math pow: A Sticky Gum Problem
Problem Statement
For this problem, we were given a situation that involved the use of probability. With the information given, we know parents with their children walk by gum ball machines and the children want gum balls that are same color to one another. This doesn't hinder the parents since each gum ball costs a penny, and with this the parents want to know the maximum amount of money they would have to spend before there is a definite match. The factors that we need to take into consideration to find the maximum cost would be the number of children and the different types of gum balls that they could possibly get (i.e. red, blue, green, etc.).
For this problem, we were given a situation that involved the use of probability. With the information given, we know parents with their children walk by gum ball machines and the children want gum balls that are same color to one another. This doesn't hinder the parents since each gum ball costs a penny, and with this the parents want to know the maximum amount of money they would have to spend before there is a definite match. The factors that we need to take into consideration to find the maximum cost would be the number of children and the different types of gum balls that they could possibly get (i.e. red, blue, green, etc.).
Process Description
The first question on this weeks POW asked why a parent had to spend at most three pennies for her twin daughters get the same colored gum ball when there are two different colored gum balls as options. I figured that since there were only two options, if the gum balls weren't the same color on the second try, it had to be on the third time. The colors of the gum balls were red and white, and on the first and second try she got red and then white, the third one had to be red or white. Which meant that the third one would match with either the first try or the second try.
The second question on this weeks POW had the same parent, with the same number of children, but with three different gum ball options. We had to find out the maximum amount of money she had to spend, and I figured that the most she would have to spend was four cents. This is so because taking into consideration that there are three different options, if on the first three tries each gum ball was a different color, and there was only three different colors, the fourth try had to be one of the previously picked colors.
The third question on this weeks POW asked another parent with three children the most amount of pennies he would need to spend if there were three different gum ball color options. The answer that I had come up with was seven pennies, or seven tries, because the possibility of getting two of the same color three time is, even though unlikely, still possible. Taking that into consideration, that in the tries to get that far, would count as six tries, one more try and each of the triplets would have the same colored gum ball.
Using the knowledge that I had, I created a table to try and calculate a formula to follow my thought process. I have two different variable occurring with the different types of gum balls and the number of children. I proceeded to continue this in a chronological process, with the number of children increasing slightly to determine how much money would be needed to spend for four and five types of gum balls with a various amount of children. With this, I created a rough draft of a formula, which states that the number of children multiplied by the different types of gum balls subtracted by the expression parenthesis the different types of gum balls subtracted by one equals the maximum amount of coins needed to get all of the children the same color gum ball. Children*Gum-(Gum-1) = Maximum Cost. On the picture below, it shows the same formula, just shown in a more advanced table, with each of the components of the problem represented by a variable.
The first question on this weeks POW asked why a parent had to spend at most three pennies for her twin daughters get the same colored gum ball when there are two different colored gum balls as options. I figured that since there were only two options, if the gum balls weren't the same color on the second try, it had to be on the third time. The colors of the gum balls were red and white, and on the first and second try she got red and then white, the third one had to be red or white. Which meant that the third one would match with either the first try or the second try.
The second question on this weeks POW had the same parent, with the same number of children, but with three different gum ball options. We had to find out the maximum amount of money she had to spend, and I figured that the most she would have to spend was four cents. This is so because taking into consideration that there are three different options, if on the first three tries each gum ball was a different color, and there was only three different colors, the fourth try had to be one of the previously picked colors.
The third question on this weeks POW asked another parent with three children the most amount of pennies he would need to spend if there were three different gum ball color options. The answer that I had come up with was seven pennies, or seven tries, because the possibility of getting two of the same color three time is, even though unlikely, still possible. Taking that into consideration, that in the tries to get that far, would count as six tries, one more try and each of the triplets would have the same colored gum ball.
Using the knowledge that I had, I created a table to try and calculate a formula to follow my thought process. I have two different variable occurring with the different types of gum balls and the number of children. I proceeded to continue this in a chronological process, with the number of children increasing slightly to determine how much money would be needed to spend for four and five types of gum balls with a various amount of children. With this, I created a rough draft of a formula, which states that the number of children multiplied by the different types of gum balls subtracted by the expression parenthesis the different types of gum balls subtracted by one equals the maximum amount of coins needed to get all of the children the same color gum ball. Children*Gum-(Gum-1) = Maximum Cost. On the picture below, it shows the same formula, just shown in a more advanced table, with each of the components of the problem represented by a variable.
Solution
The solution, or formula, that I had gotten was C x G - ( G - 1 ) = Maximum # of Pennies, with the variable C representing the number of children, and variable G representing the different types, or colors, of gum balls. I was able to test this formula with the table above, which shows different numbers representing the variables, and how when inputting the information known, can give you the maximum amount of pennies needed to spend. To find either the number of children or the different types of gum balls there are, you would need to have the variable be by itself. To find the number of children with knowledge of the different types of gum balls and the maximum amount of pennies needed, you would need to add G to both sides and subtract 1 from both sides, which leaves you with C x G = Maximum # of Pennies + G - 1, to isolate C, you need to divide both sides of the equation by G. This gives the equation C = ( Maximum # of Pennies + G - 1 ) / G. This would be how you use the basic equation to find a specific variable.
Self-Assessment and Reflection
On this problem of the week, I believe that I inhabited Looking for Patterns the most in the different Habits of a Mathematician. I showed Looking for Patterns in this problem of the week by trying to find a consistent rule that applied to all of the variables. This was shown by how many different examples I made visible through the table above. I was able to find a consistent increase from there and thus able to figure out the rule I made. This habit is going to be significant for further use because mathematics constantly use numbers in the equations and formulas, and real life example, so finding a pattern from one set to another will help solve the problem all together. This is why I believe I deserve 10/10.
The solution, or formula, that I had gotten was C x G - ( G - 1 ) = Maximum # of Pennies, with the variable C representing the number of children, and variable G representing the different types, or colors, of gum balls. I was able to test this formula with the table above, which shows different numbers representing the variables, and how when inputting the information known, can give you the maximum amount of pennies needed to spend. To find either the number of children or the different types of gum balls there are, you would need to have the variable be by itself. To find the number of children with knowledge of the different types of gum balls and the maximum amount of pennies needed, you would need to add G to both sides and subtract 1 from both sides, which leaves you with C x G = Maximum # of Pennies + G - 1, to isolate C, you need to divide both sides of the equation by G. This gives the equation C = ( Maximum # of Pennies + G - 1 ) / G. This would be how you use the basic equation to find a specific variable.
Self-Assessment and Reflection
On this problem of the week, I believe that I inhabited Looking for Patterns the most in the different Habits of a Mathematician. I showed Looking for Patterns in this problem of the week by trying to find a consistent rule that applied to all of the variables. This was shown by how many different examples I made visible through the table above. I was able to find a consistent increase from there and thus able to figure out the rule I made. This habit is going to be significant for further use because mathematics constantly use numbers in the equations and formulas, and real life example, so finding a pattern from one set to another will help solve the problem all together. This is why I believe I deserve 10/10.