\( 62 \% \) 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 lit 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.)

To solve the problem, we can use the binomial probability formula, which is given by: \[ P(X = k) = \binom{n}{k} p^k (1-p)^{n-k} \] where: - \( n \) is the number of trials (in this case, \( n = 10 \)) - \( k \) is the number of successes (the number of adults with very little confidence in newspapers) - \( p \) is the probability of success (in this case, \( p = 0.62 \)) ### Part (b): Finding \( P(X \geq 6) \) To find \( P(X \geq 6) \), we can either calculate \( P(6) + P(7) + P(8) + P(9) + P(10) \) or use the complement rule: \[ P(X \geq 6) = 1 - P(X < 6) = 1 - (P(0) + P(1) + P(2) + P(3) + P(4) + P(5)) \] After calculating \( P(0) \) through \( P(5) \): 1. **Calculating \( P(0) \) to \( P(5) \) (using the formula)** - \( P(0) = \binom{10}{0} (0.62)^0 (0.38)^{10} \) (about \( 0.000 \)) - \( P(1) = \binom{10}{1} (0.62)^1 (0.38)^{9} \) (about \( 0.000 \)) - \( P(2) = \binom{10}{2} (0.62)^2 (0.38)^{8} \) (about \( 0.001 \)) - \( P(3) = \binom{10}{3} (0.62)^3 (0.38)^{7} \) (about \( 0.013 \)) - \( P(4) = \binom{10}{4} (0.62)^4 (0.38)^{6} \) (about \( 0.076 \)) - \( P(5) = 0.183 \) (as given) Adding these probabilities together: - \( P(X < 6) \approx 0 + 0 + 0.001 + 0.013 + 0.076 + 0.183 \approx 0.273 \) Thus: \[ P(X \geq 6) = 1 - P(X < 6) = 1 - 0.273 = 0.727 \] ### Answer for Part (b): \( P(x \geq 6) \approx 0.727 \) (rounded to three decimal places) ### Part (c): Finding \( P(X < 4) \) To find \( P(X < 4) \), we can sum the probabilities of having 0, 1, 2, or 3 adults who have very little confidence in newspapers: \[ P(X < 4) = P(0) + P(1) + P(2) + P(3) \] Using the approximations made earlier: - \( P(0) \approx 0.000 \) - \( P(1) \approx 0.000 \) - \( P(2) \approx 0.001 \) - \( P(3) \approx 0.013 \) Thus: \[ P(X < 4) \approx 0 + 0 + 0.001 + 0.013 = 0.014 \] ### Answer for Part (c): \( P(x < 4) \approx 0.014 \) (rounded to three decimal places) ### Summary of Answers: (b) \( P(x \geq 6) \approx 0.727 \) (c) \( P(x < 4) \approx 0.014 \)