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61% of U.S. adults have very little confidence in newspapers. You randomly select 10 U.S. adults. Find the probability that the number of U.S. adults who have very little confidence in newspapers is
(a) exactly five, (b) at least six, and (c) less than four.
(a) P(5)= (Round to three decimal places as needed.)

1 Answer

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Answer: To find the probability of each event, we can use the binomial probability formula:

P(x) = (nCx) * p^x * (1 - p)^(n - x)

where:

P(x) = Probability of getting exactly x successes

n = Total number of trials (in this case, 10 U.S. adults)

x = Number of successes we want to find the probability for

p = Probability of success in a single trial (in this case, 0.61, since 61% have very little confidence in newspapers)

(1 - p) = Probability of failure in a single trial (1 - 0.61 = 0.39)

(a) To find the probability of exactly five U.S. adults having very little confidence in newspapers:

P(5) = (10C5) * 0.61^5 * 0.39^(10 - 5)

Using the combination formula, (nCx) = n! / (x!(n-x)!):

P(5) = (10! / (5!(10-5)!)) * 0.61^5 * 0.39^5

P(5) = (10! / (5! * 5!)) * 0.61^5 * 0.39^5

P(5) = (10 * 9 * 8 * 7 * 6 / (5 * 4 * 3 * 2 * 1)) * 0.61^5 * 0.39^5

P(5) = (30,240) * 0.61^5 * 0.39^5

P(5) ≈ 0.146 (rounded to three decimal places)

So, the probability of exactly five U.S. adults having very little confidence in newspapers is approximately 0.146 (or 14.6%).

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