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

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?\n0.08889\n0.17778\n0.16667\n0.33333
Answer
Explanation:
Step1: Calculate total number of side - dishes
There are 3 vegetable side - dishes, 2 fruit side - dishes, and 5 grain - based side - dishes. So the total number of side - dishes is $3 + 2+5=10$.
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 side - dish out of 3 is $C(3,1)=\frac{3!}{1!(3 - 1)!}=3$. 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 $3\times5 = 15$.
Step4: Calculate the probability
The probability $P$ of choosing 1 vegetable and 1 grain - based side - dish is the number of favorable outcomes divided by the number of total outcomes. So $P=\frac{15}{45}=\frac{1}{3}\approx0.33333$.
Answer:
0.33333