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 find the probability of \( x \leq 3 \) successes in a binomial probability experiment with \( n = 9 \) trials and a success probability of \( p = 0.4 \), we can use the binomial probability formula: \[ P(X = k) = \binom{n}{k} p^k (1-p)^{n-k} \] We need to calculate the probabilities for \( x = 0, 1, 2, \) and \( 3 \) and then sum these probabilities together. 1. **For \( x = 0 \)**: \[ P(X = 0) = \binom{9}{0} (0.4)^0 (0.6)^9 = 1 \cdot 1 \cdot 0.0060466176 \approx 0.0061 \] 2. **For \( x = 1 \)**: \[ P(X = 1) = \binom{9}{1} (0.4)^1 (0.6)^8 = 9 \cdot 0.4 \cdot 0.010616 \approx 0.0387 \] 3. **For \( x = 2 \)**: \[ P(X = 2) = \binom{9}{2} (0.4)^2 (0.6)^7 = 36 \cdot 0.16 \cdot 0.017576 \approx 0.1015 \] 4. **For \( x = 3 \)**: \[ P(X = 3) = \binom{9}{3} (0.4)^3 (0.6)^6 = 84 \cdot 0.064 \cdot 0.0279936 \approx 0.1884 \] Now, summing these probabilities: \[ P(X \leq 3) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) \approx 0.0061 + 0.0387 + 0.1015 + 0.1884 \approx 0.3347 \] Therefore, the probability of \( x \leq 3 \) successes is approximately \( \boxed{0.3347} \).