$$ 62 \% $$ 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 liftle confidence in newspapers is (a) exactly five, (b) at least six, and (c) less than four. (a) $$ P(5)=\square $$ (Round to three decimal places as needed.)

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We model the number of U.S. adults with very little confidence in newspapers (with probability $$ p=0.62 $$) using a binomial distribution with parameters $$ n=10 $$ and $$ p=0.62 $$.

1. For exactly five adults, the probability is given by
   $$
   P(5)=\binom{10}{5}(0.62)^5 (0.38)^5.
   $$

2. The combination term is
   $$
   \binom{10}{5} = 252.
   $$

3. Therefore, we have
   $$
   P(5)=252 \times (0.62)^5 \times (0.38)^5.
   $$

4. Upon evaluating and rounding to three decimal places, we obtain
   $$
   P(5) \approx 0.183.
   $$

Thus, the answer is $$ P(5) \approx 0.183 $$.
$$ P(5) \approx 0.183 $$

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 liftle confidence in newspapers is (a)
exactly five, (b) at least six, and © less than four.
(a) (Round to three decimal places as needed.)

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(a) Probability that exactly five have confidence:

(b) Probability that at least six have confidence:

© Probability that less than four have confidence:

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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 liftle confidence in newspapers is (a) exactly five, (b) at least six, and © less than four. (a) (Round to three decimal places as needed.)