Question

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 \(x\).

Short answer

Possible outcomes are: (1,2), (1,3), (1,4), (2,3), (2,4), (3,4), each with a probability of \(\frac{1}{6}\). The probability distribution of \(x\) (number of good bottles among the selected) is: P(\(x = 0\)) = \(\frac{1}{6}\), P(\(x = 1\)) = \(\frac{4}{6}\), P(\(x = 2\)) = \(\frac{1}{6}\).

Step-by-step solution

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).

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 \(\frac{1}{6}\).

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).

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 \(\frac{1}{6}\).