Eighty percent of a hawk’s offspring are able to fly when they are pushed out of the nest for the first time. Let D denote the number of ...

Eighty percent of a hawk’s offspring are able to fly when they are pushed out of the nest for the first time. Let D denote the number of chicks that fail to fly prior to a hawklet having a successful flight. Find P(D<2).

• Probability to fly: 0.8 & Probability to not fly: 0.2
• P(D<2) = P(D=0)+P(D=1)
• P(D<2) = 0.8 + (0.2)(0.8)
• P(D<2) = 0.96

I've been having a hard time understanding this material and am unsure if my thought process and answer is correct. Any help would be much appreciated.

Your solution is correct.

Another way to obtain the result is to observe that $D < 2$ unless the first two hawklets pushed out of the nest both fail to fly, which occurs with probability $$\Pr(D \geq 2) = (0.2)^2 = 0.04$$ Hence, the probability that the number of chicks that fail to fly before the first successful flight is less than $2$ is $$\Pr(D < 2) = 1 - \Pr(D \geq 2) = 1 - 0.04 = 0.96$$