To find the probability that at most 4 out of 6 randomly selected serves were successful, we need to add up the probabilities of having 0, 1, 2, 3, or 4 successful serves.
From the given table, the probabilities are as follows:
- The probability of 0 successful serves: [tex]\(0.0002\)[/tex]
- The probability of 1 successful serve: [tex]\(0.004\)[/tex]
- The probability of 2 successful serves: [tex]\(0.033\)[/tex]
- The probability of 3 successful serves: [tex]\(0.132\)[/tex]
- The probability of 4 successful serves: [tex]\(0.297\)[/tex]
The probability that at most 4 serves were successful is the sum of these individual probabilities:
[tex]\[
P(\text{at most 4 successes}) = P(0 \text{ successes}) + P(1 \text{ success}) + P(2 \text{ successes}) + P(3 \text{ successes}) + P(4 \text{ successes})
\][/tex]
Substitute the corresponding values from the table into the equation:
[tex]\[
P(\text{at most 4 successes}) = 0.0002 + 0.004 + 0.033 + 0.132 + 0.297
\][/tex]
Adding these probabilities together:
[tex]\[
0.0002 + 0.004 + 0.033 + 0.132 + 0.297 = 0.4662
\][/tex]
Therefore, the probability that at most 4 out of 6 randomly selected serves were successful is [tex]\(0.4662\)[/tex], which is approximately [tex]\(0.466\)[/tex] when rounded to three decimal places.
Hence, the correct answer is [tex]\(0.466\)[/tex].
