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() = 0.10
(a) Probability that a tick carries both Lyme disease (L) and HE (H) is given by
P(L H);
As we know that P(A B) = P(A) + P(B) - P(A
B)
So, in our question;
P(L H) = P(L) + P(H) - P(L
H)
0.10 = 0.16 + 0.10 - P(L H)
P(L 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) <em>Conditional Probability P(A/B) is given by</em> =
So, the conditional probability that a tick has HE given that it has Lyme disease is given by = P(H/L)
P(H/L) = =
= 1 .