Problem question: Consider a ferry that can carry both cars and campers across a waterway. Each trip costs the owner approximately $10. The fee for cars is $3 and the fee for campers is $9. Let X and Y be the number of cars and campers, respectively on a given trip. Suppose the probabilities for different values of X and Yare given by the following table.

x y=0 y=1 y=2
0 0.01 0.01 0.03
1 0.03 0.08 0.07
2 0.03 0.06 0.06
3 0.07 0.07 0.13
4 0.12 0.04 0.03
5 0.08 0.07 0.01 

The revenue for the ferry is given by R = 3X 9Y. Find the possible values of R and the associated probabilities.

From the question, I know there are 18 possibile combinations of cars and campers. I'm stuck on my function for determining the possible outcomes of R.

combos = []

def problem_three():
  for x in range(0,5):
    for y in range(0,2):
      rev = (3*int(x)   9*int(y))
  combos.append(rev)
  return combos

revenue = problem_three()
print(revenue)

This code returns: [0, 9, 3, 12, 6, 15, 9, 18, 12, 21], but that doesn't doesn't have all the values I expected - what am I missing?

CodePudding user response：

range(0, n) is the same as range(n) and will give you 0 through n-1, not 0 through n!

combos = []

def problem_three():
  for x in range(5 1):             # fixed
    for y in range(2 1):           # fixed
      rev = (3*int(x)   9*int(y))
      combos.append(rev)           # indent
  return combos

revenue = problem_three()
print(revenue)

[0, 9, 18, 3, 12, 21, 6, 15, 24, 9, 18, 27, 12, 21, 30, 15, 24, 33