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 .
Im not sure i get the question. Is that the whole picture?
They evolve alongside each other, the predator evolves to become faster, better camouflage itself, immunity to the poison of a prey, and likewise the prey evolves to become faster to avoid the predator, camouflage to hide, poison to keep the predator at bay and this cyclic relationship continues with only the strongest and fastest of each species reproducing, just increasing the strengths of both animals.
The predator is almost always larger, and a carnivore.
With the prey generally being smaller and herbivore.
<span>Answer: C. child #1 and #2 each had a 50% chance of getting HD; child #3 had a 0% chance.
Huntington’s disease is inherited in an autosomal dominant pattern which means you only need one allele from any parents to get the disease. That also means that there is no carrier, a healthy person must not carry any of the genes.
Parent of child 1 and 2 is a healthy male and female with HD. If the female has 2 HD gene, all of her children will get HD. But since she has a few normal children, then she must be heterozygous. Child of one heterozygous parent has 50% to get HD.
Since child 1 is normal, there is no chance for child 3 to get HD.</span>
