Question

If the probability of making a layup is 0.8, how many made lay-ups would you predict the player would make if they shot 100 times? *

Answer 1

To predict made layups with a 0.8 probability, multiply 0.8 by 100, resulting in 80 layups. For Helen's free throws with a 0.75 chance of making the first and a 0.85 chance of making the second given the first is made, the combined probability is 63.75%.

Step-by-step explanation:

If the probability of making a layup is 0.8, you would predict that a player would make 80 layups if they shot 100 times. This prediction is based on the expected value formula, which in this case is simply the probability of success times the number of attempts (0.8 * 100 = 80). The goal in this situation is to calculate the expected number of successful shots out of a certain number of attempts.

To answer the question regarding Helen's free throws: If the probability of Helen making the first shot (C) is 0.75, and the probability that Helen makes the second shot (D) given she made the first is 0.85, the probability that Helen makes both free throws is found by multiplying P(C) and P(D|C) together: P(C) * P(D|C) = 0.75 * 0.85 = 0.6375, or 63.75%.