A binomial probability experiment is conducted with the given parameters. Compute the probability of \( x \) successes in the \( n \) independent trials of the experiment. \( n=9, p=0.4, x \leq 3 \) 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 the given binomial probability experiment, we will 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, \( p \) is the probability of success on a single trial, and \( k \) is the number of successes. Given \( n = 9 \), \( p = 0.4 \), and \( x \leq 3 \), we need to find \( P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) \): 1. **Calculate \( P(X = 0) \)**: \[ P(X = 0) = \binom{9}{0} (0.4)^0 (0.6)^9 = 1 \cdot 1 \cdot 0.0060466176 = 0.0060 \] 2. **Calculate \( P(X = 1) \)**: \[ P(X = 1) = \binom{9}{1} (0.4)^1 (0.6)^8 = 9 \cdot 0.4 \cdot 0.0106162172 = 0.0382 \] 3. **Calculate \( P(X = 2) \)**: \[ P(X = 2) = \binom{9}{2} (0.4)^2 (0.6)^7 = 36 \cdot 0.16 \cdot 0.0173415296 = 0.0990 \] 4. **Calculate \( P(X = 3) \)**: \[ P(X = 3) = \binom{9}{3} (0.4)^3 (0.6)^6 = 84 \cdot 0.064 \cdot 0.0279936 = 0.1890 \] Now, sum these probabilities for \( x \leq 3 \): \[ P(X \leq 3) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) = 0.0060 + 0.0382 + 0.0990 + 0.1890 = 0.3322 \] So, the probability of \( x \leq 3 \) successes is \( \boxed{0.3322} \) (rounded to four decimal places).