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In binomial probability, the term 'probability of success' refers to the likelihood of a single trial resulting in a desired outcome. This is denoted by the symbol \( p \). In the context of our crystal goblet problem, the success is defined as selecting a goblet with a cosmetic flaw. The company has determined that each goblet has a \( 10\% \) chance of having a flaw. Therefore, the probability of success \( p \) is \( 0.10 \).

This probability remains constant for every goblet you examine since each one is independent of the others. Therefore, regardless of how many goblets you select, the chance of finding one with a defect is always \( 0.10 \). This uniform probability is foundational to solving binomial probability problems. The real challenge comes when you need to calculate the probability of multiple successes across several trials.

Understanding this concept allows us to use the binomial formula effectively to find probabilities for different scenarios, such as exactly one goblet being flawed among six or finding at least two flawed goblets.

Binomial trials refer to experiments that have a fixed number of repeated trials with two possible outcomes for each trial: success or failure. In our exercise involving goblets, each time we select a goblet, either it will be classified as a 'second' (success) if it has a flaw, or it will not (failure).

When solving for this type of problem, we often employ the binomial probability formula, which helps determine the probability of a certain number of successes in a given set of trials. The formula is: \[P(X = k) = \binom{n}{k} p^k (1-p)^{n-k}\]where:

• \( n \) is the total number of trials (e.g., 6 goblets selected)
• \( k \) is the number of successes we are interested in (e.g., exactly one flawed goblet)
• \( p \) is the probability of success on a single trial
• \( 1-p \) is the probability of failure

The binomial coefficient \( \binom{n}{k} \) calculates the number of ways to pick \( k \) successes out of \( n \) trials. This concept helps us understand how to approach varying situations, such as in part a, where we calculate the probability of exactly one goblet being flawed.

In probability, independent events are events where the occurrence of one does not affect the occurrence of another. This is crucial in our goblet problem because the state of one goblet (whether flawed or not) does not impact the state of another. Each selection is an independent event.

For example, if you have already selected one flawed goblet, the probability of the next having a flaw remains \( 0.10 \). This independence is fundamental to the application of the binomial distribution. It allows each trial to be considered separately, maintaining a consistent probability of success throughout.

This property of independence is important when combining probabilities for multiple scenarios, such as calculating the probability of several different outcomes. It simplifies the computation as each selection is a fresh, independent event. Recognizing this allows for the combination of probabilities to find the answer to parts like b and c, where you calculate the probability of multiple successes in sequential trials.