Final answer:
The probability of a basketball player making the first successful free throw on the second attempt with a 71% success rate is found by multiplying the probability of missing the first shot by the probability of making the second shot, which is 0.2059
Step-by-step explanation:
The student is asking about determining the probability of a basketball player making her first successful free throw on the second attempt, given a certain success rate for free throw shots.
To calculate this, we use the concept of independent events in probability.
Assuming the basketball player has a success rate of 71% for her free throws, the probability of her missing the first shot (which is necessary for the second shot to be the first successful one) is 29% or 0.29. This is 1 minus the success rate (1 - 0.71 = 0.29). The probability of then making the second shot is the success rate of 0.71.
Thus, the probability of the first successful shot being the second attempt is calculated as the product of these two probabilities:
Probability of missing first shot = 1 - P(success)
= 1 - 0.71
= 0.29
Probability of making second shot = P(success)
= 0.71
Probability of first success on second shot = P(missing first) × P(making second)
= 0.29 × 0.71
= 0.2059