An athletic league does drug testing of its athletes, 10 percent of whom use drugs. this test, however, is only 95 percent reliable. that is, a drug user will test positive with probability 0.95 and negative with probability 0.05, and a nonuser will test negative with probability 0.95 and positive with probability 0.05. develop a probability tree diagram to determine the posterior probability of each of the following outcomes of testing an athlete.

(a) the athlete is a drug user, given that the test is positive.

(b) the athlete is not a drug user, given that the test is positive.

(c) the athlete is a drug user, given that the test is negative.

(d) the athlete is not a drug user, given that the test is negative.

(e) use the corresponding excel template to check your answers in the preceding parts.

0.6786

0.3214

1 / 172

171 / 172

Step-by-step explanation:

Note: The tree diagram is attached below.

a) The athlete is a drug user, given that the test is positive P(D/+)

Using conditional Probability:

P ( D / + ) = P ( D and + ) / P ( +)

= (0.1 * 0.95) / (0.1 * 0.95 + 0.9*0.05)

= 0.6786

P ( D / + ) = 0.6786

b) The athlete is not a drug user, given that the test is positive P(D' / + )

Using conditional Probability:

P ( D' / + ) = P ( D' and + ) / P (+)

= (0.9 * 0.05) / (0.1 * 0.95 + 0.9*0.05)

= 0.3214

P ( D' / + ) = 0.3214

c) The athlete is a drug user, given that the test is negative P (D / - )

Using conditional Probability:

P ( D / - ) = P ( D and - ) / P (-)

= (0.1 * 0.05) / (0.9 * 0.95 + 0.1*0.05)

= 1 / 172

P ( D / - ) = 1 / 172

(d) The athlete is not a drug user, given that the test is negative P (D' / - )

Using conditional Probability:

P ( D' / - ) = P ( D' and - ) / P (-)

= (0.9 * 0.95) / (0.9 * 0.95 + 0.1*0.05)

= 171 / 172

P ( D / - ) = 171 / 172