## Libniz Group Title 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? one year ago one year ago

1. Libniz Group Title

@Zarkon

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@eliassaab

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@phi

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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 $$P(H|L) = \frac{P(H\cap L)}{P(L)}$$

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thank you

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You're welcome! :)

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I believe it was a typo? $P(L\cap H) = \frac{(P(L) + P(H)) }{10}$

8. mathmate Group Title

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.

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Thank again

10. mathmate Group Title

You're welcome!