Probability Theory - conditional

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SUMMARY

The discussion focuses on calculating the probability P[LH] that a deer tick carries both Lyme disease (L) and human granulocytic ehrlichiosis (HGE) (H). The probabilities provided are P[L] = 0.16 for Lyme disease and P[H] = 0.1 for HGE. It is established that 10% of ticks carrying either disease also carry both, leading to the interpretation P[LH | (L U H)] = 0.1 as the correct formulation for the probability of interest. The participant expresses uncertainty in their calculations but ultimately aligns with this interpretation.

PREREQUISITES
  • Understanding of basic probability concepts, including conditional probability
  • Familiarity with notation such as P[L], P[H], and P[LH | (L U H)]
  • Knowledge of set theory, particularly union and intersection of sets
  • Basic statistics, especially in the context of epidemiology
NEXT STEPS
  • Study conditional probability and its applications in epidemiology
  • Learn about joint probability distributions and how to calculate them
  • Explore set theory concepts, particularly union and intersection in probability
  • Investigate real-world applications of probability theory in disease modeling
USEFUL FOR

Students of probability theory, epidemiologists, and anyone interested in understanding the statistical analysis of disease transmission among populations.

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Question:
Deer ticks can carry both Lyme disease and human granulocytic ehrilichiosis (HGE). IN a study of ticks in the Midwest, it was found that 16% carried Lyme disease, 10% had HGE, and that 10% of the ticks that had either Lyme disease or HGE carried both diseases.

(a) What is the probability P[LH] that a tick carries both Lyme disease (L) and HGE (H)?

My Part:
I don't know how to interpret this problem. Let me show my work thus far.

P[L] = 0.16
P[H] = 0.1

Now this last part, how do I interpret this..."10% of the ticks that had either Lyme disease or HGE carried both diseases."

Like this? given that the ticks had L or HGE, then the ticks that both diseases. So I would write this as:

P[LH | (L U H)] = 0.1

I don't know if this is right though. Any help would be awesome, thanks!

another thought I had was:
P[LH|L] + P[LH|H] = 0.1

This was the first one I did, but I got a probability of something like 0.006 which seems WAY to low.
 
Last edited:
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P[LH | (L U H)] = 0.1 sounds correct.

It is the probability that a deer has L AND H knowing that it has L OR H. It is basically what the problem is saying but it speaks in percentage of a quantity instead of in probability.
 

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