Question 206164
To find the answer to this problem, you first have to make a list of all the possible outcomes. Let's say Bob's meal is A, Mary's meal is B, and Jen's meal is C. The 6 possible outcomes for the order in which the waiter hands out their meals are as follows:


ABC
ACB
BAC
BCA
CAB
CBA


If we let "ABC" represent the outcome where all 3 diners receive the correct meal, we can calculate the probability for the other outcomes (nobody gets the right meal, 1/3 get the correct meal, 2/3 get the correct meal).


ABC ... (everyone gets the correct meal)
ACB ... (only Bob gets the correct meal)
BAC ... (only Jen gets the correct meal)
BCA ... (nobody gets the correct meal)
CAB ... (nobody gets the correct meal)
CBA ... (only Mary gets the correct meal)


As you can see, it's not possible for only 2 people to receive the correct meal (if someone has the wrong meal then that means they received someone else's meal...making at least 2/3 meals wrong).


We were asked to calculate the probability that:
A. No diner gets the correct meal (2/6 = 33.3%)
B. Exactly one diner gets the correct meal (3/6 = 50%)
C. Exactly two diners get the correct meal (0/6 = 0.00%)
D. All diners get the correct meal (1/6 = 16.7%)