| Date | May 2011 | Marks available | 2 | Reference code | 11M.1.hl.TZ1.6 |
| Level | HL only | Paper | 1 | Time zone | TZ1 |
| Command term | Find | Question number | 6 | Adapted from | N/A |
Question
In a population of rabbits, \(1\%\) are known to have a particular disease. A test is developed for the disease that gives a positive result for a rabbit that does have the disease in \(99\%\) of cases. It is also known that the test gives a positive result for a rabbit that does not have the disease in \(0.1\%\) of cases. A rabbit is chosen at random from the population.
Find the probability that the rabbit tests positive for the disease.
Given that the rabbit tests positive for the disease, show that the probability that the rabbit does not have the disease is less than 10 %.
Markscheme
R is ‘rabbit with the disease’
P is ‘rabbit testing positive for the disease’
\({\text{P}}(P) = P(R \cap P) + P(R' \cap P)\)
\( = 0.01 \times 0.99 + 0.99 \times 0.001\) M1
\( = 0.01089( = 0.0109)\) A1
Note: Award M1 for a correct tree diagram with correct probability values shown.
[2 marks]
R is ‘rabbit with the disease’
P is ‘rabbit testing positive for the disease’
\(P(R'|P) = \frac{{0.001 \times 0.99}}{{0.001 \times 0.99 + 0.01 \times 0.99}}\left( { = \frac{{0.00099}}{{0.01089}}} \right)\) M1A1
\(\frac{{0.00099}}{{0.01089}} < \frac{{0.001}}{{0.01}} = 10\% \) (or other valid argument) R1
[3 marks]
Examiners report
There was a mixed performance in this question with some candidates showing good understanding of probability and scoring well and many others showing no understanding of conditional probability and difficulties in working with decimals. Very few candidates were able to provide a valid argument to justify their answer to part (b).
There was a mixed performance in this question with some candidates showing good understanding of probability and scoring well and many others showing no understanding of conditional probability and difficulties in working with decimals. Very few candidates were able to provide a valid argument to justify their answer to part (b).
