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 HE, and that 10% of the ticks that had either Lyme disease or HGE carried both diseases.

(a) What is the probability that a tick carries both Lyme disease (L) and HE (H)?

(b) What is the conditional probability that a tick has HE given that it has Lyme disease?

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Alcaa Alcaa · Answer 1 · 2020-03-09T09:33:00+01:00

Answer:

(a) 0.16

(b) 1

Explanation:

Let Probability that ticks in the Midwest carried Lyme disease, P(L) = 0.16

Probability that ticks in the Midwest carried HGE disease, P(H) = 0.10

Probability that ticks in the Midwest carried either Lyme disease or HGE disease, P([tex]L \bigcup H[/tex]) = 0.10

(a) Probability that a tick carries both Lyme disease (L) and HE (H) is given by

P(L [tex]\bigcap[/tex] H);

As we know that P(A [tex]\bigcup[/tex] B) = P(A) + P(B) - P(A [tex]\bigcap[/tex] B)

So, in our question;

P(L [tex]\bigcup[/tex] H) = P(L) + P(H) - P(L [tex]\bigcap[/tex] H)

0.10 = 0.16 + 0.10 - P(L [tex]\bigcap[/tex] H)

P(L [tex]\bigcap[/tex] H) = 0.16 + 0.10 - 0.10 = 0.16

Therefore, the probability that a tick carries both Lyme disease (L) and HE (H) is 0.16 or 16% .

(b) Conditional Probability P(A/B) is given by = [tex]\frac{P(A\bigcap B)}{P(B)}[/tex]

So, the conditional probability that a tick has HE given that it has Lyme disease is given by = P(H/L)

P(H/L) = [tex]\frac{P(H\bigcap L)}{P(L)}[/tex] = [tex]\frac{0.16}{0.16}[/tex] = 1 .