Check here for instructional material to complete this problem. Evaluate ${ }_{n} C_{x} p^{x}(1-p)^{n-x}$ for $n=4, p=0.3, x=3$. The answer is (Round to four decimal places as needed.)

Step 1 ：We are given the formula for binomial probability: ${ }_{n} C_{x} p^{x}(1-p)^{n-x}$, where ${ }_{n} C_{x}$ is the number of combinations of n items taken x at a time, $p^{x}$ is the probability of success raised to the power of the number of successes, and $(1-p)^{n-x}$ is the probability of failure raised to the power of the number of failures.

Step 2 ：We are asked to find the probability of having exactly 3 successes in 4 trials, where the probability of success on each trial is 0.3. So, we have $n=4$, $p=0.3$, and $x=3$.

Step 3 ：First, we calculate the 'n choose x' part, which is ${ }_{4} C_{3} = 4$.

Step 4 ：Next, we calculate $p^{x}$, which is $0.3^{3} = 0.027$.

Step 5 ：Then, we calculate $(1-p)^{n-x}$, which is $(1-0.3)^{4-3} = 0.7$.

Step 6 ：Finally, we multiply these three values together to get the result: $4 * 0.027 * 0.7 = 0.0756$.

Step 7 ：Final Answer: The value of ${ }_{n} C_{x} p^{x}(1-p)^{n-x}$ for $n=4, p=0.3, x=3$ is \(\boxed{0.0756}\).