A particular college basketball player is a 80% free throw shooter. In a game, he took 20 free throws. Find the probability he made 18 or more of his free throws.

P(18) = .8^18 * .2^2 = ?

P(19) = .8^19 * .2 = ?
P(20) = .8^20 = ?

Either-or probability is found by adding the above three answers.

To find the probability that the college basketball player made 18 or more of his free throws, we can use the binomial probability formula.

The binomial probability formula is given by:
P(x) = (nCx) * p^x * q^(n-x)

Where:
P(x) is the probability of getting exactly x successes,
n is the total number of trials,
x is the number of desired successes,
p is the probability of a single success,
q is the probability of a single failure (1 - p),
nCx is the binomial coefficient.

Let's apply this formula to the given problem.

n = 20 (since he took 20 free throws)
x = 18 (to find the probability of making 18 or more free throws)
p = 0.8 (80% free throw shooter, so the probability of making a free throw is 0.8)
q = 1 - p = 1 - 0.8 = 0.2 (the probability of missing a free throw)

Now, let's calculate the required probability:
P(18 or more) = P(18) + P(19) + P(20)

P(18) = (20C18) * (0.8^18) * (0.2^2)
P(19) = (20C19) * (0.8^19) * (0.2^1)
P(20) = (20C20) * (0.8^20) * (0.2^0)

To compute the binomial coefficients (20Cx), you can use the formula:
nCx = n! / (x! * (n-x)!)

Now, let's substitute the values and calculate the minimum free throws made:
P(18 or more) = [(20! / (18! * (20-18)!)] * (0.8^18) * (0.2^2) + [(20! / (19! * (20-19)!)] * (0.8^19) * (0.2^1) + [(20! / (20! * (20-20)!)] * (0.8^20) * (0.2^0)

Simplifying this equation will give us the probability that the college basketball player made 18 or more of his free throws.