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
  • Stacey Warren - Expert brainly.com
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SOLVED
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schrodinger
  • schrodinger
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anonymous
  • anonymous
@Zarkon
anonymous
  • anonymous
@eliassaab
anonymous
  • anonymous
@phi

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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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