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?
0.08889
0.17778
0.16667
0.33333

Answer 1

The answer on E2020 is D: 0.33333

Answer 2

Answer: 0.33333

Step-by-step explanation:

Given: Number of Vegetable side dishes = 3

Number of Fruit side dishes = 2

Number of Grain-based side dishes = 5

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

Now, the number of ways for choosing two side dishes from total 10 dishes will be :

$^{10}C_2=\frac{10!}{(10-2)!2!=45}$

Then the number of ways to randomly choose one vegetable and one grain-based side dish is given by :-

$^{3}C_1\times^{5}C_1=3\times5=15$

Thus, the probability of randomly choosing one vegetable and one grain-based side dish =$\frac{15}{45}=\frac{1}{3}=0.33333333...\approx0.33333$