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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 the Given Probabilities

Let \(L\) represent the event that a tick carries Lyme disease and \(H\) represent the event that a tick carries HGE. From the problem statement:- \(P(L) = 0.16\)- \(P(H) = 0.10\)- \(P(L \cap H | L \cup H) = 0.10\)These probabilities provide information about the percentages of ticks carrying each disease and both diseases together.
02

Calculate the Probability of the Union

We need to calculate \(P(L \cup H)\), the probability that a tick carries at least one of the two diseases. We know that:\[P(L \cup H) = P(L) + P(H) - P(L \cap H)\]Let's isolate \(P(L \cap H)\). Given \(P(L \cap H | L \cup H) = 0.10\), we know:\[P(L \cap H) = 0.10 \times P(L \cup H)\]Substitute back into the union formula:\[P(L \cup H) = 0.16 + 0.10 - 0.10 \times P(L \cup H)\]
03

Solve for \(P(L \cap H)\) and \(P(L \cup H)\)

Rewriting the equation from the union step:\[P(L \cup H) = 0.26 - 0.10 \times P(L \cup H)\]We rearrange it to solve for \(P(L \cup H)\):\[1.10 \times P(L \cup H) = 0.26\]\[P(L \cup H) = \frac{0.26}{1.10} \approx 0.2364\]Now, using \(P(L \cap H) = 0.10 \times P(L \cup H)\):\[P(L \cap H) = 0.10 \times 0.2364 = 0.02364\]
04

Calculate Conditional Probability

We need to calculate \(P(L | H)\), the probability that a tick carries Lyme disease given that it carries HGE. The formula is:\[P(L | H) = \frac{P(L \cap H)}{P(H)}\]We already know \(P(L \cap H) = 0.02364\) and \(P(H) = 0.10\). Substitute these values in:\[P(L | H) = \frac{0.02364}{0.10} = 0.2364\]
05

Conclusion

Thus, 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.

Joint Probability
Joint probability is all about understanding how two events can occur simultaneously. In simple terms, it's the chance of both events happening at the same time. For instance, in this problem, it's the likelihood that a tick carries both Lyme disease and HGE. To calculate joint probability, you look at the intersection of two events, which is represented as \(P(A \cap B)\). This means we're interested in the ticks that carry both diseases. We use information like \(P(L \cap H | L \cup H)\), which indicates the probability of a tick carrying both diseases in relation to ticks carrying at least one disease.

Using the provided probability, \(0.10\), for ticks carrying both diseases, joint probability becomes pivotal because it allows us to understand and calculate the likelihood of two health issues existing in the same tick.
  • It gives insights into the interaction between two conditions.
  • Helps in understanding the probability of co-occurrences.

Understanding joint probability is crucial for analyzing situations where multiple outcomes are possible simultaneously, especially in disease research and other fields dealing with overlapping conditions.
Probability Theory
Probability theory is the math that helps us make sense of uncertainty. It provides the foundation for dealing with chance and random events. At its core, probability theory allows us to assign a numerical value between 0 and 1 to an event, indicating its likelihood. If an event is certain, the probability is 1. If it's impossible, the probability is 0. Everything in between reflects degrees of chance.

In this exercise, probability theory enables us to take known values, like the percent chance a tick has Lyme disease \(P(L) = 0.16\) or HGE \(P(H) = 0.10\), and combine them to solve more complex questions, such as the probability a tick has Lyme disease given it has HGE \(P(L | H)\). It's crucial to:
  • Understand individual probabilities before tackling combinations.
  • Apply set theory to calculate unions \(P(L \cup H)\) and intersections \(P(L \cap H)\).
Probability theory simplifies complex realities, providing tools to shut out guesswork and manage randomness, which is immensely useful in health studies, forecasting, risk assessment, and decision making.
Lyme Disease and HGE Probability
Understanding Lyme disease and HGE probabilities helps us deal with real-world disease concerns in a systematic way. In this scenario, we know certain percentages of ticks carry Lyme disease and/or HGE. Through conditional probability, we get to answer questions about how these diseases coexist. Conditional probability \(P(A | B)\) tells us the probability of an event happening given that another event has already occurred. Here, \(P(L | H) = \frac{0.02364}{0.10} = 0.2364\), frames the issue of how likely a tick with HGE also has Lyme disease.

This type of analysis is vital for:
  • Developing targeted interventions.
  • Understanding the spread and co-existence of diseases.
  • Informed decision-making in medical and environmental studies.
By interpreting these probabilities, researchers and public health officials can better plan strategies for disease control and prevention. It's a statistical glance at the likelihood that helps guide actions to combat these diseases.

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