| Date | May 2010 | Marks available | 3 | Reference code | 10M.2.sl.TZ2.1 |
| Level | SL only | Paper | 2 | Time zone | TZ2 |
| Command term | Find | Question number | 1 | Adapted from | N/A |
Question
Phoebe chooses a biscuit from a blue tin on a shelf. The tin contains one chocolate biscuit and four plain biscuits. She eats the biscuit and chooses another one from the tin. The tree diagram below represents the situation with the four possible outcomes where A stands for chocolate biscuit and B for plain biscuit.
On another shelf there are two tins, one red and one green. The red tin contains three chocolate biscuits and seven plain biscuits and the green tin contains one chocolate biscuit and four plain biscuits. Andrew randomly chooses either the red or the green tin and randomly selects a biscuit.
Write down the value of a.
Write down the value of b.
Find the probability that both biscuits are plain.
Copy and complete the tree diagram below.
Find the probability that he chooses a chocolate biscuit.
Find the probability that he chooses a biscuit from the red tin given that it is a chocolate biscuit.
Markscheme
a = 0 \(\left( {\frac{0}{4}} \right)\) (A1)
[1 mark]
\(b = \frac{3}{4}(0.75,{\text{ }}75\% )\) (A2)(G2)
[2 marks]
\(\frac{4}{5} \times \frac{3}{4}\) (M1)(A1)
\(\frac{{12}}{{20}}\left( {\frac{3}{5},{\text{ }}0.6,{\text{ 60% }}} \right)\) (A1)(ft)(G2)
Note: Award (M1) for multiplying two probabilities, (A1) for using their probabilities, (A1) for answer.
[3 marks]
(A1)(A1)(A1)
Note: Award (A1) for each pair.
[3 marks]
\(\frac{1}{2} \times \frac{3}{{10}} + \frac{1}{2} \times \frac{1}{5}\) (M1)(M1)
\( = \frac{5}{{20}}\left( {\frac{1}{4},{\text{ }}0.25,{\text{ 25% }}} \right)\) (A1)(ft)(G2)
Note: Award (M1) for two products seen with numbers from the problem, (M1) for adding two products. Follow through from their tree diagram.
[3 marks]
\(\frac{{\frac{1}{2} \times \frac{3}{{10}}}}{{\frac{1}{4}}}\) (M1)(A1)
\( = \frac{3}{5}{\text{ }}\left( {0.6,{\text{ 60 }}\% } \right)\) (A1)(ft)(G2)
Note: Award (M1) for substituted conditional probability formula, (A1) for correct substitution.
Follow through from their part (b) and part (c) (i).
[3 marks]
Examiners report
This question was well handled by most of the candidates except for (c)(ii) in which they had to find a conditional probability. Some candidates did not copy the second tree diagram in the answer sheets and instead wrote their answers in the exam booklet thus losing the 3 marks allocated to part (b).
This question was well handled by most of the candidates except for (c)(ii) in which they had to find a conditional probability. Some candidates did not copy the second tree diagram in the answer sheets and instead wrote their answers in the exam booklet thus losing the 3 marks allocated to part (b).
This question was well handled by most of the candidates except for (c)(ii) in which they had to find a conditional probability. Some candidates did not copy the second tree diagram in the answer sheets and instead wrote their answers in the exam booklet thus losing the 3 marks allocated to part (b).
This question was well handled by most of the candidates except for (c)(ii) in which they had to find a conditional probability. Some candidates did not copy the second tree diagram in the answer sheets and instead wrote their answers in the exam booklet thus losing the 3 marks allocated to part (b).
This question was well handled by most of the candidates except for (c)(ii) in which they had to find a conditional probability. Some candidates did not copy the second tree diagram in the answer sheets and instead wrote their answers in the exam booklet thus losing the 3 marks allocated to part (b).
This question was well handled by most of the candidates except for (c)(ii) in which they had to find a conditional probability. Some candidates did not copy the second tree diagram in the answer sheets and instead wrote their answers in the exam booklet thus losing the 3 marks allocated to part (b).
