Question

A cafeteria manager can choose from among six side dishes for the lunch menu: applesauce, broccoli, corn, dumplings, egg

rolls, or French fries. He uses a computer program to randomly select three dishes for Monday's lunch.
What is the theoretical probability that applesauce and broccoli will both be offered on Monday?​

Answer 1

20% it would be 10% if there was 10 items. 5 items mean every item has 20%.

Explanation:

Answer 2

Answer: 0.20

Explanation:

Given : The number of side dishes for the lunch menu =6

The number of ways to select 3 dishes from 6 :

Total outcomes :
`^6C_3=(6!)/(3!(6-3!)) \ \ [\because\ ^nC_r=(n!)/(r!(n-r)!)\ ]`

If applesauce and broccoli is already selected , then we need to select only one dish out of remaining 4 dishes.

Number of ways to select 1 dish from 4 :

Favorable outcomes:
`^4C_1=(4!)/(1!(4-1)!))=4`

Now, the theoretical probability that applesauce and broccoli will both be offered on Monday :-

`\frac{\text{Favorable outcomes}}{\text{Total outcomes}}\\\\=(4)/(20)=0.20`

Hence, the theoretical probability that applesauce and broccoli will both be offered on Monday = 0.20