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 mondays lunch. what is the theoretical probability that applesauce and broccoli will both be offered on monday? 10% 20% 50% 80%

Explanation:

1. First, find the total number of ways to select 3 dishes out of 6:

• Use the combination formula \(C(n,r)=\frac{n!}{r!(n - r)!}\), where \(n = 6\) and \(r=3\).
• \(C(6,3)=\frac{6!}{3!(6 - 3)!}=\frac{6!}{3!3!}=\frac{6\times5\times4}{3\times2\times1}=20\).

1. Then, find the number of ways to select apples - auce and broccoli along with one - more dish:

• Since apples - auce and broccoli are already selected, we need to choose 1 more dish from the remaining \(6-2 = 4\) dishes.
• Using the combination formula with \(n = 4\) and \(r = 1\), \(C(4,1)=\frac{4!}{1!(4 - 1)!}=\frac{4!}{1!3!}=4\).

1. Calculate the probability:

• The probability \(P\) that apples - auce and broccoli are both selected is the number of favorable outcomes divided by the number of total outcomes.
• \(P=\frac{C(4,1)}{C(6,3)}=\frac{4}{20}=0.2 = 20\%\).

Answer:

20%