Question

Question: Suppose that 4% of the patients tested in a clinic are infected with avian influenza. Furthermore, suppose that when a blood test for avian influenza is given, 97% of the patients infected with avian influenza test positive and that 2% of the patients not infected with avian influenza test positive. What is the probability that:

a) a patient testing positive for avian influenza with this test is infected with it?

b) a patient testing positive for avian influenza with this test is not infected with it?

c) a patient testing negative for avian influenza with this test is infected with it?

d) a patient testing negative for avian influenza with this test is not infected with it?

Short answer

Answer:

Probability that a patient testing positive for avian influenza with this test is infected with it .6690

Probability that a patient testing positive for avian influenza with this test is not infected with it .331

Probability that a patient testing negative for avian influenza with this test is infected with it .001274

Probability that a patient testing negative for avian influenza with this test is not infected with it .9987

Step-by-step solution

Given data  

Percentage of patient tested infected with avian influenza = 4%

Percentage of patients infected tested positive = 97%

Percentage of patients not infected tested positive = 2%

Formula used   

\(p\left( {\frac{E}{{{E_1}}}} \right) = \frac{{p\left( {\frac{{{E_1}}}{E}} \right)p\left( E \right)}}{{p\left( {\frac{{{E_1}}}{E}} \right)p\left( E \right) + p\left( {\frac{{{E_2}}}{F}} \right)p\left( F \right)}}\)

Calculating (a)

Consider the following event

E= test positive

F= has disease

\(p\left( F \right) = 4\% = 0.04\)

\(p\left( {\frac{E}{F}} \right) = 97\% = 0.97\)

\(p\left( {\frac{E}{{\overline F }}} \right) = 2\% = 0.02\)

Using Bayes’ Theorem we get

\(p\left( {\frac{F}{E}} \right) = \frac{{p\left( {\frac{E}{F}} \right)p\left( F \right)}}{{p\left( {\frac{E}{F}} \right)p\left( F \right) + p\left( {\frac{E}{{\overline F }}} \right)p\left( {\overline F } \right)}}\)

\(\begin{array}{l} = \frac{{(0.97)(0.04)}}{{(0.97)(0.04) + (0.02)(0.96)}}\\ = \frac{{97}}{{145}}\\ = 0.6690\end{array}\)

Calculating (b)  

Here we are asked for the probability of the complementary event

\(p\left( {\frac{{\overline F }}{E}} \right) = 1 - p\left( {\frac{F}{E}} \right)\)

\( = 1 - 0.6690\)

\( = 0.331\)

Calculating (c)  

\(p\left( {\overline F } \right) = 1 - 0.04 = 0.96\)

\(p\left( {\frac{{\overline E }}{F}} \right) = 1 - 0.97 = 0.03\)

\(p\left( {\frac{{\overline E }}{{\overline F }}} \right) = 1 - 0.02 = 0.98\)

We are asked to find \(p\left( {\frac{F}{{\overline E }}} \right)\)

Using Bayes’ theorem we get

\(p\left( {\frac{F}{{\overline E }}} \right) = \frac{{p\left( {\frac{{\overline E }}{F}} \right)p\left( F \right)}}{{p\left( {\frac{{\overline E }}{F}} \right)p\left( F \right) + p\left( {\frac{{\overline E }}{{\overline F }}} \right)p\left( {\overline F } \right)}}\)

\(\begin{array}{l} = \frac{{(0.03)(0.04)}}{{(0.03)(0.04) + (0.98)(0.96)}}\\ = 0.001274\end{array}\)

Calculation (d)       

Here we are asked for the probability of the complementary event.

\(p\left( {\frac{F}{E}} \right) = 1 - p\left( {\frac{F}{{\overline E }}} \right)\)

\( = 1 - 0.001274\)

\( = 0.9987\)