| Date | May 2012 | Marks available | 1 | Reference code | 12M.1.sl.TZ1.12 |
| Level | SL only | Paper | 1 | Time zone | TZ1 |
| Command term | Complete | Question number | 12 | Adapted from | N/A |
Question
Merryn plans to travel to a concert tomorrow. Due to bad weather, there is a 60 % chance that all flights will be cancelled tomorrow. If the flights are cancelled Merryn will travel by car.
If she travels by plane the probability that she will be late for the concert is 10 %.
If she travels by car, the probability that she will not be late for the concert is 25 %.
Complete the tree diagram below.
Find the probability that Merryn will not be late for the concert.
Merryn was not late for the concert the next day.
Given that, find the probability that she travelled to the concert by car.
Markscheme
(A1) (C1)
Note: Award (A1) for 0.9 and 0.75.
[1 mark]
0.4 × 0.9 + 0.6 × 0.25 (M1)(M1)
Note: Award (M1) for their two relevant products, (M1) for adding their two products.
\(0.51\left( {\frac{{51}}{{100}},{\text{ }}51\% } \right)\) (A1)(ft) (C3)
Note: Follow through from their answers to part (a).
[3 marks]
\(\frac{{0.6 \times 0.25}}{{0.51}}\) (M1)
Note: Award (M1) for correctly substituted conditional probability formula.
\(0.294\left( {\frac{5}{{17}}{\text{, }}0.294117...} \right)\) (A1)(ft) (C2)
Note: Follow through from their tree diagram and their part (b).
[2 marks]
Examiners report
It was pleasing to see many correct answers in parts (a) and (b) with many writing their answer to part (b) in the context of the question and writing down a percentage.
It was pleasing to see many correct answers in parts (a) and (b) with many writing their answer to part (b) in the context of the question and writing down a percentage.
Conditional probability is not an easy topic for candidates to understand and many simply wrote down 0.6 × 0.25 = 0.15(15%) for part (c).
