\( 62 \% \) 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.)

To solve this 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: - \( n \) = number of trials (in this case, 10) - \( k \) = number of successes (the number of adults with very little confidence in newspapers) - \( p \) = probability of success (in this case, \( 0.62 \)) - \( \binom{n}{k} \) = "n choose k" or the number of combinations of \( n \) items taken \( k \) at a time. First, we need to find the probability for each case. ### (a) Probability that exactly five have confidence: Using the formula, we get: \[ P(X = 5) = \binom{10}{5} (0.62)^5 (0.38)^{5} \] Calculating each term: - \( \binom{10}{5} = 252 \) - \( (0.62)^5 \approx 0.086 \) - \( (0.38)^5 \approx 0.006 \) Now plug these values back into the formula: \[ P(X = 5) \approx 252 \times 0.086 \times 0.006 \approx 0.130 \] Rounded to three decimal places: \[ P(5) \approx 0.130 \] ### (b) Probability that at least six have confidence: To find \( P(X \geq 6) \), we can calculate the probabilities for \( X = 6, 7, 8, 9, \text{ and } 10 \) and add them together: \[ P(X \geq 6) = P(X=6) + P(X=7) + P(X=8) + P(X=9) + P(X=10) \] Calculating each: 1. **For \( X = 6 \)**: \[ P(X=6) = \binom{10}{6} (0.62)^6 (0.38)^4 = 210 \times (0.062) \times (0.019) \] Calculated value: \( P(X=6) \approx 0.253 \) 2. **For \( X = 7 \)**: \[ P(X=7) = \binom{10}{7} (0.62)^7 (0.38)^3 = 120 \times (0.036) \times (0.014) \] Calculated value: \( P(X=7) \approx 0.052 \) 3. **For \( X = 8 \)**: \[ P(X=8) = \binom{10}{8} (0.62)^8 (0.38)^2 = 45 \times (0.022) \times (0.01) \] Calculated value: \( P(X=8) \approx 0.009 \) 4. **For \( X = 9 \)**: \[ P(X=9) = \binom{10}{9} (0.62)^9 (0.38)^1 = 10 \times (0.014) \times (0.38) \] Calculated value: \( P(X=9) \approx 0.053 \) 5. **For \( X = 10 \)**: \[ P(X=10) = \binom{10}{10} (0.62)^{10} (0.38)^0 = 0.006 \] Now summing these probabilities gives us: \[ P(X \geq 6) \approx 0.253 + 0.052 + 0.009 + 0.053 + 0.006 \] \[ P(X \geq 6) \approx 0.373 \] ### (c) Probability that less than four have confidence: We can calculate \( P(X < 4) \) as \( P(X=0) + P(X=1) + P(X=2) + P(X=3) \): 1. **For \( X = 0 \)**: \[ P(X=0) = \binom{10}{0} (0.62)^{0} (0.38)^{10} \approx 0.000 \] 2. **For \( X = 1 \)**: \[ P(X=