Three cents is the most Ms. Hernandez might have to spend to get both her twins the same colored gumballs if there are only white and red gumballs. This is because for the first penny she uses there is a 50% Ms. Hernandez can get a red gumball and a 50% chance she can get a red one. For the second and third she has the same chances. The chart below shows all the possible combinations of gumballs Ms. Hernandez could have gotten. PenniesColor 1st PennyRed 2nd PennyWhite 3rd Penny Red st PennyWhite 2nd PennyRed 3rd PennyWhite 2. The next day Ms. Hernandez and her twins pass another gumball machine with three colors, red, white, and blue and again her twins want the same color. The most Ms. Hernandez might have to spend is 4 cents. This is because she could get the following: PenniesColor 1stRed 2ndWhite 3rdBlue 4thWhite 3. Seven cents is the most Mr. Hodges might have to spend to get his triplets the same color gumballs at the same three-color gumball machine as Ms.

Hernandez. This is because he could get the following: PenniesColors 1st pennyBlue 2nd pennyRed 3rd pennyWhite 4th pennyWhite 5th pennyBlue 6th penny Red 7th pennyBlue 4. The next day Mr. Hodges passes a two-color (red and white) gumball machine with his triplets again, they each want the same color. The most Mr. Hodges would have to spend is 5 cents. This is because he can get the following: PenniesColor 1st pennyRed 2nd pennyWhite 3rd pennyWhite 4th pennyRed 5th pennywhite . The formula I found to solve these problems is: [(# of colors)(of kids)]- [(#of colors)-1]= how much money they need to spend. Ex of formula for question #1 is: [(2 colors)(2 kids)]-[(2 colors)-1]= 3 cents Ex of formula for question #2 is: [(3 colors)(2 kids)]- [(3 colors)-1]= 4 cents Ex of formula for question #3 is: [(3 colors)(3kids)]-[(3 colors)-1]= 7 cents Ex of formula for question #4 is: [(2 colors)(3 kids)]-[(2 colors)-1]= 5 cents

I figured this formula out by writing out how kids and colors in each problem and then the maximum amount spent after I found each answer. Ex: Problem 1: 2 kids, 2 colors, max spent: 4. I knew that two this needed to be multiplied and this needed to be subtracted so I did a guess and check until I found the answer.