Question

1. you randomly choose a shape from the set of unshaded shapes, and then randomly choose a shape from the set of shaded shapes. what is the probability that you will choose an unshaded cube, followed by a shaded cube?

Explanation:

Step1: Count unshaded shapes

There are 9 unshaded shapes, and 3 of them are cubes. So the probability of choosing an unshaded cube is $\frac{3}{9}$.

Step2: Count shaded shapes

There are 9 shaded shapes, and 2 of them are cubes. So the probability of choosing a shaded cube is $\frac{2}{9}$.

Step3: Calculate combined probability

Since these are independent - events, the probability of both events occurring is the product of their individual probabilities. So $P=\frac{3}{9}\times\frac{2}{9}=\frac{6}{81}=\frac{2}{27}$ (There seems to be an error in the provided options as the correct answer based on the steps is $\frac{2}{27}$). But if we assume there was a counting or problem - setup error and we recalculate:
If we assume the correct number of unshaded shapes is 9 and number of unshaded cubes is 3, and number of shaded shapes is 9 and number of shaded cubes is 2. The probability of choosing an unshaded cube first and then a shaded cube is $P=\frac{3}{9}\times\frac{2}{9}=\frac{2}{27}$. If we assume the problem was set up with different counts, say 10 unshaded shapes with 3 unshaded cubes and 9 shaded shapes with 2 shaded cubes, the probability is $\frac{3}{10}\times\frac{2}{9}=\frac{1}{15}$. If we assume 9 unshaded shapes with 3 unshaded cubes and 10 shaded shapes with 2 shaded cubes, $P = \frac{3}{9}\times\frac{2}{10}=\frac{1}{15}$. If we assume 10 unshaded shapes with 3 unshaded cubes and 10 shaded shapes with 2 shaded cubes, $P=\frac{3}{10}\times\frac{2}{10}=\frac{3}{50}$. Based on the most likely interpretation from the given problem setup with 9 unshaded and 9 shaded shapes:

Answer:

None of the provided options are correct. If we assume standard counting from the problem - setup description, the probability is $\frac{2}{27}$.