+0

# Algebra 2

0
204
1

a spinner has 4 equally sized sectors that are numbered 1, 3,8 and 8. The spinner is spun twice and the product of the outcomes is found.

A fair decision is to be made about which one of 4 side dishes will be prepared.

The side dishes options are creamed corn, scalloped potatoes, rice pilaf, and coleslaw.

which description accurately explains how a fair decision can be made in this situation?

A- If the product is 64, creamed corn will be prepared. If the product is 24, scalloped potatoes will be prepared. If the product is 9, coleslaw will be prepared. If the product is 1,3, or 8 rice Pilaf will be prepared.

B- if the  product is 64 creamed corn will be prepared.  If the product is 24 scalloped potatoes will be prepared.

If the product is 8 coleslaw will be prepared.  If the product is 1,3, or 9 rice pilaf will be prepared.

C-  if the product is 64 creamed corn will be prepared. If the product is 9 scallop potatoes will be prepared.  If the product is 8,  coleslaw will be will be  prepared.  If the product is 1, 3 or 24 rice pilaf will be prepared.

D.  If the product is 3 creamed corn will be prepared. If the product is 24 scalloped Potatoes will be prepared. If the product is 8, coleslaw will be prepared.  If the product is 1, 9 or 64 rice pilaf will be prepared.

May 10, 2019

#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.