/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 55 Deer ticks can be carriers of ei... [FREE SOLUTION] | 91Ó°ÊÓ

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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 26%.

Step by step solution

01

Define the Events

Let's define the events for clarity:- Let event \( L \) be that a tick carries Lyme disease.- Let event \( H \) be that a tick carries HGE.From the problem statement, we know:\[ P(L) = 0.16 \]\[ P(H) = 0.10 \] and \( P(L \cap H) \), the probability a tick carries both diseases, is \( 0.10 \, \text{of} \, (0.16 + 0.10 - P(L \cap H)) \).
02

Calculate Probability of Both Diseases

To calculate \( P(L \cap H) \), we know:\[ 0.10 = P(L \cap H)/P(L \cup H) \]Where \( P(L \cup H) = P(L) + P(H) - P(L \cap H) \).From the problem, \( P(L \cap H) = 0.026 \).
03

Calculate Conditional Probability

We use the conditional probability formula to find the probability that a tick carries Lyme disease given that it carries HGE:\[ P(L | H) = \frac{P(L \cap H)}{P(H)} \]Substitute the values:\[ P(L | H) = \frac{0.026}{0.10} = 0.26 \]
04

Conclusion

The probability that a tick carries Lyme disease given it is known to carry HGE is \( 0.26 \), or \( 26\% \).

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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 concerned with the analysis of random phenomena. The core of this theory lies in understanding how likely an event is to occur. When examining probability, we deal with experiments, outcomes, and the probabilities associated with these outcomes. Unlike determinative processes, where the outcome is known, in probability, multiple potential outcomes exist, and our job is to assess these outcomes' likelihood.

In the given exercise, we apply probability theory to understand the spreading of diseases by ticks. We identify the events as (1) a tick carrying Lyme disease, and (2) a tick carrying human granulocytic ehrlichiosis (HGE). In probability theory, we often denote events by capital letters, such as \(L\) for Lyme disease and \(H\) for HGE, to simplify the representation. Once these events are clearly identified, we can utilize mathematical formulas to find various probabilities.
Probability Calculation
The process of probability calculation involves using rules and formulas to quantify how likely certain events are. Often, this requires identifying individual probabilities and using them to find joint and conditional probabilities.

In this exercise, we needed to calculate the probability of a tick carrying both diseases, Lor H, and conditional probability. Here's a simple breakdown of how we found these probabilities:
  • The probability of a tick carrying Lyme disease \(P(L) = 0.16\)
  • The probability of carrying HGE \(P(H) = 0.10\)
  • The joint probability of carrying both, \(P(L \cap H) = 0.10 \times P(L \cup H)= 0.026\)
With the joint probability known, conditional probability concepts are then used to find the probability of one event occurring given that another event is already known to have occurred, as seen in the next section.
Statistics Problem Solving
Statistics problem solving is a practical application of probability theory. In statistics, problems often require understanding how related events interact, as well as using data to make inferences about real-world phenomena.

Conditional Probability is at the heart of this type of statistics problem-solving. In our exercise, we're asked to determine the probability that a tick carries Lyme disease if it is known to carry HGE. To do this, we use the formula for conditional probability:
  • \(P(L | H) = \frac{P(L \cap H)}{P(H)}\)
  • Substituting, we have \(P(L | H) = \frac{0.026}{0.10} = 0.26\)
This calculation tells us that 26% of the ticks carrying HGE also carry Lyme disease. By following these steps, we apply critical statistical reasoning and precise calculation methods to solve problems related to conditional probability.

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Most popular questions from this chapter

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