Before a captive breeding program and subsequent reintroduction of the black-footed ferret it is believed that their total population size was only around 18 individuals. If the annual survival rate in the wild was 0.4.

a. What is the probability that all 18 would have died in a single year?
b. Because of the presence of canine distemper disease, the annual mortality rate might have been as high as 0.89. What is the probability that all 18 would have died in a single year if canine distemper was present?

a) 99.93%

b) 99.98%

Explanation:

Given that:

Total population size was only around 18 individuals ( for  black-footed ferret)

If the annual survival rate in the wild was 0.4%.

a)

What is the probability that all 18 would have died in a single year?

To find that; we multiply the annual survival rate in the wild with the total population; which is:

(0.4%×18) = 0.072%

Then; we subtract it from a total of 100% in order to determine the probability that all 18 would have died in a single year.

= (100.00% - 0.072%)

= 99.93%

b)

What is the probability that all 18 would have died in a single year if canine distemper was present?

Given that;

Because of the presence of canine distemper disease, the annual mortality rate might have been as high as 0.89.

To determine the probability; we have:

1 - Annual mortality rate = annual survival rate

1 - 0.89 = 0.11 %

Therefore, 0.11% of 18 individuals = (0.0011 × 18)

= 0.0198%

Probability that all 18 would have died in a single year if canine distemper was present = 100 - probability of annual survival rate

= (100 - 0.0198)%

= 99.98%

not sure if this is what you'r looking for eubacteria and archaebacteria

multiply it by 5 in order for you to understand i cant give an exact answer.