A binomial experiment with probability of success \( p=0.3 \) and \( n=6 \) trials is conducted. What is the probability that the experiment results in exactly 1 success? Do not round your intermediate computations, and round your answer to three decimal places. (If necessary, consult a list of formulas.)

To find the probability of exactly 1 success in a binomial experiment with \( n = 6 \) trials and a probability of success \( p = 0.3 \), we can use the binomial probability formula: \[ P(X = k) = \binom{n}{k} p^k (1-p)^{n-k} \] Where: - \( \binom{n}{k} \) is the binomial coefficient, - \( k \) is the number of successes, - \( p \) is the probability of success, - \( (1-p) \) is the probability of failure. In this case, \( n = 6 \), \( k = 1 \), and \( p = 0.3 \). First, we calculate the binomial coefficient: \[ \binom{6}{1} = \frac{6!}{1!(6-1)!} = \frac{6}{1} = 6 \] Next, we calculate \( p^k \) and \( (1-p)^{n-k} \): \[ p^1 = (0.3)^1 = 0.3 \] \[ (1-p)^{n-k} = (0.7)^{6-1} = (0.7)^5 \] Calculate \( (0.7)^5 \): \[ (0.7)^5 = 0.16807 \] Now we can plug everything back into the formula: \[ P(X = 1) = \binom{6}{1} \cdot p^1 \cdot (0.7)^5 = 6 \cdot 0.3 \cdot 0.16807 \] Calculating that step: \[ P(X = 1) = 6 \cdot 0.3 \cdot 0.16807 = 6 \cdot 0.050421 = 0.302526 \] Finally, rounding to three decimal places gives: \[ P(X = 1) \approx 0.303 \] So, the probability that the experiment results in exactly 1 success is approximately **0.303**.