Final answer:
The probability that Michael Jordan makes the next four free throws is 0.299, misses the next four is 0.005, and makes at least one is 0.995. These calculations are based on his free throw percentage of 74% and rounded to three decimal places.
Step-by-step explanation:
The task involves calculating probabilities related to Michael Jordan's free throw chances, expressed as percentages. We will consider his probability of making the next four free throws, missing the next four, and making at least one of the next four.
A. To calculate the probability that Jordan makes the next four free throws when his free throw percentage is 74%, we multiply the probability of making each shot together because each shot is an independent event:
P(makes the next 4) = 0.74 × 0.74 × 0.74 × 0.74 = 0.2990 (rounded to four decimal places)
B. The probability that he misses the next four free throws is calculated by first finding the probability of missing one shot, which is 1 - 0.74 = 0.26, and then multiplying it four times:
P(misses the next 4) = 0.26 × 0.26 × 0.26 × 0.26 = 0.0046 (rounded to four decimal places)
C. The probability of making at least one of the next four free throws is the complement of the probability that he misses all four. Therefore:
P(makes at least one of the next 4) = 1 - P(misses the next 4) = 1 - 0.0046 = 0.9954 (rounded to four decimal places)
When we round these probabilities to three decimal places as requested:
- P(makes the next 4) = 0.299
- P(misses the next 4) = 0.005
- P(makes at least one of the next 4) = 0.995
Comparing our calculated probabilities to the provided options, it appears that option 4 (0.300, 0.023, and 0.995) is the closest to our calculations, but it differs slightly because the second value should be 0.005, not 0.023.