anonymous
  • anonymous
Deer ticks can carry both Lyme disease and human granulocytic ehrlichiosis (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 PLH that a tick carries both Lyme disease (L) and HGE (H)? (b) What is the conditional probability that a tick has HGE given that it has Lyme disease?
Mathematics
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anonymous
  • anonymous
Deer ticks can carry both Lyme disease and human granulocytic ehrlichiosis (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 PLH that a tick carries both Lyme disease (L) and HGE (H)? (b) What is the conditional probability that a tick has HGE given that it has Lyme disease?
Mathematics
katieb
  • katieb
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At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.

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anonymous
  • anonymous
anonymous
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anonymous
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mathmate
  • mathmate
Let L=event of carrying Lyme's H=event of carrying HGE Then \( P(L\cup H) = P(L) + P(H) - P(L\cap H) \) But we also know that \( P(L \cap H) = 0.1P(L\cup H)\) So, substituting, \( 10P(L\cap H) = P(L) + P(H) - P(L\cap H) \) giving \( P(L\cap H) = \frac{(P(L) + P(H)) }{11} \) and \( P(H|L) = \frac{P(H\cap L)}{P(L)} \)
anonymous
  • anonymous
thank you
mathmate
  • mathmate
You're welcome! :)
anonymous
  • anonymous
I believe it was a typo? \[P(L\cap H) = \frac{(P(L) + P(H)) }{10}\]
mathmate
  • mathmate
It comes from: \( P(L\cap H)=0.1P(L\cup H) \) or \( 10P(L\cap H)=(L\cup H) \) So \( 10 P(L∩H)=P(L)+P(H)−P(L∩H) \) If we transpose \( -P(L\cap H) \) to the left, we get \( 11P(L∩H)=P(L)+P(H) \) and the rest follows.
anonymous
  • anonymous
Thank again
mathmate
  • mathmate
You're welcome!

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