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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## More answers

- 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

thank you

- mathmate

You're welcome! :)

- anonymous

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

- 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

Thank again

- mathmate

You're welcome!

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