#1**+1 **

To start this problem, we should get a general idea of the probabilities of each product of the outcomes.

Let's make a table of the products.

1 | 3 | 8 | 8 | |
---|---|---|---|---|

1 | 1 | 3 | 8 | 8 |

3 | 3 | 9 | 24 | 24 |

8 | 8 | 24 | 64 | 64 |

8 | 8 | 24 | 64 | 64 |

The probability of each product then becomes:

\(P(1) = \frac{1}{16}, P(3) = \frac{2}{16}, P(8) = \frac{4}{16}, P(9) = \frac{1}{16}, P(24) = \frac{4}{16}, P(64) = \frac{4}{16}\)

For the decision of side dishes to be fair, we would want each dish to have a \(\frac{1}{4} \) or \(\frac{4}{16}\) probability of being prepared.

Since three products (8, 24, and 64) already have probabilities of \(\frac{4}{16}\), we would want those probabilities being assigned to only **one dish each**, and the other three products (1, 3, and 9) add up to \(\frac{4}{16}\) for the **last dish**.

To rephrase, product of 8 goes to one dish, product of 24 goes to another, product of 64 goes to another, and products 1, 3, and 9 go to the last dish.

**Answer B** satisfies this.

AnthraxMay 10, 2019