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.)

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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

$$
\boxed{0.4825}.
$$
The probability of at most 3 successes is approximately 0.4825.

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