Question

A college basket ball player makes 80% of her free throws. Suppose this probability is the same for each free throw she attempts. The probability that she makes all of her first four free throws and then misses her fifth attempt this season is

Answer 1

prob = (.8)^4 (.2) = .08192

Answer 2

Three

Answer 3

.00128

Answer 4

[AI-generated answer — Explain Bot]

To find the probability that the college basketball player makes all of her first four free throws and then misses her fifth attempt, we need to multiply the probabilities together.

Given that the player makes 80% of her free throws, the probability of making a free throw is 0.8, and the probability of missing a free throw is 0.2 (1 - 0.8).

Since each free throw attempt is independent of each other, we can multiply the probabilities together to find the overall probability.

P(making all four free throws and then missing the fifth)
= P(making the first free throw) * P(making the second free throw) * P(making the third free throw) * P(making the fourth free throw) * P(missing the fifth free throw)

= 0.8 * 0.8 * 0.8 * 0.8 * 0.2

= (0.8)^4 * 0.2

= 0.4096 * 0.2

= **0.08192** (or approximately 8.19%)

Therefore, the probability that the college basketball player makes all of her first four free throws and then misses her fifth attempt this season is approximately 8.19%.