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? 10% 20% 50% 80%

Explanation:

Step1: Calculate total combinations

The formula for combinations is $C(n,r)=\frac{n!}{r!(n - r)!}$, where $n = 6$ (total side - dishes) and $r=3$ (number of dishes to be selected). So, $C(6,3)=\frac{6!}{3!(6 - 3)!}=\frac{6!}{3!3!}=\frac{6\times5\times4}{3\times2\times1}=20$.

Step2: Calculate favorable combinations

If applesauce and broccoli are already selected, we need to select 1 more dish from the remaining 4 dishes. Using the combination formula with $n = 4$ and $r = 1$, we get $C(4,1)=\frac{4!}{1!(4 - 1)!}=\frac{4!}{1!3!}=4$.

Step3: Calculate probability

The probability $P$ is the number of favorable outcomes divided by the number of total outcomes. So, $P=\frac{4}{20}=0.2 = 20\%$.

Answer:

20%