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 Kwan guessing the correct answer on a multiple-choice question is 0.25, then guessing the answer on a multiple-choice question can be modeled by a toss of an unfair coin for which Heads is associated with Correct and Tails with Incorrect. Thus, the probability of Kwan guessing correctly 2 answers on a multiple-choice test with 8 questions is exactly the same as the probability of getting 2 heads from tossing the coin 8 times and can be computed using the following formula P(kH/n) Chp(1-p)"-k with n=8, k-2, and p = 0.25: P(2H/8)=C-0.252-0.756-28-0.0625-0.177979 -0.3115 Find the probability of Kwan guessing correctly 3 answers on a multiple-choice test with 9 questions: (Round the answer to 4 decimal places.)
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 Kwan guessing the correct answer on a multiple-choice question is 0.25, then guessing the answer on a multiple-choice question can be modeled by a toss of an unfair coin for which Heads is associated with Correct and Tails with Incorrect. Thus, the probability of Kwan guessing correctly 2 answers on a multiple-choice test with 8 questions is exactly the same as the probability of getting 2 heads from tossing the coin 8 times and can be computed using the following formula P(kH/n) Chp(1-p)"-k with n=8, k-2, and p = 0.25: P(2H/8)=C-0.252-0.756-28-0.0625-0.177979 -0.3115 Find the probability of Kwan guessing correctly 3 answers on a multiple-choice test with 9 questions: (Round the answer to 4 decimal places.)
Chapter8: Sequences, Series,and Probability
Section8.7: Probability
Problem 4ECP: Show that the probability of drawing a club at random from a standard deck of 52 playing cards is...
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Given that the binomial experiment is performed in order to determine the number of possible success.
We need to find P(X=3) given n=9 and p=0.25.
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