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Libniz

  • 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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  1. Libniz
    • 3 years ago
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    @Zarkon

  2. Libniz
    • 3 years ago
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    @eliassaab

  3. Libniz
    • 3 years ago
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    @phi

  4. mathmate
    • 3 years ago
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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 \( P(H|L) = \frac{P(H\cap L)}{P(L)} \)

  5. Libniz
    • 3 years ago
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    thank you

  6. mathmate
    • 3 years ago
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    You're welcome! :)

  7. Libniz
    • 3 years ago
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    I believe it was a typo? \[P(L\cap H) = \frac{(P(L) + P(H)) }{10}\]

  8. mathmate
    • 3 years ago
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    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.

  9. Libniz
    • 3 years ago
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    Thank again

  10. mathmate
    • 3 years ago
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    You're welcome!

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