Question

. A chemist wishes to detect an impurity in a certain compound that she is making. There is a test that detects an impurity with probability 0.92; however,

This test indicates that an impurity is there when it is not about 5% of the time. The chemist produces compounds with the impurity about 15% of the time. A compound is selected at random from the chemist’s output. The test indicates that an impurity is present. What is the conditional probability that the compound actually has the impurity?

Answer 1

• Chemist produces compound with impurities about 15% of time i.e,

P ( Impurity Actually Present ) = 0.15

P ( Impurity Not Actually Present ) = 0.85

• Test detects impurity with probability 0.92 i.e. ,

P ( Detecting Impurity | Impurity Actually Present ) = 0.92

• Test indicate Impurity when it is not about 5 % of the time i.e ,

P ( Detecting Impurity | Impurity Actually Not Present ) = 0.05

We want probability that selected compound actually has the Impurity given that test has detected impurity i.e. ,

P ( Impurity Actually Present | Detecting Impurity ) = ?

We use Bayes Theorem in this Case :

   P ( Impurity Actually Present | Detecting Impurity )

= [ P ( Detecting Impurity | Impurity Actually Present ) * P ( Impurity Actually Present ) ] / { P ( Detecting Impurity | Impurity Actually Present ) * P ( Impurity Actually Present ) +  P ( Detecting Impurity | Impurity Actually Not Present ) * P ( Impurity Actually Not Present ) }

= ( 0.92 * 0.15 ) / ( ( 0.92 * 0.15 ) + ( 0.05 * 0.85 ) ) = 276 / 361 = 0.764543