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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 (Round to four decimal places as needed.)

Step 1 ：The problem is asking for the probability of getting 3 or fewer successes in a binomial experiment with 9 trials and a success probability of 0.6.

Step 2 ：The binomial probability formula is: \(P(X = k) = C(n, k) * (p^k) * ((1-p)^(n-k))\), where \(P(X = k)\) is the probability of k successes, \(C(n, k)\) is the number of combinations of n items taken k at a time, p is the probability of success on a single trial, n is the number of trials, and k is the number of successes.

Step 3 ：Since we want the probability of 3 or fewer successes, we need to calculate the sum of the probabilities of 0, 1, 2, and 3 successes.

Step 4 ：Given n = 9 and p = 0.6, we calculate the probabilities as follows: \(P_0 = 0.0002621440000000001\), \(P_1 = 0.0035389440000000013\), \(P_2 = 0.021233664000000006\), and \(P_3 = 0.07431782400000002\).

Step 5 ：Adding these probabilities together gives us \(P = 0.09935257600000003\).

Step 6 ：Rounding to four decimal places, the final answer is: The probability of getting 3 or fewer successes in a binomial experiment with 9 trials and a success probability of 0.6 is approximately \(\boxed{0.0994}\).