a cafeteria manager can choose from among six side dishes for the lunch menu: applesauce, broccoli, corn…

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%

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%

Answer

Explanation:

Step1: Calculate total combinations

The number of ways to choose 3 dishes out of 6 is given by the combination formula $C(n,r)=\frac{n!}{r!(n - r)!}$, where $n = 6$ and $r=3$. So $C(6,3)=\frac{6!}{3!(6 - 3)!}=\frac{6\times5\times4}{3\times2\times1}=20$.

Step2: Calculate favorable combinations

If applesauce and broccoli are already chosen, we need to choose 1 more dish out of the remaining 4. Using the combination formula with $n = 4$ and $r = 1$, we get $C(4,1)=\frac{4!}{1!(4 - 1)!}=\frac{4}{1}=4$.

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%