The goal is to come up with two ways to represent this probability:
An archer shoots a bulls-eye with probability $0.4$. If the archer shoots ten arrows, what's the probability that at least 3 are bullseyes?
The first representation I came up with is $X\sim$binomial$(10,0.4)$. To find the probability, I did $1-F(2)$.
What other distributions can be used to represent this question? I suspect the Negative binomial may work, but I can't seem to get the computations to match up.
OK, I think I understand your question now.
As you guessed, you can use the negative binomial:
Let $Y_i\sim Geometric(0.4)$ is the number of shots to get one bulls eye.
Then $Z=Y_1+Y_2+Y_3\sim NegBinom(3,0.4)$
Then, you want to know $P(Z\leq 10)$
From Excel, the $P(Z\leq 10)$ is the probability of getting 0 to 7 failures and three successes, which give a probability of $83.2\%$ whcih is the same as $1-F_X(2)$ for $X\sim binomial(10,0.4)$