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Chapter 7: Problem 47

A study was conducted among a certain group of union members whose health insurance policies required second opinions prior to surgery. Of those members whose doctors advised them to have surgery, \(20 \%\) were informed by a second doctor that no surgery was needed. Of these, \(70 \%\) took the second doctor's opinion and did not go through with the surgery. Of the members who were advised to have surgery by both doctors, \(95 \%\) went through with the surgery. What is the probability that a union member who had surgery was advised to do so by a second doctor?

Short Answer

Expert verified
The probability that a union member who had surgery was advised to do so by a second doctor is approximately \(45.56\% \).

Step by step solution

01

Calculate P(B | A) and P(~B | A)

Based on the information given, we can directly find P(B | A) and P(~B | A). P(B | A) = 1 - P(~B | A) = 1 - 0.20 = 0.80 P(~B | A) = 0.20
02

Calculate P(S | A&B) and P(S | A&~B)

Using the given information, we can obtain the probability of having surgery given both doctors' opinions. P(S | A&B) = 0.95 P(S | A&~B) = 1 - P(~S | A&~B) = 1 - 0.70 = 0.30
03

Apply Bayes' Theorem to find P(B | S)

We will use Bayes' Theorem to find P(B | S), which is given by: P(B | S) = P(S | B) * P(B) / P(S) However, we do not have the values for P(S | B), P(B), and P(S) directly. We can apply the law of total probability to find these values.
04

Find P(S | B) using the law of total probability

The probability P(S | B) can be found using the law of total probability, given by: P(S | B) = P(S | A&B) * P(A | B) Here, we need to find P(A | B), which is equal to P(B | A) under the assumption that the two doctor's opinions are independent. P(A | B) = P(B | A) = 0.80 P(S | B) = P(S | A&B) * P(A | B) = 0.95 * 0.80 = 0.76
05

Find P(B) using the law of total probability

To find P(B), we can use the law of total probability: P(B) = P(B | A) * P(A) + P(B | ~A) * P(~A) We do not have the values of P(A) and P(~A), but under the assumption of the independence of the two doctors' opinions, we can make the following statement: P(B) = P(B | A) * P(A) + P(B | ~A) * P(~A) = P(B | A) = 0.80
06

Find P(S) using the law of total probability

Using the law of total probability, we can find P(S): P(S) = P(S | A) * P(A) + P(S | ~A) * P(~A) We do not have P(S | ~A), but we can rewrite P(S) using the law of total probability and the fact that P(A | B) = P(B | A) under the assumption of independence: P(S) = P(S | B) * P(B) + P(S | ~B) * P(~B) = 0.76 * 0.80 + 0.30 * 0.20 = 0.608 + 0.06 = 0.668
07

Compute P(B | S) using Bayes' Theorem

Now, we can substitute the values found in Steps 4, 5, and 6 into the Bayes' Theorem formula: P(B | S) = P(S | B) * P(B) / P(S) = 0.76 * 0.80 / 0.668 = 0.45564 So, the probability that a union member who had surgery was advised to do so by a second doctor is approximately 45.56%.

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Most popular questions from this chapter

A medical test has been designed to detect the presence of a certain disease. Among those who have the disease, the probability that the disease will be detected by the test is \(.95\). However, the probability that the test will erroneously indicate the presence of the disease in those who do not actually have it is .04. It is estimated that \(4 \%\) of the population who take this test have the disease. a. If the test administered to an individual is positive, what is the probability that the person actually has the disease? b. If an individual takes the test twice and both times the test is positive, what is the probability that the person actually has the disease? (Assume that the tests are independent.)

In a past presidential election, it was estimated that the probability that the Republican candidate would be elected was \(\frac{3}{5}\), and therefore the probability that the Democratic candidate would be elected was \(\frac{2}{5}\) (the two Independent candidates were given no chance of being elected). It was also estimated that if the Republican candidate were elected, the probability that a conservative, moderate, or liberal judge would be appointed to the Supreme Court (one retirement was expected during the presidential term) was \(\frac{1}{2}, \frac{1}{3}\), and \(\frac{1}{6}\), respectively. If the Democratic candidate were elected, the probabilities that a conservative, moderate, or liberal judge would be appointed to the Supreme Court would be \(\frac{1}{8}, \frac{3}{8}\), and \(\frac{1}{2}\), respectively. A conservative judge was appointed to the Supreme Court during the presidential term. What is the probability that the Democratic candidate was elected?

In an online survey of 1962 executives from 64 countries conducted by Korn/Ferry International between August and October 2006, the executives were asked if they would try to influence their children's career choices. Their replies: A (to a very great extent), B (to a great extent), \(\mathrm{C}\) (to some extent), D (to a small extent), and \(\mathrm{E}\) (not at all) are recorded below: $$\begin{array}{lccccc} \hline \text { Answer } & \text { A } & \text { B } & \text { C } & \text { D } & \text { E } \\ \hline \text { Respondents } & 135 & 404 & 1057 & 211 & 155 \\ \hline \end{array}$$ What is the probability that a randomly selected respondent's answer was \(\mathrm{D}\) (to a small extent) or \(\mathrm{E}\) (not at all)?

In a recent senatorial election, \(50 \%\) of the voters in a certain district were registered as Democrats, \(35 \%\) were registered as Republicans, and \(15 \%\) were registered as Independents. The incumbent Democratic senator was reelected over her Republican and Independent opponents. Exit polls indicated that she gained \(75 \%\) of the Democratic vote, \(25 \%\) of the Republican vote, and \(30 \%\) of the Independent vote. Assuming that the exit poll is accurate, what is the probability that a vote for the incumbent was cast by a registered Republican?

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