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In probability, we're often interested in finding how likely a certain event is to occur. When dealing with scenarios involving multiple outcomes, such as whether a phone line is occupied or not, we can use a mathematical framework to calculate these probabilities.
For scenario (a) in the problem, we're asked to determine the likelihood that exactly three out of ten phone calls will find the lines occupied. This situation is modeled using a binomial distribution, which is ideal for cases where there are two possible outcomes (like 'occupied' or 'not occupied').
The probability for any specific number of successes (in this case, occupied lines) is given by the formula: \[ P(X = k) = \binom{n}{k} p^k (1-p)^{n-k} \] where \( n \) is the number of trials (ten calls), \( k \) is the number of successful trials we are interested in (three occupied calls), and \( p \) is the probability of success on an individual trial (0.4 for an occupied line).
Understanding this calculation allows us to accurately determine the chance of specific outcomes in similar real-world scenarios.

Expectation, or expected value, gives us the average outcome of a probability event if we were to repeat the experiment many times. It tells us what we should anticipate, on average, over the long run.
For any binomial distribution, the expected number of successes can be easily calculated using the formula: \[ E(X) = n \times p \] where \( n \) is the total number of trials, and \( p \) is the probability of success in each trial.
In our exercise, we are determining how many calls out of ten are expected to find the phone lines occupied, which is calculated as \( 10 \times 0.4 = 4 \). Hence, on average, four calls will find the lines busy.
This concept is crucial for making informed decisions based on probabilistic events, helping predict future occurrences based on current data.

Independent events are those where the outcome of one event does not affect the outcome of another. In the context of our exercise, each call made to the airline reservation system is independent of the others.
This means that no matter the outcome of one call (occupied or not), it doesn’t change the probability of the outcome of the next call. This property simplifies probability calculations, as the likelihood of an event occurring remains constant across trials.
If events were not independent, we would have to consider conditional probabilities, which involve adjusting probabilities based on the outcome of previous events. Understanding independence helps us accurately apply the binomial distribution method without unnecessary complications.
In our exercise, assuming independence ensures that each call has the same 40% chance of finding an occupied line, consistent across all calls.

The Complement Rule in probability is a powerful tool that allows you to find the probability of an event occurring by instead calculating the probability of it not occurring, and then subtracting that value from 1. This is particularly useful for calculating events that are easier to consider by their opposite.
In the exercise, part (b) asks for the probability that at least one call finds the line not occupied. Rather than calculating the probability for numerous different combinations where the lines are not occupied, we can use the complement rule.
First, calculate the probability of the opposite event (that all 10 calls find the lines occupied), then subtract from 1 to find the probability of at least one unoccupied line. Mathematically, it looks like: \[ P( ext{at least one not occupied}) = 1 - P( ext{all occupied}) \] This calculation simplifies the process significantly, making problems that seem complex much more manageable and is a classic strategy in probability theory to deal with large sets of possible outcomes.