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anonymous
 3 years ago
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?
anonymous
 3 years ago
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?

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mathmate
 3 years ago
Best ResponseYou've already chosen the best response.3Let 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 \( P(HL) = \frac{P(H\cap L)}{P(L)} \)

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0I believe it was a typo? \[P(L\cap H) = \frac{(P(L) + P(H)) }{10}\]

mathmate
 3 years ago
Best ResponseYou've already chosen the best response.3It 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.
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