Question

slader Suppose that 8% of all bicycle racers use steroids, that a bicyclist who uses steroids tests positive for steroids 96% of the times, and that a bicyclist who does not use steroids tests positive for steroids 9% of the time. What is the probability that a randomly selected bicyclist who tests positive for steroids actually uses steroids?

Answer 1

Answer:

P = 0.4812

Step-by-step explanation:

First, we need to use here two expressions and then do the calculations.

The first one is the conditional probability which is:

P(B|A) = P(A∩B)/P(A)   (1)

The second expression to use has relation with the Bayes's theorem which is the following:

P(D|C) = P(C|D)*P(D) / P(C|D)*P(D) + P(C|d)*P(d)   (2)

Now, the expression (2) is the one that we will use to calculate the probability of a selected random bicyclist who tests positive for steroids.

So, in this case, we will call C for positive and D that is using steroids and d is the opposite of d, which means do not use steroids.

Then, the probabilities are the following:

P(D) = 8% or 0.08

P(C|D) = 96% or 0.96

P(C|d) = 9% or 0.09

P(d) = 1 - 0.08 = 0.92

With these data, let's replace in expression 2

P(D|C) = 0.96 * 0.08 /0.96 * 0.08 + 0.09*0.92

P(D|C) = 0.0768 / 0.1596

P(D|C) = 0.4812 or 48.12%