Question

Suppose that a test for opium use has a 2% false positive rate and a 5% false negative rate. That is, 2% of people who do not use opium test positive for opium, and 5% of opium users test negative for opium. Furthermore, suppose that 1% of people actually use opium.

a)Find the probability that someone who tests negative for opium use does not use opium.

b) Find the probability that someone who tests positive for opium use actually uses opium.

Short answer

Answer

a) The probability that someone who tests negative for opium use does not use opium is\(0.9995\)

b) The probability that someone who tests positive for opium use actually uses opium is 0.3242

Step-by-step solution

Step-1 Given Information

Let,

A: use opium

B: test positive

\(\begin{array}{l}P(B|\overline A ) = 0.02\\P(\overline B |A) = 0.05\\P(A) = 0.01\end{array}\)

Step-2 Definition and Formula

A: use opium

B: test positive

\(P(\overline A |\overline B ) = \frac{{P(\overline B |\overline A )*P(\overline A )}}{{P(\overline B |\overline A )*P(\overline A ) + P(\overline B |A)*P(A)}}\)

Step-3 Calculate Probability

\(P(\overline A |\overline B ) = \frac{{P(\overline B |\overline A )*P(\overline A )}}{{P(\overline B |\overline A )*P(\overline A ) + P(\overline B |A)*P(A)}}\)

\(\begin{array}{l} = \frac{{(1 - 0.02)(1 - 0.01)}}{{(1 - 0.02)(1 - 0.01) + (0.05)(0.01)}}\\ = 0.9995\end{array}\)

Hence,

The probability that someone who tests negative for opium use does not use opium is\(0.9995\)

\(P(A|B) = \frac{{P(B|A)*P(A)}}{{P(B|A)*P(A) + P(B|\overline A )*P(\overline A )}}\)

\(\begin{array}{l} = \frac{{(1 - 0.05)(0.01)}}{{(1 - 0.05)(0.01) + (0.02)(1 - 0.01)}}\\ = 0.3242\end{array}\)

Hence,

The probability that someone who tests positive for opium use actually uses opium is\(0.3242\)