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Probability calculation is a fundamental concept in statistics that helps us understand the likelihood of different outcomes in uncertain situations. In the context of the given exercise, we use probability to determine how likely it is for the phone lines to be occupied or not during several calls.
To solve problems with probabilities, one of the key tools we use is the binomial distribution. This distribution is applicable when there are a fixed number of independent trials, each with two possible outcomes: success or failure. In the case of the airline phone lines, a "success" can be considered when the lines are occupied.

• First, identify the total number of trials, which in this case is 10 calls.
• The probability of success for each trial, or the lines being occupied, is 0.4.

For instance, to find the probability that exactly three calls have occupied lines, we apply the binomial probability formula: \[ P(X = k) = \binom{n}{k} p^k (1-p)^{n-k} \]This formula helps calculate the likelihood of exactly 'k' successes in 'n' trials. For the exercise problem: calculate \[ P(X = 3) = \binom{10}{3} \times (0.4)^3 \times (0.6)^7 \].
This mathematical setup gives us the exact probability for three occupied calls, further showing practical applications of probability calculations.

Understanding independent events is crucial when working with probabilities. In the context of our problem, each phone call is an independent event. This means that the outcome of one call does not influence the outcome of another.
This independence is essential for the use of the binomial distribution. Each call to the airline system maintains its own probability of being either occupied or not occupied. Cross-call relationship doesn't exist, ensuring consistent application across all trials.

• An independent event in probability means the result of any single trial does not affect others.
• When computing the probability of multiple outcomes, like all calls being occupied, we multiply the probabilities of each independent event.

A practical implication of independence in this exercise is seen when calculating the likelihood that at least one of 10 calls is not occupied. We first need to understand what happens if they were all occupied:\[ P(X = 10) = p^{10} \]Now, find its complement:\[ P(\text{at least 1 not occupied}) = 1 - P(\text{all occupied}) \].This highlights the role of independent events in achieving accurate probability outcomes.

Expectation in probability provides a way to measure the average outcome of a random event. For a binomial distribution, like the one in this exercise with 10 calls, we can calculate the expected number of successes rather efficiently.
The formula for expectation in a binomial distribution is:\[ E(X) = n \times p \]Here, 'n' represents the number of trials, and 'p' stands for the probability of success. Applying to our scenario:\[ E(X) = 10 \times 0.4 = 4 \]

• This calculation indicates that, on average, we can expect four calls to be occupied out of the ten tries.
• Expectation provides the central value to anticipate; it sums up the most probable outcome over many instances.

Expectation is particularly useful in risk assessments and predicting outcomes in real-world scenarios, like forecasting the number of services needed based on a predictable number of busy lines. This bridges theoretical statistics with everyday practical application.