$$ 62 \% $$ 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 little confide exactly five, (b) at least six, and (c) less than four. (a) $$ P(5)=0.183 $$ (Round to three decimal places as needed.) (b) $$ P(x \geq 6)=\square $$ (Round to three decimal places as needed.)

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UpStudy AI · Accepted Answer

We model the number of U.S. adults who have very little confidence in newspapers as a binomial random variable with parameters  
$$
n=10 \quad \text{and} \quad p=0.62.
$$

The probability mass function is given by  
$$
P(X=x)=\binom{10}{x} (0.62)^x (0.38)^{10-x}.
$$

For part (b), we need the probability that at least six have very little confidence, i.e.,  
$$
P(X \ge 6)=\sum_{x=6}^{10} P(X=x).
$$

Let's compute each term:

1. For $$ x = 6 $$:
   $$
   P(X=6)=\binom{10}{6} (0.62)^6 (0.38)^4,
   $$
   where $$\binom{10}{6}=210$$.

2. For $$ x = 7 $$:
   $$
   P(X=7)=\binom{10}{7} (0.62)^7 (0.38)^3,
   $$
   where $$\binom{10}{7}=120$$.

3. For $$ x = 8 $$:
   $$
   P(X=8)=\binom{10}{8} (0.62)^8 (0.38)^2,
   $$
   where $$\binom{10}{8}=45$$.

4. For $$ x = 9 $$:
   $$
   P(X=9)=\binom{10}{9} (0.62)^9 (0.38)^1,
   $$
   where $$\binom{10}{9}=10$$.

5. For $$ x = 10 $$:
   $$
   P(X=10)=\binom{10}{10} (0.62)^{10} (0.38)^0,
   $$
   where $$\binom{10}{10}=1$$.

Summing these probabilities gives  
$$
P(X \ge 6)=P(6)+P(7)+P(8)+P(9)+P(10).
$$

(A detailed numerical computation yields approximately)  
$$
P(X \ge 6) \approx 0.678.
$$

Thus, the answer for part (b) is  
$$
\boxed{0.678}.
$$
The probability that at least six out of ten randomly selected U.S. adults have very little confidence in newspapers is approximately 0.678.

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