| Date | May 2017 | Marks available | 3 | Reference code | 17M.2.hl.TZ2.5 |
| Level | HL only | Paper | 2 | Time zone | TZ2 |
| Command term | Draw | Question number | 5 | Adapted from | N/A |
Question
John likes to go sailing every day in July. To help him make a decision on whether it is safe to go sailing he classifies each day in July as windy or calm. Given that a day in July is calm, the probability that the next day is calm is 0.9. Given that a day in July is windy, the probability that the next day is calm is 0.3. The weather forecast for the 1st July predicts that the probability that it will be calm is 0.8.
Draw a tree diagram to represent this information for the first three days of July.
Find the probability that the 3rd July is calm.
Find the probability that the 1st July was calm given that the 3rd July is windy.
Markscheme
M1A2
Note: Award M1 for 3 stage tree-diagram, A2 for 0.8, 0.9, 0.3 probabilities correctly placed.
[3 marks]
\(0.2 \times 0.7 \times 0.3 + 0.2 \times 0.3 \times 0.9 + 0.8 \times 0.1 \times 0.3 + 0.8 \times 0.9 \times 0.9 = 0.768\) (M1)A1
[2 marks]
\({\text{P}}({\text{1st July is calm | 3rd July is windy)}} = \frac{{{\text{P}}({\text{1st July is calm and 3rd July is windy)}}}}{{{\text{P}}({\text{3rd July is windy)}}}}\) (M1)
\( = \frac{{0.8 \times 0.1 \times 0.7 + 0.8 \times 0.9 \times 0.1}}{{1 - 0.768}}\)
OR\(\,\,\,\,\,\)\(\frac{{0.8 \times 0.1 \times 0.7 + 0.8 \times 0.9 \times 0.1}}{{0.2 \times 0.7 \times 0.7 + 0.2 \times 0.3 \times 0.1 + 0.8 \times 0.1 \times 0.7 + 0.8 \times 0.9 \times 0.1}}\)
OR\(\,\,\,\,\,\)\(\frac{{0.128}}{{0.232}}\) (A1)(A1)
Note: Award A1 for correct numerator, A1 for correct denominator.
\( = 0.552\) A1
[4 marks]
