Question

A restaurant offers three vegetable side dishes, two fruit side dishes, and five grain-based side dishes. Each entree comes with two side dishes.
What is the approximate probability of randomly choosing one vegetable and one grain-based side dish?
a.0.08889
b.0.17778
c.0.16667
d.0.33333

Answer 1

The approximate probability of randomly choosing one vegetable and one grain-based side dish is:   d.  0.33333

Explanation

Vegetable side dishes = 3,  Fruit side dishes = 2 and Grain-based side dishes = 5

So, total number of side dishes $= 3+2+5= 10$

Number of ways for choosing two side dishes from total 10 dishes will be : $^1^0C_{2}$

We need to randomly choose one vegetable and one grain-based side dish, so the number of ways $= ^3C_{1} * ^5C_{1}$

Thus, the probability $= \frac{^3C_{1} * ^5C_{1}}{^1^0C_{2}} = \frac{3*5}{45}= \frac{15}{45}= \frac{1}{3}= 0.3333...$