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Mike makes 46% of his free throw attempts. if he attempts 3 free throws, find the probability that he makes at least one.

User Thar
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Explanation:

To find the probability that Mike makes at least one free throw, we can calculate the probability of the complement event (the event of him missing all three free throws) and then subtract it from 1.

The probability of Mike missing a free throw is given by 1 - 46% = 54%.

Assuming each free throw attempt is independent, the probability of him missing all three free throws is given by:

P(missing all three) = (0.54)^3 ≈ 0.1665

Therefore, the probability that he makes at least one free throw is:

P(making at least one) = 1 - P(missing all three) = 1 - 0.1665 ≈ 0.8335

So, the probability that Mike makes at least one free throw is approximately 0.8335 or 83.35%.

User Bynx
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