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Date May 2009 Marks available 1 Reference code 09M.2.sl.TZ1.2
Level SL only Paper 2 Time zone TZ1
Command term Write down Question number 2 Adapted from N/A

Question

Sharon and Lisa share a flat. Sharon cooks dinner three nights out of ten. If Sharon does not cook dinner, then Lisa does. If Sharon cooks dinner the probability that they have pasta is 0.75. If Lisa cooks dinner the probability that they have pasta is 0.12.

A survey was carried out in a year 12 class. The pupils were asked which pop groups they like out of the Rockers (R), the Salseros (S), and the Bluers (B). The results are shown in the following diagram.

Copy and complete the tree diagram to represent this information.

[3]
i, a.

Find the probability that Lisa cooks dinner and they do not have pasta.

[2]
i, b.

Find the probability that they do not have pasta.

[3]
i, c.

Given that they do not have pasta, find the probability that Lisa cooked dinner.

[3]
i, d.

Write down \(n(R \cap S \cap B)\).

[1]
ii, a.

Find \(n(R')\).

[2]
ii, b.

Describe which groups the pupils in the set \(S \cap B\) like.

[2]
ii, c.

Use set notation to describe the group of pupils who like the Rockers and the Bluers but do not like the Salseros.

[2]
ii, d.

There are 33 pupils in the class.

Find x.

[2]
ii, e, i.

There are 33 pupils in the class.

Find the number of pupils who like the Rockers.

[1]
ii, e, ii.

Markscheme



Note: Award (A1) for each correct pair.     (A3)

 

[3 marks]

i, a.

\(0.7 \times 0.88 = 0.616{\text{   }}\left( {\frac{{77}}{{125}},{\text{ }}61.6{\text{ }}\% } \right)\)    (M1)(A1)(ft)(G2)

 

Note: Award (M1) for multiplying the correct probabilities.

 

[2 marks]

i, b.

\(0.3 \times 0.25 + 0.7 \times 0.88\)     (M1)(M1)


Notes: Award (M1) for a relevant two-factor product, could be \(S \times NP\) OR \(L \times NP\).

Award (M1) for summing 2 two-factor products.


\({\text{P}} = 0.691{\text{   }}\left( {\frac{{691}}{{1000}},{\text{ }}69.1{\text{ }}\% } \right)\)    
(A1)(ft)(G2)

 

Notes: (ft) from their answer to (b).

 

[3 marks]

i, c.

\(\frac{{0.616}}{{0.691}}\)     (M1)(A1)


Note: Award (M1) for substituted conditional probability formula, (A1) for correct substitution.

\({\text{P}} = 0.891{\text{   }}\left( {\frac{{616}}{{691}},{\text{ }}89.1{\text{ }}\% } \right)\)     (A1)(ft)(G2)

[3 marks]

i, d.

3     (A1)

[1 mark]

ii, a.

For 5, 4, 7 (0) seen with no extra values     (A1)

16     (A1)(G2)

[2 marks]

ii, b.

They like (both) the Salseros (S) and they like the Bluers (B)     (A1)(A1)

 

Note: Award (A1) for “and”, (A1) for the correct groups.

 

[2 marks]

ii, c.

\(R \cap B \cap S'\)     (A1)(A1)

 

Note: Award (A1) for \(R \cap B\), (A1) for \( \cap S'\)

 

[2 marks]

ii, d.

\(21+ 3x = 33\)     (M1)

\(x = 4\)     (A1)(G2)

[2 marks]

ii, e, i.

17     (A1)(ft)

[1 mark]

ii, e, ii.

Examiners report

The tree diagram was quite well answered by many students, but sometimes it was missing on many papers. It seemed they had it on their examination paper because the subsequent questions were answered accurately. Conditional probability was of great difficulty to many candidates.

i, a.

The tree diagram was quite well answered by many students, but sometimes it was missing on many papers. It seemed they had it on their examination paper because the subsequent questions were answered accurately. Conditional probability was of great difficulty to many candidates.

i, b.

The tree diagram was quite well answered by many students, but sometimes it was missing on many papers. It seemed they had it on their examination paper because the subsequent questions were answered accurately. Conditional probability was of great difficulty to many candidates.

i, c.

The tree diagram was quite well answered by many students, but sometimes it was missing on many papers. It seemed they had it on their examination paper because the subsequent questions were answered accurately. Conditional probability was of great difficulty to many candidates.

i, d.

This question was well handled although part (d) proved too difficult for many candidates and demonstrated, overall, a poor level of understanding of basic set notation. Students generally had the algebraic skills required to solve for x in part (e)(ii).

ii, a.

This question was well handled although part (d) proved too difficult for many candidates and demonstrated, overall, a poor level of understanding of basic set notation. Students generally had the algebraic skills required to solve for x in part (e)(ii).

ii, b.

This question was well handled although part (d) proved too difficult for many candidates and demonstrated, overall, a poor level of understanding of basic set notation. Students generally had the algebraic skills required to solve for x in part (e)(ii).

ii, c.

This question was well handled although part (d) proved too difficult for many candidates and demonstrated, overall, a poor level of understanding of basic set notation. Students generally had the algebraic skills required to solve for x in part (e)(ii).

ii, d.

This question was well handled although part (d) proved too difficult for many candidates and demonstrated, overall, a poor level of understanding of basic set notation. Students generally had the algebraic skills required to solve for x in part (e)(ii).

ii, e, i.

This question was well handled although part (d) proved too difficult for many candidates and demonstrated, overall, a poor level of understanding of basic set notation. Students generally had the algebraic skills required to solve for x in part (e)(ii).

ii, e, ii.

Syllabus sections

Topic 3 - Logic, sets and probability » 3.5 » Basic concepts of set theory: elements \(x \in A\), subsets \(A \subset B\); intersection \(A\mathop \cap \nolimits B\); union \(A\mathop \cup \nolimits B\); complement \({A'}\) .
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