Questions: Bob is asked to construct a probability model for rolling a pair of fair dice. He lists the outcomes as 2,3,4,5,6,7,8,9,10,11,12. Because there are 11 outcomes, he reasoned, the probability of rolling a five must be 1/11. What is wrong with Bob's reasoning? Choose the correct answer below. A. The probability of an event is greater than 1. B. The sum of the probabilities of all outcomes does not equal 1. C. The experiment does not have equally likely outcomes. D. The probability of an event is less than 0. E. There is nothing wrong with Bob's reasoning.

Solution Steps

To determine what is wrong with Bob's reasoning, we need to consider the nature of rolling two dice. Each die has 6 faces, and the total number of outcomes when rolling two dice is 6 * 6 = 36. Bob's reasoning assumes that each sum (from 2 to 12) is equally likely, but this is incorrect because different sums have different numbers of combinations that can result in them. For example, there is only one way to roll a 2 (1+1), but there are multiple ways to roll a 7 (1+6, 2+5, 3+4, etc.). Therefore, the outcomes are not equally likely, which makes option C the correct answer.

When rolling two fair dice, the total number of outcomes is given by \( 6 \times 6 = 36 \).

The possible sums when rolling two dice range from \( 2 \) to \( 12 \). The frequency of each sum is as follows:

• \( 2 \): \( 1 \) way (1+1)
• \( 3 \): \( 2 \) ways (1+2, 2+1)
• \( 4 \): \( 3 \) ways (1+3, 2+2, 3+1)
• \( 5 \): \( 4 \) ways (1+4, 2+3, 3+2, 4+1)
• \( 6 \): \( 5 \) ways (1+5, 2+4, 3+3, 4+2, 5+1)
• \( 7 \): \( 6 \) ways (1+6, 2+5, 3+4, 4+3, 5+2, 6+1)
• \( 8 \): \( 5 \) ways (2+6, 3+5, 4+4, 5+3, 6+2)
• \( 9 \): \( 4 \) ways (3+6, 4+5, 5+4, 6+3)
• \( 10 \): \( 3 \) ways (4+6, 5+5, 6+4)
• \( 11 \): \( 2 \) ways (5+6, 6+5)
• \( 12 \): \( 1 \) way (6+6)

The probability of rolling a specific sum \( x \) is given by: \[ P(x) = \frac{\text{Number of ways to achieve } x}{\text{Total outcomes}} = \frac{\text{Frequency of } x}{36} \] For example, the probability of rolling a \( 5 \) is: \[ P(5) = \frac{4}{36} = \frac{1}{9} \]

Bob incorrectly assumed that there are \( 11 \) equally likely outcomes (the sums from \( 2 \) to \( 12 \)). However, the outcomes are not equally likely, as demonstrated by the varying frequencies of each sum. Therefore, the correct answer to the question is that the experiment does not have equally likely outcomes.

Final Answer