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.0.08889b.0.17778c.0.16667d.0.33333

The approximate probability of randomly choosing one vegetable and one grain-based side dish is:   d.  0.33333ExplanationVegetable side dishes = 3,  Fruit side dishes = 2 and Grain-based side dishes = 5 So, total number of side dishes [tex]= 3+2+5= 10[/tex]Number of ways for choosing two side dishes from total 10 dishes will be : [tex]^1^0C_{2}[/tex]We need to randomly choose one vegetable and one grain-based side dish, so the number of ways [tex]= ^3C_{1} * ^5C_{1}[/tex]Thus, the probability [tex]= \frac{^3C_{1} * ^5C_{1}}{^1^0C_{2}} = \frac{3*5}{45}= \frac{15}{45}= \frac{1}{3}= 0.3333...[/tex]