 marks]

c.

P (B wins both) \(\frac{4}{5} \times \frac{5}{6}{\text{ }}\left( { = \frac{2}{3}} \right)\)     A1

evidence of recognizing conditional probability     (R1)

eg   \({\text{P}}(A\left| B \right.),{\text{ P (Bill wins both}}\left| {{\text{Bill wins at least one), tree diagram}}} \right.\)

correct substitution     (A2)

eg   \(\frac{{\frac{4}{5} \times \frac{5}{6}}}{{\frac{{14}}{{15}}}}\)

\(\frac{{20}}{{28}}{\text{ }}\left( { = \frac{5}{7}} \right)\)     A1     N3

[5 marks]

d.

Examiners report

[N/A]
a.
[N/A]
b.
[N/A]
c.
[N/A]
d.

Syllabus sections

Topic 5 - Statistics and probability » 5.6 » Conditional probability; the definition \(P\left( {\left. A \right|B} \right) = \frac{{P\left( {A\mathop \cap \nolimits B} \right)}}{{P\left( B \right)}}\) .
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