Question

A small independent physicians' practice has three doctors. Dr. Sarabia sees 41% of the patients, Dr. Tran sees 32% , and Dr. Jackson sees the rest. Dr. Sarabia requests blood tests on 5% of her patients, Dr. Tran requests blood tests on 8% of his patients, and Dr. Jackson requests blood tests on 6% of her patients. An auditor randomly selects a patient from the past week and discovers that the patient had a blood test as a result of the physician visit. Knowing this information, what is the probability that the patient saw Dr. Sarabia? For what percentage of all patients at this practice are blood tests requested?

(Round your answer to 4 decimal places.)

p = ______________

(enter the probability that a randomly selected patient was seen by Dr. Sarabia last week rounded to 4 decimal places)

For what percentage of all patients at this practice are blood tests requested?

(Round your answer to 2 decimal places.)

p = _______________

(enter the share of all the patients for which blood tests are requested in percentages rounded to 2 decimal places  %)

Answer 1

Consider the events

A₁ : Dr. Sarabia see the patients.

A₂ : Dr. Tran see the patients.

A₃: Dr. Jackson see the patients.

B : patients requires blood test.

From the information

P ( A₁) = 0.41 , P ( A₂) = 0.32 and P ( A₃) = 0.27.

P(B / A₁) = 0.05 , P(B / A₂) = 0.08 and P(B / A₃) = 0.06.

i) P ( A randomly selected patient was seen by Dr. Sarabia) = P ( A₁/ B)

Since A₁ , A₂ and A₃ forms a partition of sample space.

By Baye's Theorem

= 0.0205 / 0.0623

=0.3291

P ( A randomly selected patient was seen by Dr. Sarabia) =0.3291

ii) P ( Patients required blood test) = P(B)

= 0.41 * 0.05 + 0.32 *0.08 + 0.27 * 0.06

=0.0205 + 0.0256 +0.0162

= 0.0623

P ( Patients required blood test) = 0.0623

Percentage of all patients at this practice are blood test required = 6.23%.