Assume that a procedure yields a binomial distribution with \( n=6 \) trials and a probability of success of \( p=0.05 \). Use a binomial probability table to find the probability that the number of successes \( x \) is exactly 2 . \( \mathrm{P}(2)=\square \) (Round to three decimal places as needed.)

To find the probability of exactly 2 successes in a binomial distribution with \( n = 6 \) trials and a probability of success \( p = 0.05 \), we can use the binomial probability formula: \[ P(X = x) = \binom{n}{x} p^x (1-p)^{n-x} \] Where: - \( n \) is the number of trials (6 in this case), - \( x \) is the number of successes (2 in this case), - \( p \) is the probability of success (0.05), - \( \binom{n}{x} \) is the binomial coefficient, calculated as \( \frac{n!}{x!(n-x)!} \). Now, let's calculate \( P(X = 2) \): 1. Calculate the binomial coefficient \( \binom{6}{2} \): \[ \binom{6}{2} = \frac{6!}{2!(6-2)!} = \frac{6 \times 5}{2 \times 1} = 15 \] 2. Calculate \( p^x \): \[ p^2 = (0.05)^2 = 0.0025 \] 3. Calculate \( (1-p)^{n-x} \): \[ (1-p)^{6-2} = (0.95)^4 \] Calculating \( (0.95)^4 \): \[ (0.95)^4 \approx 0.8145 \] 4. Now, substitute these values into the binomial probability formula: \[ P(X = 2) = \binom{6}{2} (0.05)^2 (0.95)^4 \] \[ P(X = 2) = 15 \times 0.0025 \times 0.8145 \] Calculating this step-by-step: \[ 15 \times 0.0025 = 0.0375 \] \[ 0.0375 \times 0.8145 \approx 0.0306 \] Thus, rounding to three decimal places, we find: \[ P(2) \approx 0.031 \] So, the final answer is: \[ \mathrm{P}(2) = 0.031 \]