Answer to Question #233390 in Statistics and Probability for RJA

Question

1. A company is trialling a new COVID test. They nd that the probabil-

ity that the test is positive given that the patient does have COVID is

0.94, whereas the probability that the test is positive given that they

don't have COVID is 0.03. In each case give answers to 4 decimal

places.

(a) Suppose the test is used in a population where it is estimated

that 4% of people tested have COVID. What is the probability

of a test being negative?

(b) In the same population as above, what is the probability that

someone does not have COVID if the test is negative?

(c) Now the test is used in a population with an unknown infection

rate. If 10% of this population tests positive, what percent of the

population is likely to actually have COVID?

Answer 1

a)

Let A- patient has COVID,

B- result of test is positive;

We have P(A)=0.04;P(B/A)=0.94, P(B/$\bar {A}$)=0.03;

P(B)=P(A) $\cdot P(B/A)+P(\bar {A})\cdot P(B/\bar{ A}) =$

=0.04$\cdot$ 0.94+0.96$\cdot$ 0.03=0.0664;

$P(\bar {B})=1-p(B)=1-0.0664=0.9336=93.36$% -

probability of negative test;

б)

$P(\bar{B}/\bar{A})=1-P(B/\bar{A})=1-0.03=0.97;$

$P(\bar{A} /\bar{B})=\frac{P(\bar{B}/\bar{A})\cdot {P(\bar{A})}} {P(\bar {B})}=$

= $\frac{0.97\cdot 0.96}{0.9336}=0.9974=99.74$%

So, probability not to have COVID if test negative equals

0.9974.

c)

Let X be unknown infection rate or P(A)=X;

P(B)=10%=0.1=P(A) $\cdot P(B/A)+P(\bar {A})\cdot P(B/\bar{ A}) =$

=X$\cdot P(B/A)+(1-X) \cdot P(B/\bar{A})=$

$=X\cdot 0.94+(1-X)\cdot 0.03=0.91\cdot X+0.03;$

$0.91\cdot X=0.1-0.03=0.07;$

$X=\frac{0.07}{0.91}=0.0769$ =7.69%

Thus infection rate is 7,69%