a restaurant offers three vegetable side dishes, two fruit side dishes, and five grain - based side dishes…

Explanation:

Step1: Calculate total number of side - dishes

There are 3 vegetable + 2 fruit+ 5 grain - based = 10 side - dishes.

Step2: Calculate number of ways to choose 2 side - dishes

The number of ways to choose 2 side - dishes out of 10 is given by the combination formula $C(n,r)=\frac{n!}{r!(n - r)!}$, where $n = 10$ and $r=2$. So $C(10,2)=\frac{10!}{2!(10 - 2)!}=\frac{10\times9}{2\times1}=45$.

Step3: Calculate number of ways to choose 1 vegetable and 1 grain - based side - dish

The number of ways to choose 1 vegetable out of 3 is $C(3,1)=\frac{3!}{1!(3 - 1)!}=3$, and the number of ways to choose 1 grain - based side - dish out of 5 is $C(5,1)=\frac{5!}{1!(5 - 1)!}=5$. By the multiplication principle, the number of ways to choose 1 vegetable and 1 grain - based side - dish is $C(3,1)\times C(5,1)=3\times5 = 15$.

Step4: Calculate the probability

The probability $P$ of choosing 1 vegetable and 1 grain - based side - dish is $P=\frac{15}{45}=\frac{1}{3}\approx0.33333$.

Answer:

0.33333