C. Each trial depends on the previous trial. D. The experiment is performed a fixed number of times. E. The probability of success is different for each trial of the experiment. G. The trials are independent. H. The probability of success is the same for each trial of the experiment. (b) Using the binomial distribution, determine the values of $$ n $$ and $$ p $$. $$ n=17 $$ (Type an integer or a decimal. Do not round.) $$ p=0.8 $$ (Type an integer or a decimal. Do not round.) (c) Using the binomial distribution, the probability that exactly 11 flights are on time is (Round to four decimal places as needed.)

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To solve this problem, we need to determine the values of $$ n $$ and $$ p $$ using the binomial distribution and then calculate the probability that exactly 11 flights are on time.

Given:
- The experiment is performed a fixed number of times.
- The probability of success is the same for each trial of the experiment.
- The probability of success is 0.8.

From the given information:
- The probability of success ($$ p $$) is 0.8.
- The number of trials ($$ n $$) is 17.

Now, we can calculate the probability that exactly 11 flights are on time using the binomial distribution formula:

$$ P(X = k) = \binom{n}{k} p^k (1-p)^{n-k} $$

where:
- $$ n $$ is the number of trials,
- $$ k $$ is the number of successful trials,
- $$ p $$ is the probability of success.

Substitute the given values into the formula:

$$ P(X = 11) = \binom{17}{11} (0.8)^{11} (1-0.8)^{17-11} $$

Now, we can calculate the probability.
Calculate the value by following steps:
- step0: Calculate:
  $$ { }_{17}C_{11}\times 0.8^{11}\left(1-0.8\right)^{17-11}$$
- step1: Subtract the numbers:
  $$ { }_{17}C_{11}\times 0.8^{11}\times 0.2^{17-11}$$
- step2: Subtract the numbers:
  $$ { }_{17}C_{11}\times 0.8^{11}\times 0.2^{6}$$
- step3: Expand the expression:
  $$\frac{17!}{11!\times \left(17-11\right)!}\times 0.8^{11}\times 0.2^{6}$$
- step4: Subtract the numbers:
  $$\frac{17!}{11!\times 6!}\times 0.8^{11}\times 0.2^{6}$$
- step5: Reduce the fraction:
  $$\frac{17\times 16\times 15\times 14\times 13\times 12}{6!}\times 0.8^{11}\times 0.2^{6}$$
- step6: Multiply the terms:
  $$\frac{8910720}{6!}\times 0.8^{11}\times 0.2^{6}$$
- step7: Divide the terms:
  $$12376\times 0.8^{11}\times 0.2^{6}$$
- step8: Convert the expressions:
  $$12376\left(\frac{4}{5}\right)^{11}\times 0.2^{6}$$
- step9: Convert the expressions:
  $$12376\left(\frac{4}{5}\right)^{11}\left(\frac{1}{5}\right)^{6}$$
- step10: Simplify:
  $$12376\times \frac{4^{11}}{5^{11}}\times \left(\frac{1}{5}\right)^{6}$$
- step11: Multiply the terms:
  $$\frac{12376\times 4^{11}}{5^{11}}\times \left(\frac{1}{5}\right)^{6}$$
- step12: Evaluate the power:
  $$\frac{12376\times 4^{11}}{5^{11}}\times \frac{1}{5^{6}}$$
- step13: Multiply the fractions:
  $$\frac{12376\times 4^{11}}{5^{11}\times 5^{6}}$$
- step14: Multiply:
  $$\frac{12376\times 4^{11}}{5^{17}}$$
The probability that exactly 11 flights are on time is approximately 0.0680 (rounded to four decimal places).
$$ n = 17 $$, $$ p = 0.8 $$, and the probability that exactly 11 flights are on time is approximately 0.0680.

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