Final answer:
The probability that a basketball player with a 60% chance of making each free throw will make exactly three out of six free throws is calculated using the binomial probability formula and is found to be approximately 27.648%.
Step-by-step explanation:
The student has a 60% chance of making each free throw, and we want to know the probability that the player makes exactly three out of six free throws. The scenario can be described by the binomial probability formula, which is used for calculating the probability of getting a fixed number of successful outcomes in a fixed number of independent trials, with the successful outcome of each trial having the same probability.
To calculate this, we use the formula:
P(X = k) = C(n, k) * p^k * (1 - p)^(n - k)
Where:
- P(X = k) is the probability of k successes in n trials
- C(n, k) is the combination of n things taken k at a time
- p is the probability of success on a single trial
- n is the number of trials
- k is the number of successes
Given that the probability of making a free throw (p) is 0.60, and the number of trials (n) is 6, we want to calculate the probability of getting exactly 3 successes (k = 3).
P(X = 3) = C(6, 3) * 0.60^3 * (1 - 0.60)^(6 - 3)
Calculating the combination part, we get:
C(6, 3) = 6! / (3! * (6 - 3)!) = 20
Therefore, the probability is:
P(X = 3) = 20 * (0.60)^3 * (0.40)^3
Performing the calculations:
P(X = 3) = 20 * 0.216 * 0.064 = 20 * 0.013824 = 0.27648
So, the probability that the player makes exactly three out of six free throws is 0.27648, or 27.648%.