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Medical: Tuberculosis The state medical school has discovered a new test for tuberculosis. (If the test indicates a person has tuberculosis, the test is positive.) Experimentation has shown that the probability of a positive test is 0.82, given that a person has tuberculosis. The probability is 0.09 that the test registers positive, given that the person does not have tuberculosis. Assume that in the general population, the probability that a person has tuberculosis is 0.04. What is the probability that a person chosen at random will (a) have tuberculosis and have a positive test? (b) not have tuberculosis? (c) not have tuberculosis and have a positive test?

Short Answer

Expert verified
(a) Probability = 0.0328 (b) Probability = 0.96 (c) Probability = 0.0864

Step by step solution

01

Understand the Probability Definitions

We have several pieces of information:- Probability of a positive test given tuberculosis, \( P(T+ | TB) = 0.82 \).- Probability of a positive test without tuberculosis, \( P(T+ | eg TB) = 0.09 \).- Probability of having tuberculosis, \( P(TB) = 0.04 \).- Probability of not having tuberculosis is the complement, \( P(eg TB) = 1 - P(TB) = 0.96 \).
02

Calculate Probability for Part (a)

We need to find \( P(TB \cap T+) \), the probability that a person has tuberculosis and the test is positive. Use the formula for conditional probability:\[ P(TB \cap T+) = P(T+) \ | \ TB) \times P(TB) \]Substitute the given values:\[ P(TB \cap T+) = 0.82 \times 0.04 = 0.0328 \]
03

Calculate Probability for Part (b)

We need to find \( P(eg TB) \), the probability that a person does not have tuberculosis. Since it's given that the probability of having tuberculosis is 0.04:\[ P(eg TB) = 1 - P(TB) = 0.96 \]
04

Calculate Probability for Part (c)

We are asked to find \( P(eg TB \cap T+) \), the probability that a person does not have tuberculosis and the test is positive. We again use the formula for joint probability:\[ P(eg TB \cap T+) = P(T+ | eg TB) \times P(eg TB) \]Substituting in the given values:\[ P(eg TB \cap T+) = 0.09 \times 0.96 = 0.0864 \]

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Bayes' theorem
Bayes' theorem is a pivotal concept in conditional probability that allows us to reverse conditional probabilities. It answers the question: If we know certain outcomes' probabilities given their causes, how can we determine the probability of the causes given the outcomes? In more simple terms, Bayes' theorem lets you update the probability of a hypothesis as more evidence or information becomes available.

In its standard form, Bayes’ theorem is given as:
  • \( P(A | B) = \frac{P(B | A) P(A)}{P(B)} \)
- \( P(A | B) \): The probability of event A occurring given that B is true.
- \( P(B | A) \): The probability of event B occurring given that A is true.
- \( P(A) \): The prior probability of event A occurring.
- \( P(B) \): The probability of event B.

Applying Bayes' theorem helps in medical diagnostics, spam filters, and other areas where deducing the likelihood of causes from the effects is crucial. In this context, with tuberculosis testing, Bayes' theorem could be used to determine how likely it is for someone to have tuberculosis when they test positive, contrasting the pre-test probability (base rate) with the test reliability.
joint probability
Joint probability deals with the likelihood of two events happening at the same time. It is the probability of the intersection of two events. In mathematical terms, the joint probability of event A and event B occurring together is denoted as \( P(A \cap B) \).

To calculate joint probability when dealing with conditional scenarios, we often use the formula:
  • \( P(A \cap B) = P(A | B) \times P(B) \)
This is derived from the idea of conditional probability, where \( P(A \cap B) \) is considered as finding the probability of B happening and then A happening given B has already happened.

In the tuberculosis example, one aspect of joint probability is calculating \( P(TB \cap T+) \), the probability that someone has tuberculosis and also tests positive. This can be interpreted as the test's reliability, multiplied by the disease's prevalence. Recognizing joint probabilities is essential in scenarios like disease testing, marketing statistics, and risk assessment.
complement rule
The complement rule is a fundamental principle that involves calculating the probability of the non-occurrence of an event. It is often used to simplify probability problems by focusing on the "outside" probability – that is, things not happening.

Mathematically, if you have an event A, its complement (not A) is denoted as \( \bar{A} \) or \( A^c \). The complement rule is expressed as:
  • \( P(\bar{A}) = 1 - P(A) \)
This simplifies finding probabilities since you can focus on the only two possibilities – the event happening or not happening, the sum of which is always 1.

In the context of the tuberculosis problem, the complement rule helps find the probability that someone does not have tuberculosis. With the known probability of having tuberculosis at 0.04, the probability of not having it is just \( 1 - 0.04 = 0.96 \). This approach is incredibly useful when it's easier to compute one probability than its complement, making calculations more straightforward and intuitive.

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Most popular questions from this chapter

Critical Thinking Consider the following events for a driver selected at random from the general population: \(A=\) driver is under 25 years old \(B=\) driver has received a speeding ticket Translate each of the following phrases into symbols. (a) The probability the driver has received a speeding ticket and is under 25 years old (b) The probability a driver who is under 25 years old has received a speeding ticket (c) The probability a driver who has received a speeding ticket is 25 years old or older (d) The probability the driver is under 25 years old or has received a speeding ticket (e) The probability the driver has not received a speeding ticket or is under 25 years old

Involve a standard deck of 52 playing cards. In such a deck of cards there are four suits of 13 cards each. The four suits are: hearts, diamonds, clubs, and spades. The 26 cards included in hearts and diamonds are red. The 26 cards included in clubs and spades are black. The 13 cards in each suit are: \(2,3,4,5,6,7,8,9,10,\) Jack, Queen, King, and Ace. This means there are four Aces, four Kings, four Queens, four \(10 \mathrm{s},\) etc., down to four \(2 \mathrm{s}\) in each deck. You draw two cards from a standard deck of 52 cards, but before you draw the second card, you put the first one back and reshuffle the deck. (a) Are the outcomes on the two cards independent? Why? (b) Find \(P(\text { Ace on } 1 \text { st card and King on } 2 \text { nd })\) (c) Find \(P(\text { King on } 1 \text { st card and Ace on } 2\)nd). (d) Find the probability of drawing an Ace and a King in either order.

(a) Draw a tree diagram to display all the possible head-tail sequences that can occur when you flip a coin three times. (b) How many sequences contain exactly two heads? (c) Probability Extension Assuming the sequences are all equally likely, what is the probability that you will get exactly two heads when you toss a coin three times?

Involve a standard deck of 52 playing cards. In such a deck of cards there are four suits of 13 cards each. The four suits are: hearts, diamonds, clubs, and spades. The 26 cards included in hearts and diamonds are red. The 26 cards included in clubs and spades are black. The 13 cards in each suit are: \(2,3,4,5,6,7,8,9,10,\) Jack, Queen, King, and Ace. This means there are four Aces, four Kings, four Queens, four \(10 \mathrm{s},\) etc., down to four \(2 \mathrm{s}\) in each deck. You draw two cards from a standard deck of 52 cards without replacing the first one before drawing the second. (a) Are the outcomes on the two cards independent? Why? (b) Find \(P(\text { Ace on } 1 \text { st card and } \text { King on } 2 n d)\) (c) Find \(P(\text { King on 1st card and Ace on } 2 \mathrm{nd}\) ). (d) Find the probability of drawing an Ace and a King in either order.

Involve a standard deck of 52 playing cards. In such a deck of cards there are four suits of 13 cards each. The four suits are: hearts, diamonds, clubs, and spades. The 26 cards included in hearts and diamonds are red. The 26 cards included in clubs and spades are black. The 13 cards in each suit are: \(2,3,4,5,6,7,8,9,10,\) Jack, Queen, King, and Ace. This means there are four Aces, four Kings, four Queens, four \(10 \mathrm{s},\) etc., down to four \(2 \mathrm{s}\) in each deck. You draw two cards from a standard deck of 52 cards, but before you draw the second card, you put the first one back and reshuffle the deck. (a) Are the outcomes on the two cards independent? Why? (b) Find \(P(3 \text { on } 1 \text { st card and } 10 \text { on } 2\)nd). (c) Find \(P(10 \text { on } 1 \text { st card and } 3 \text { on } 2 \text { nd })\) (d) Find the probability of drawing a 10 and a 3 in either order.

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