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.)

We want the probability of at most 3 successes in 9 trials when the success probability is \( p = 0.4 \). This probability is given by \[ P(X \leq 3) = \sum_{x=0}^{3} \binom{9}{x} (0.4)^x (0.6)^{9-x}. \] We now compute each term step by step. 1. For \( x = 0 \): \[ \binom{9}{0} (0.4)^0 (0.6)^9 = 1 \cdot 1 \cdot (0.6)^9. \] Calculating \[ (0.6)^9 \approx 0.0100777. \] 2. For \( x = 1 \): \[ \binom{9}{1} (0.4)^1 (0.6)^8 = 9 \cdot 0.4 \cdot (0.6)^8. \] Since \[ (0.6)^8 \approx \frac{(0.6)^9}{0.6} \approx \frac{0.0100777}{0.6} \approx 0.0167962, \] we have \[ 9 \cdot 0.4 \cdot 0.0167962 \approx 0.0604662. \] 3. For \( x = 2 \): \[ \binom{9}{2} (0.4)^2 (0.6)^7 = 36 \cdot 0.16 \cdot (0.6)^7. \] Estimating \[ (0.6)^7 \approx \frac{(0.6)^9}{(0.6)^2} \approx \frac{0.0100777}{0.36} \approx 0.0279942, \] we get \[ 36 \cdot 0.16 \cdot 0.0279942 \approx 0.161071. \] 4. For \( x = 3 \): \[ \binom{9}{3} (0.4)^3 (0.6)^6 = 84 \cdot 0.064 \cdot (0.6)^6. \] Noting that \[ (0.6)^6 \approx 0.046656, \] the calculation gives \[ 84 \cdot 0.064 \cdot 0.046656 \approx 0.2509056. \] Now, sum the values: \[ P(X \leq 3) \approx 0.0100777 + 0.0604662 + 0.161071 + 0.2509056 \approx 0.4825205. \] Rounded to four decimal places, the probability is \[ \boxed{0.4825}. \]