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A restaurant has four bottles of a certain wine in stock. Unbeknownst to the wine steward, two of these bottles (Bottles 1 and 2 ) are bad. Suppose that two bottles are ordered, and let \(x\) be the number of good bottles among these two. a. One possible experimental outcome is \((1,2)\) (Bottles 1 and 2 are the ones selected) and another is \((2,4)\). List all possible outcomes. b. Assuming that the two bottles are randomly selected from among the four, what is the probability of each outcome in Part (a)? c. The value of \(x\) for the \((1,2)\) outcome is 0 (neither selected bottle is good), and \(x=1\) for the outcome \((2,4)\). Determine the \(x\) value for each possible outcome. Then use the probabilities in Part (b) to determine the probability distribution of \(x\).

Step by step solution

01

Identification of all possible outcomes

Using combinations, we can list all possible outcomes when two bottles are selected out of four. We denote the bottles as 1, 2, 3, and 4, where 1 and 2 are bad bottles and 3 and 4 are good bottles. The possible outcomes are: (1,2), (1,3), (1,4), (2,3), (2,4), and (3,4).

02

Calculation of the probability of each outcome

Since the bottles are randomly selected, each pair of bottles has an equal chance of being selected. The total number of combinations of 4 items taken 2 at a time is \( \binom{4}{2} = 6\). So, the probability of each outcome is \(\frac{1}{6}\).

03

Determination of the x value for each possible outcome

We let \(x\) be the number of good bottles among the selected two. Thus, \(x = 0\) for outcome (1,2); \(x = 1\) for outcomes (1,3), (1,4), (2,3), and (2,4); and \(x = 2\) for outcome (3,4).

04

Establish the probability distribution of x

Based on the above steps, we already know the outcome for each value of \(x\) and the corresponding probabilities. We can present this in a tabular probability distribution: For \(x = 0\) there is 1 outcome, (1,2), so the probability is \(\frac{1}{6}\); for \(x = 1\) there are 4 outcomes, so the probability is \(\frac{4}{6}\); for \(x = 2\) there is 1 outcome, (3,4), so the probability is \(\frac{1}{6}\).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Combinatorics

Combinatorics is the branch of mathematics that deals with counting combinations and permutations. It helps us determine how many ways we can choose a set of items from a larger pool. For example, when selecting two bottles from four, we are dealing with combinations because the order of selection does not matter.
To find the number of possible combinations of two bottles out of four, we can use the combination formula: \[ \binom{n}{r} = \frac{n!}{r!(n-r)!} \]Where \(n\) is the total number of items and \(r\) is the number of items to be chosen. In our case, \(n = 4\) and \(r = 2\), which gives us 6 possible outcomes:

• (1,2)
• (1,3)
• (1,4)
• (2,3)
• (2,4)
• (3,4)

Combinatorics thus provides a systematic way of listing all possible outcomes when random selection is involved.

Random Selection

Random selection is a process where every item in a set has an equal chance of being chosen. In our wine bottle example, the selection of two bottles out of four is done randomly.
This means each pair has an equal probability of being selected. Understanding random selection is crucial for analyzing experimental outcomes in probability.

• It ensures no bias in choosing the bottles.
• Facilitates objective calculation of probabilities.

Random selection leads to fair representation of all possibilities, allowing us to use probability theory accurately to predict outcomes.

Probability Calculation

Probability calculation involves determining the likelihood of an event occurring. In our scenario, we calculate the probability of selecting each pair of bottles. Since selection is random, each of the 6 combinations has the same chance.
To find the probability of any specific outcome, we apply the formula:\[ P(\text{outcome}) = \frac{1}{\text{total outcomes}} \]For our case, there are 6 possible combinations, so each has a probability of \(\frac{1}{6}\).

• Each outcome has an equal probability due to randomness.
• Total probability across all outcomes needs to sum up to 1.

By calculating these probabilities, we can better understand the distribution and make valid predictions about the possible outcomes.

Probability Distribution Table

A probability distribution table displays all possible values of a random variable and their respective probabilities. For instance, in the wine bottle problem, we want to know the number of good bottles (\(x\)) in our selection of two bottles.
The possible values of \(x\) depend on whether 0, 1, or 2 of the selected bottles are good.

• \(x = 0\): Probability is \(\frac{1}{6}\) (outcome (1,2))
• \(x = 1\): Probability is \(\frac{4}{6}\) (outcomes: (1,3), (1,4), (2,3), (2,4))
• \(x = 2\): Probability is \(\frac{1}{6}\) (outcome (3,4))

The table summarizes this probability distribution, providing a visual and numerical way to understand the likelihood of each scenario occurring. This is a helpful tool for interpreting experimental results and predicting future events.