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Deer ticks can be carriers of either Lyme disease or human granulocytic ehrlichiosis (HGE). Based on a recent study, suppose that \(16 \%\) of all ticks in a certain location carry Lyme disease, \(10 \%\) carry HGE, and \(10 \%\) of the ticks that carry at least one of these diseases in fact carry both of them. If a randomly selected tick is found to have carried HGE, what is the probability that the selected tick is also a carrier of Lyme disease?

Short Answer

Expert verified
The probability is approximately 0.2364.

Step by step solution

01

Define Events

Let \( L \) be the event that a tick carries Lyme disease, and \( H \) be the event that a tick carries HGE. We are given the following probabilities: \( P(L) = 0.16 \), \( P(H) = 0.10 \), and \( P(L \cap H | L \cup H) = 0.10 \).
02

Apply Inclusion-Exclusion Principle

Find the probability of the union of events, \( P(L \cup H) \). We use the formula: \[ P(L \cup H) = P(L) + P(H) - P(L \cap H) \]However, we need to find \( P(L \cap H) \). We know \( P(L \cap H | L \cup H) = 0.10 \), implying \( P(L \cap H) = 0.10 \cdot P(L \cup H) \).
03

Solve for Intersection

Substitute \( P(L \cup H) \) from Step 2:\[ P(L \cap H) = 0.10 \cdot P(L \cup H) \]Plug it back into the inclusion-exclusion formula: \[ P(L \cup H) = 0.16 + 0.10 - 0.10 \cdot P(L \cup H) \] Solving this equation will give us \( P(L \cup H) \).
04

Calculate \( P(L \cup H) \)

Simplify the equation:\[ 0.1 \cdot P(L \cup H) = 0.26 - P(L \cup H) \]\[ 1.1 \cdot P(L \cup H) = 0.26 \]\[ P(L \cup H) = \frac{0.26}{1.1} \approx 0.2364 \]
05

Calculate \( P(L \cap H) \)

Using \( P(L \cup H) = 0.2364 \), calculate \( P(L \cap H) \):\[ P(L \cap H) = 0.10 \times 0.2364 = 0.02364 \]
06

Apply Conditional Probability Formula

To find the probability that a tick carries Lyme disease given it carries HGE, use:\[ P(L | H) = \frac{P(L \cap H)}{P(H)} \]Substitute the known values:\[ P(L | H) = \frac{0.02364}{0.10} = 0.2364 \]
07

Conclusion

The probability that a tick carries Lyme disease, given that it carries HGE, is approximately 0.2364.

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Key Concepts

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

Probability Theory
Probability theory is a branch of mathematics that deals with uncertainty. It's an essential tool for understanding events that occur by chance. In probability theory, an event is any outcome or a combination of outcomes from an experiment or situation. For example, in our exercise of examining ticks, events include a tick carrying Lyme disease or a tick carrying HGE. Probabilities are values between 0 and 1, representing how likely an event is to occur.
  • A probability of 0 means the event never happens.
  • A probability of 1 means the event always happens.
All probabilities must sum up to 1 when considering all possible outcomes of an event. Studying these probabilities helps us make informed decisions in uncertain conditions.
Inclusion-Exclusion Principle
The Inclusion-Exclusion Principle is a technique used to find the probability of the union of multiple events. In simpler terms, it calculates the probability that at least one of several events happens. The basic formula for two events, say A and B, is: \[ P(A \cup B) = P(A) + P(B) - P(A \cap B) \] Here, we prevent double-counting the probability of both events happening simultaneously. In the context of our exercise, we applied this principle to find \( P(L \cup H) \), the probability that a tick carries either Lyme disease or HGE. This principle ensures that we account for cases where ticks carry both diseases only once, giving us a more accurate probability measure.
Intersection of Events
The intersection of events, denoted as \( L \cap H \), represents the situation where both events occur simultaneously. In our tick example, the intersection is the probability that a tick carries both Lyme disease and HGE. The formula used involves multiplying the conditional probability by the probability of the union of events: \[ P(L \cap H) = 0.10 \cdot P(L \cup H) \] This step in the solution helps identify how often both conditions (diseases) are present at the same time in a tick. Intersection probabilities are useful when considering joint outcomes in probability theory.
Bayes' Theorem
Bayes' Theorem is a powerful tool in probability theory, used to update the probability estimate for an event based on new evidence. The theorem relates the conditional and marginal probabilities of random events and is expressed as: \[ P(A | B) = \frac{P(B | A) \cdot P(A)}{P(B)} \] In our scenario, we want to find \( P(L | H) \), the probability that a tick carries Lyme disease given that it carries HGE. We use the relationship: \[ P(L | H) = \frac{P(L \cap H)}{P(H)} \] This allows us to incorporate the information about one event (HGE) into the probability assessment of another event (Lyme disease), providing a more informed and precise probability measure. This aspect of probability helps make predictions and informed decisions in fields such as medical diagnostics and risk assessment.

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Most popular questions from this chapter

At a large university, in the never-ending quest for a satisfactory textbook, the Statistics Department has tried a different text during each of the last three quarters. During the fall quarter, 500 students used the text by Professor Mean; during the winter quarter, 300 students used the text by Professor Median; and during the spring quarter, \(200 \mathrm{stu}-\) dents used the text by Professor Mode. A survey at the end of each quarter showed that 200 students were satisfied with Mean's book, 150 were satisfied with Median's book, and 160 were satisfied with Mode's book. If a student who took statistics during one of these quarters is selected at random and admits to having been satisfied with the text, is the student most likely to have used the book by Mean, Median, or Mode? Who is the least likely author? [Hint: Draw a tree diagram or use Bayes' theorem.]

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