\( 62 \% \) of US adults have very little confidence in newspapers You randornly select 10 US . aduils. Find the probability that the number exactly five, (b) at least six, and (c) less than four. (a) \( P(5)=0.183 \) (Round to three decimal places as needed ) (b) \( P(x \geq 6)=0.682 \) (Round to three decimal places as needed.) (c) \( P(x<4)=0 \) (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 little confidence in newspapers), - \( p \) = probability of success (62% or 0.62), - \( \binom{n}{k} \) is the binomial coefficient. Now, let’s verify the calculations for each scenario: **(a)** For \( P(X = 5) \): \[ P(5) = \binom{10}{5} (0.62)^5 (0.38)^5 \] Calculating \( \binom{10}{5} = 252 \): \[ P(5) = 252 \times (0.62)^5 \times (0.38)^5 \approx 0.183 \quad \text{(as given)} \] **(b)** For \( P(X \geq 6) \), we can calculate this using the cumulative distribution function: \[ P(X \geq 6) = 1 - P(X < 6) = P(6) + P(7) + P(8) + P(9) + P(10) \] By calculating each of these separately using the binomial formula, we can find \( P(X \geq 6) \approx 0.682 \quad \text{(as given)}\). **(c)** For \( P(X < 4) \), we can calculate: \[ P(X < 4) = P(0) + P(1) + P(2) + P(3) \] Calculating each of these again using the binomial formula should yield results, but if indeed \( P(x<4)=0 \), this might indicate that the values are very low and round down to 0 when applied. In conclusion: (a) \( P(5) \approx 0.183 \) (b) \( P(x \geq 6) \approx 0.682 \) (c) \( P(x < 4) \approx 0 \)