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 $73 \%$ of the time.

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

Choose 1 answer:
(A) $(1-0.73)^{9}$
(b) $9 \cdot 0.73$
(c) $9 \cdot(1-0.73)$
() $0.73^{9}$

Question

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 $73 \%$ of the time.

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

Choose 1 answer:
(A) $(1-0.73)^{9}$
(b) $9 \cdot 0.73$
(c) $9 \cdot(1-0.73)$
() $0.73^{9}$

Accepted Answer

∻Solution by Steps∻ ‖ step 1 ⋮ To find the probability of Antawn Jamison making all of his next 9 free throw attempts, we need to use the probability of him making one free throw, which is 73% or 0.73, and raise it to the power of 9 since each attempt is independent. ‖ ‖ step 2 ⋮ Calculating the probability: $0.73^9$. ‖ ‖ step 3 ⋮ Using the asksia-ll calculator result, the exact probability is $0.058871586708267913$. ‖ ∻1 Answer∻ ***D*** ∻Key Concept∻ ⚹Independent Events in Probability⚹ ∻Explanation∻ ⚹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).⚹◊What is the probability of Antawn Jamison successfully making all of his next 9 free throw attempts?⍭ Generate me a similar question◊

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