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%

the father in the first generation probably had the recessive colorblind trait in his x chromosome. if he and his wife both had the colorblind trait, even if she had 1 recessive and 1 dominant, then that trait would be passed on to their kids.

(since you didn't attach the family tree this is just an assumption)