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Chapter 7: Problem 91

Design an experiment based on rolling a fair die for which there are exactly three outcomes with the same probabilities.

Short Answer

Expert verified
Design an experiment by rolling a fair die once and group the outcomes into three groups with equal probabilities: Group 1 (rolling a 1 or 2), Group 2 (rolling a 3 or 4), and Group 3 (rolling a 5 or 6). Each group has a probability of \(\frac{1}{3}\) as required.

Step by step solution

01

Identify all possible outcomes

List all the possible outcomes of rolling a fair die: \(1, 2, 3, 4, 5,\) and \(6\).
02

Combine outcomes in three groups

We need to group the outcomes in such a way that the probabilities for each group are equal. As there are six outcomes and three groups, there should be two outcomes in each group. We can represent these groups as follows: - Group 1: Outcome 1 or 2 - Group 2: Outcome 3 or 4 - Group 3: Outcome 5 or 6
03

Determine the probability for each group

Since there are six possible outcomes with each outcome having a probability of \(\frac{1}{6}\), and we combined two outcomes in each group, we find that the probability for each group is: - Group 1: Probability = \(\frac{1}{6} + \frac{1}{6} = \frac{1}{3}\) - Group 2: Probability = \(\frac{1}{6} + \frac{1}{6} = \frac{1}{3}\) - Group 3: Probability = \(\frac{1}{6} + \frac{1}{6} = \frac{1}{3}\)
04

Design the experiment

The experiment will consist of rolling a fair die once and observing the outcome to determine which group it belongs to. The experiment will have three possible outcomes: - Outcome 1: Rolling a \(1\) or \(2\) (Probability = \(\frac{1}{3}\)) - Outcome 2: Rolling a \(3\) or \(4\) (Probability = \(\frac{1}{3}\)) - Outcome 3: Rolling a \(5\) or \(6\) (Probability = \(\frac{1}{3}\)) The designed experiment meets the condition of having exactly three outcomes, each with the same probabilities of \(\frac{1}{3}\).

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