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Combinatorics is a branch of mathematics that deals with counting and arranging objects. It helps us understand how to count the number of ways an event can occur, given specific conditions. In the context of the exercise, combinatorics is used to determine how many different pairs of suspects the woman can choose from a lineup. Since she needs to pick 2 suspects out of a total of 5, we use a formula for combinations which is expressed as \({n \choose k}\) where \(n\) is the total number of items, and \(k\) is the number of items to choose.

• The formula for combinations, without repetition, is given by \( {n \choose k} = \frac{n!}{k!(n-k)!} \).
• Here, \(!\) symbolizes a factorial, which is the product of an integer and all the integers below it. For instance, \(5! = 5 \times 4 \times 3 \times 2 \times 1\).

Applying this formula, we find that the woman has 10 different possible pairs to choose from the 5 suspects, which frame our sample space later on.

The sample space in probability theory is the set of all possible outcomes of an experiment. It forms the foundation for determining probabilities and is crucial for analyzing any probabilistic scenario. In our exercise, the sample space consists of all different combinations of two suspects the woman could choose from a group of five.

• This includes every possible pair of suspects, considering there are no repetitions, and order does not matter.
• Since we calculated in the previous section that there are 10 ways to choose 2 suspects from 5, our sample space consists of 10 outcomes.

These outcomes cover every scenario, from picking both perpetrators to choosing two innocent bystanders. Understanding the sample space is essential as it sets the stage for identifying specific events within it.

Event identification involves characterizing specific outcomes within a sample space that meet certain criteria. An event can be any subset of the sample space. In our scenario, we are particularly interested in the event \(A\), where the woman makes at least one incorrect identification.

• Identifying conditions of event \(A\), the woman could either:
  • Choose one perpetrator and one innocent person (one incorrect identification).
  • Choose two innocent people (two incorrect identifications).
• To find these outcomes:
  • First, calculate pairs with one thief. There are two ways to choose one thief and three ways to choose one innocent, resulting in 6 combinations.
  • Second, calculate pairs with both suspects being innocent, choosing 2 from 3 innocents gives us 3 combinations.

In total, this means 9 outcomes make up event \(A\), where there are either one or two incorrect identifications. Event identification allows us to focus on the part of the sample space relevant to probability questions, like determining how likely the event is to occur.