A binomial probability experiment is conducted with the given parameters. Compute the probability of \( x \) successes in the \( n \) independent trials of the experiment. \( n=9, p=0.6, x \leq 3 \) The probability of \( x \leq 3 \) successes is \( \square \). (Round to four decimal places as needed.)

To compute the probability of \( x \leq 3 \) successes in a binomial experiment with \( n = 9 \) trials and a probability of success \( p = 0.6 \), we need to sum up the probabilities of \( x = 0, 1, 2, \) and \( 3 \) successes. The formula for the binomial probability is given by: \[ P(X = x) = \binom{n}{x} p^x (1 - p)^{n - x} \] Calculating for each \( x \): 1. For \( x = 0 \): \[ P(X = 0) = \binom{9}{0} (0.6)^0 (0.4)^9 = 1 \cdot 1 \cdot 0.000262144 \approx 0.0003 \] 2. For \( x = 1 \): \[ P(X = 1) = \binom{9}{1} (0.6)^1 (0.4)^8 = 9 \cdot 0.6 \cdot 0.00065536 = 0.0035 \] 3. For \( x = 2 \): \[ P(X = 2) = \binom{9}{2} (0.6)^2 (0.4)^7 = 36 \cdot 0.36 \cdot 0.0016384 \approx 0.0212 \] 4. For \( x = 3 \): \[ P(X = 3) = \binom{9}{3} (0.6)^3 (0.4)^6 = 84 \cdot 0.216 \cdot 0.004096 \approx 0.6936 \] Now summing these probabilities: \[ P(X \leq 3) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) = 0.0003 + 0.0035 + 0.0212 + 0.6936 \approx 0.7186 \] Therefore, the final answer, rounded to four decimal places, is: The probability of \( x \leq 3 \) successes is \( 0.7186 \).