User interface language: English | Español

Date November 2011 Marks available 6 Reference code 11N.1.hl.TZ0.3
Level HL only Paper 1 Time zone TZ0
Command term Find Question number 3 Adapted from N/A

Question

In a particular city 20 % of the inhabitants have been immunized against a certain disease. The probability of infection from the disease among those immunized is \(\frac{1}{{10}}\), and among those not immunized the probability is \(\frac{3}{4}\). If a person is chosen at random and found to be infected, find the probability that this person has been immunized.

Markscheme

tree diagram     (M1)

\({\text{P(I|D)}} = \frac{{{\text{P(D|I)}} \times {\text{P(I)}}}}{{{\text{P(D)}}}}\)     (M1)

\( = \frac{{0.1 \times 0.2}}{{0.1 \times 0.2 + 0.8 \times 0.75}}\)     A1A1A1

\(\left( { = \frac{{0.02}}{{0.62}}} \right) = \frac{1}{{31}}\)     A1

Note: Alternative presentation of results: M1 for labelled tree; A1 for initial branching probabilities, 0.2 and 0.8; A1 for at least the relevant second branching probabilities, 0.1 and 0.75; A1 for the ‘infected’ end-point probabilities, 0.02 and 0.6; M1A1 for the final conditional probability calculation.

 

[6 marks]

Examiners report

Candidates who drew a tree diagram, the majority, usually found the correct answer.

Syllabus sections

Topic 5 - Core: Statistics and probability » 5.4 » Conditional probability; the definition \(P\left( {\left. A \right|P} \right) = \frac{{P\left( {A\mathop \cap \nolimits B} \right)}}{{P\left( B \right)}}\) .

View options