Question 8 One of the reasons why the concept of an unfair coin is important in probability is that many real-life experiments can be modeled by a toss of an unfair coin. For example, if the probability of Ana Paula saving a penalty kick is 0.21, then saving a penalty kick can be modeled by a toss of an unfair coin for which Heads is associated with Saving and Tails with Missing. Thus, the probability of Ana Paula saving 3 penalty kicks out of 6 attempts is exactly the same as the probability of getting 3 heads from tossing the coin 6 times and can be computed using the following formula P(kH/n) = Cp (1 - p)n-k with n = 6, k = 3, and p = 0.21: P(3H/6)= C -0.21³-0.79³ 20-0.009261-0.493039 = 0.0913 = Find the probability of Ana Paula saving 4 penalty kicks out of 7 attempts: (Round the answer to 4 decimal places.)

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Question 8
One of the reasons why the concept of an unfair coin is important in probability is that many real-life
experiments can be modeled by a toss of an unfair coin. For example, if the probability of Ana Paula saving
a penalty kick is 0.21, then saving a penalty kick can be modeled by a toss of an unfair coin for which
Heads is associated with Saving and Tails with Missing. Thus, the probability of Ana Paula saving 3 penalty
kicks out of 6 attempts is exactly the same as the probability of getting 3 heads from tossing the coin 6
times and can be computed using the following formula
P(kH/n) = Cp (1 - p)n-k
with n = 6, k = 3, and p = 0.21:
P(3H/6)= C -0.21³-0.79³ 20-0.009261-0.493039 = 0.0913
=
Find the probability of Ana Paula saving 4 penalty kicks out of 7 attempts:
(Round the answer to 4 decimal places.)
Transcribed Image Text:Question 8 One of the reasons why the concept of an unfair coin is important in probability is that many real-life experiments can be modeled by a toss of an unfair coin. For example, if the probability of Ana Paula saving a penalty kick is 0.21, then saving a penalty kick can be modeled by a toss of an unfair coin for which Heads is associated with Saving and Tails with Missing. Thus, the probability of Ana Paula saving 3 penalty kicks out of 6 attempts is exactly the same as the probability of getting 3 heads from tossing the coin 6 times and can be computed using the following formula P(kH/n) = Cp (1 - p)n-k with n = 6, k = 3, and p = 0.21: P(3H/6)= C -0.21³-0.79³ 20-0.009261-0.493039 = 0.0913 = Find the probability of Ana Paula saving 4 penalty kicks out of 7 attempts: (Round the answer to 4 decimal places.)
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