\( |\rightarrow| \begin{array}{l}\text { 49\% of U.S. adults have very little confidence in } \\ \text { newspapers. You randomly select 10 U.S. adults. Find the } \\ \text { probability that the number of U.S. adults who have very } \\ \text { little confidence in newspapers is (a) exactly five, (b) at } \\ \text { least six, and (c) less than four. }\end{array} \)

To solve this problem, we can model the situation using the binomial distribution. The binomial distribution is appropriate here because we have a fixed number of trials (10 U.S. adults), two possible outcomes (having very little confidence in newspapers or not), and a constant probability of success (49% or 0.49). Let \( n = 10 \) (the number of trials) and \( p = 0.49 \) (the probability of success). The probability mass function of a binomial distribution is given by: \[ P(X = k) = \binom{n}{k} p^k (1-p)^{n-k} \] where \( \binom{n}{k} \) is the binomial coefficient, which can be calculated as: \[ \binom{n}{k} = \frac{n!}{k!(n-k)!} \] ### (a) Probability that exactly five adults have very little confidence in newspapers We want to find \( P(X = 5) \): \[ P(X = 5) = \binom{10}{5} (0.49)^5 (0.51)^{10-5} \] Calculating \( \binom{10}{5} \): \[ \binom{10}{5} = \frac{10!}{5!5!} = \frac{10 \times 9 \times 8 \times 7 \times 6}{5 \times 4 \times 3 \times 2 \times 1} = 252 \] Now substituting the values: \[ P(X = 5) = 252 \times (0.49)^5 \times (0.51)^5 \] Calculating \( (0.49)^5 \) and \( (0.51)^5 \): \[ (0.49)^5 \approx 0.049 \] \[ (0.51)^5 \approx 0.037 \] Now substituting these values: \[ P(X = 5) \approx 252 \times 0.049 \times 0.037 \approx 252 \times 0.001813 \approx 0.457 \] ### (b) Probability that at least six adults have very little confidence in newspapers We want to find \( P(X \geq 6) \), which can be calculated as: \[ P(X \geq 6) = P(X = 6) + P(X = 7) + P(X = 8) + P(X = 9) + P(X = 10) \] Calculating each term: 1. **For \( P(X = 6) \)**: \[ P(X = 6) = \binom{10}{6} (0.49)^6 (0.51)^4 \] \[ \binom{10}{6} = 210 \] \[ P(X = 6) = 210 \times (0.49)^6 \times (0.51)^4 \approx 210 \times 0.024 \times 0.067 \approx 0.327 \] 2. **For \( P(X = 7) \)**: \[ P(X = 7) = \binom{10}{7} (0.49)^7 (0.51)^3 \] \[ \binom{10}{7} = 120 \] \[ P(X = 7) = 120 \times (0.49)^7 \times (0.51)^3 \approx 120 \times 0.012 \times 0.132 \approx 0.191 \] 3. **For \( P(X = 8) \)**: \[ P(X = 8) = \binom{10}{8} (0.49)^8 (0.51)^2 \] \[ \binom{10}{8} = 45 \] \[ P(X = 8) = 45 \times (0.49)^8 \times (0.51)^2 \approx 45 \times 0.006 \times 0.260 \approx 0.070 \] 4. **For \( P(X = 9) \)**: \[ P(X = 9) = \binom{10}{9} (0.49)^9 (0.51)^1 \] \[ \binom{10}{9} = 10 \] \[ P(X = 9) = 10 \times (0.49)^9 \times (0.51)^1 \approx 10 \times 0.003 \times 0.51 \approx 0.015 \] 5. **For \( P(X = 10) \)**: \[ P(X = 10) = \binom{10}{10} (0.49)^{10} (0.51)^0 \] \[ P(X = 10) = 1 \times (0.49)^{10} \approx 0.001 \] Now summing these probabilities: \[ P(X \geq 6) \approx 0.327 + 0.191 + 0.070 + 0.015 + 0.001 \approx 0.604 \] ### (c) Probability that less than four adults have very little confidence in newspapers We want to find \( P(X < 4) \), which can be calculated as: \[ P(X < 4) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) \] Calculating each term: 1. **For \( P(X = 0) \)**: \[ P(X = 0) = \binom{10}{0} (0.49)^0 (0.51)^{10} = 1 \times 1 \times (0.51)^{10} \approx 0.002 \] 2. **For \( P(X = 1) \)**: \[ P(X = 1) = \binom{10}{1} (0.49)^1 (0.51)^9 = 10 \times 0.49 \times (0.51)^9 \approx 10 \times 0.49 \times 0.004 \approx 0.020 \] 3. **For \( P(X = 2) \)**: \[ P(X = 2) = \binom{10}{2} (0.49)^2 (0.51)^8 = 45 \times (0.49)^2 \times (0.51)^8 \approx 45 \times 0.2401 \times 0.008 \approx 0.086 \] 4. **For \( P(X = 3) \)**: \[ P(X = 3) = \binom{10}{3} (0.49)^3 (0.51)^7 = 120 \times (0.49)^3 \times (0.51)^7 \approx 120 \times 0.117649 \times 0.016 \approx 0.225 \] Now summing these probabilities: \[ P(X < 4) \approx 0.002 + 0.020 + 0.086 + 0.225 \approx 0.333 \] ### Summary of Results - (a) \( P(X = 5) \approx 0.457 \) - (b) \( P(X \geq 6) \approx 0.604 \) - (c) \( P(X < 4) \approx 0.333 \)