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Posted 2 months ago

Antawn Jamison is shooting free throws. Making or missing free throws doesn't change the probability that he will make his next one, and he makes his free throws 73%73 \% of the time.

What is the probability of Antawn Jamison making all of his next 9 free throw attempts?

Choose 1 answer:
(A) (10.73)9(1-0.73)^{9}
(b) 90.739 \cdot 0.73
(c) 9(10.73)9 \cdot(1-0.73)
() 0.7390.73^{9}
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Answer from Sia
Posted 2 months ago
Solution by Steps
step 2
Calculating the probability: 0.7390.73^9
step 3
Using the asksia-ll calculator result, the exact probability is 0.0588715867082679130.058871586708267913
1 Answer
Key Concept
Independent Events in Probability
When events are independent, the probability of all events occurring is the product of their individual probabilities. In this case, raising the single event probability to the power of the number of events (9 free throws).

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