Question

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

Answer 1

Answer:  0.20

Step-by-step explanation:

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

The number of ways to select 3 dishes from 6 :

Total outcomes : $^6C_3=\dfrac{6!}{3!(6-3!)} \ \ [\because\ ^nC_r=\dfrac{n!}{r!(n-r)!}\ ]$                  

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: $^4C_1=\dfrac{4!}{1!(4-1)!)}=4$

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

$\dfrac{\text{Favorable outcomes}}{\text{Total outcomes}}\\\\=\dfrac{4}{20}=0.20$

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

Answer 2

Answer:

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

Step-by-step explanation: