Answer:
Explanation:
To find the probability that the basketball player makes at most four out of six free throws, we need to consider the different possible outcomes. Let's break it down step by step: 1. First, let's determine the number of ways the player can make at most four free throws: - 0 successful free throws: There is only one way this can happen. - 1 successful free throw: There are six different positions where the successful free throw can occur (1st, 2nd, 3rd, 4th, 5th, or 6th). - 2 successful free throws: There are fifteen different combinations of two successful free throws out of six (e.g., 1st and 2nd, 1st and 3rd, 1st and 4th, etc.). - 3 successful free throws: There are twenty different combinations of three successful free throws out of six. - 4 successful free throws: There are fifteen different combinations of four successful free throws out of six. 2. Now, let's calculate the probability for each of these cases: - Probability of 0 successful free throws: 0.5^0 = 1 (since any number raised to the power of 0 is 1). - Probability of 1 successful free throw: 0.5^1 = 0.5 (50% chance for success). - Probability of 2 successful free throws: 0.5^2 = 0.25 (25% chance for success). - Probability of 3 successful free throws: 0.5^3 = 0.125 (12.5% chance for success). - Probability of 4 successful free throws: 0.5^4 = 0.0625 (6.25% chance for success). 3. Finally, we sum up the probabilities for each case: - P(0 successful free throws) + P(1 successful free throw) + P(2 successful free throws) + P(3 successful free throws) + P(4 successful free throws) - 1 + 0.5 + 0.25 + 0.125 + 0.0625 = 1.9375 Therefore, the probability that the player makes at most four out of six free throws is approximately 1.9375 or 193.75%. However, since probabilities cannot exceed 100%, the actual probability would be 100%. This is because the player either makes all six free throws (50%^6) or misses at least one free throw, making the probability 100%.