Question

A study of how people are using online services for medical consulting is described in the paper "Internet Based Consultation to Transfer Knowledge for Patients Requiring Specialized Care" (British Medical Journal [2003]: \(696-699)\). Patients using a particular online site could request one or both (or neither) of two services: specialist opinion and assessment of pathology results. The paper reported that \(98.7 \%\) of those using the service requested a specialist opinion, \(35.4 \%\) requested the assessment of pathology results, and \(34.7 \%\) requested both a specialist opinion and assessment of pathology results. Consider the following two events: \(S=\) event that a specialist opinion is requested \(A=\) event that an assessment of pathology results is requested a. What are the values of \(P(S), P(A)\), and \(P(S \cap A)\) ? b. Use the given probability information to set up a "hypothetical 1000 " table with columns corresponding to \(S\) and not \(S\) and rows corresponding to \(A\) and \(\operatorname{not} A .\) c. Use the table to find the following probabilities: i. the probability that a request is for neither a specialist opinion nor assessment of pathology results. ii. the probability that a request is for a specialist opinion or an assessment of pathology results.

Short answer

a. \(P(S) = 0.987\), \(P(A) = 0.354\), \(P(S \cap A) = 0.347\) b. Numbers per category: \(S \cap A = 347\), \(S \text{ and not } A = 640\), \(A \text{ and not } S = 7\), Neither \(S\) nor \(A = 6 \) c. i. \(P(\text{not } S \cap \text{not } A) = 0.006 \) ii. \(P(S \cup A) = 0.994 \)

Step-by-step solution

Identify Given Probabilities

From the exercise, the given probabilities are: \(P(S) = 0.987\) , \(P(A) = 0.354\) , and \(P(S \cap A) = 0.347\)

Construct a Hypothetical 1000 Table

Construct a hypothetical 1000 table based on these probabilities. The four sections of the table would be: both S and A (this is \(S \cap A\)), S and not A, A and not S, and neither S nor A. Using the given probabilities, the number of people in each category if there were a hypothetical 1000 people is: \((S \cap A) = 0.347 \times 1000 = 347\), \(S\) and not \(A = P(S) - P(S \cap A) = 0.987 - 0.347 = 0.64 \times 1000 = 640\), \(A\) and not \(S = P(A) - P(S \cap A) = 0.354 - 0.347 = 0.007 \times 1000 = 7 \)and neither \(S\) nor \(A = 1000 - ((S \cap A) + (S \text{ and not } A) + (A \text{ and not } S) = 1000 - (347+640+7) = 6\).

Calculate Request for Neither Specialist Opinion Nor Pathology Result Assessment

The probability that a request is for neither a specialist opinion nor an assessment of pathology results is \(P(\text{not} S \cap \text{not} A) = \text{Number of neither S nor A / Total Number} = 6/1000 = 0.006\).

Calculate Request for Either Specialist Opinion or Pathology Result Assessment

The probability that a request is for a specialist opinion or an assessment of pathology results (this implies either one or both) is \(P(S \cup A) = P(S) + P(A) - P(S \cap A) = 0.987 + 0.354 - 0.347 = 0.994 \).