During this week for our Open-Ended Problem, we were able to explore real-life situations on probability, that could potentially be used sometime in our future lives. We explored 3 problems inside "The Gumball Dilemma" problem, which were about probability. The first problem was about a woman named Ms. Hernandez, who had a pair of twins that wanted a gumball. Each gumball cost one penny (1 cent), and of course the twins wanted the same color of gumball. There were two colors of gumball (red and white). This question is asking why 3 cents may be the most Ms. Hernandez will have to spend in order to satisfy her twins. The second problem was about Ms. Hernandez farther into the future, where her twins want the same color of gumball once again, but instead of 2 colors, the gumball factory introduced a 3rd color (blue). The final question is about a man named Mr. Hodges, who has triplets, who pass a 3 color gumball machine and all want the same color gumball. He also wants to know the maximum amount of money he may have to spend in order to satisfy his triplets.
I began the first problem by writing down the amount of twins and the amount of cents that the initial problem gave (3 cents). Once the class shared out, I realized that the ratio really didn't matter because if one twin receives both colors (in the worst case), no matter what, the other twin will receive a matching color. The second problem was very similar because I used the same tactic of using the worst case scenario. For example, if one twin got all 3 colors, the other twin MUST receive a matching color. The same tactic is also used in the last problem but there are 3 kids and 4 colors. This means we can still use the same tactic, but we also have to keep in mind that there are 3 children. That means in the worst case scenario, 2 children receive all the colors, which creates the final gumball to match.
1.) 3 cents because if one twin receives both colors, the other twin is a definite match
2.) 4 cents
3.) 7 cents
Kids Gumballs Money 2 2 +1 3 +1 2 3 +1 4 +1 2 4 +1 5 +1 2 5 6 Kids
Gumballs Money 3 2 +1 5 +2 3 3 +1 7 +2 3 4 +1 9 +2 3 5 11 Kids Gumballs Money 4 2 +1 7 +3 4 3 +1 10 +3 4 4 +1 13 +3 4 5 16
I definitely learned how to expand upon a problem efficiently and correctly because I usually will not expand on a problem. This definitely gave me a sense of challenge in math and gave me confidence in the field of probability and math as a whole. (Sharing in class and seeing multiple classmates present, help give me confidence.) I used the habit starting small because I definitely began to try and find the ratios, fractions etc in the beginning. Once seeing classmates give simple answers to somewhat simple problems, I began to start small and see what else I could build on. The problems were actually very simple and through using this habit, I definitely gained a sense of where to start when even building upon larger problems, that can potentially be challenging in the future. I feel like I deserve a 10/10 because I collaborated with others, and I thoroughly understood the circumstances and what I needed to complete. Overall this problem has given me a sense of confidence in math, and I am able to confidently present and discuss the topic of probability.
I began the first problem by writing down the amount of twins and the amount of cents that the initial problem gave (3 cents). Once the class shared out, I realized that the ratio really didn't matter because if one twin receives both colors (in the worst case), no matter what, the other twin will receive a matching color. The second problem was very similar because I used the same tactic of using the worst case scenario. For example, if one twin got all 3 colors, the other twin MUST receive a matching color. The same tactic is also used in the last problem but there are 3 kids and 4 colors. This means we can still use the same tactic, but we also have to keep in mind that there are 3 children. That means in the worst case scenario, 2 children receive all the colors, which creates the final gumball to match.
1.) 3 cents because if one twin receives both colors, the other twin is a definite match
2.) 4 cents
3.) 7 cents
Kids Gumballs Money 2 2 +1 3 +1 2 3 +1 4 +1 2 4 +1 5 +1 2 5 6 Kids
Gumballs Money 3 2 +1 5 +2 3 3 +1 7 +2 3 4 +1 9 +2 3 5 11 Kids Gumballs Money 4 2 +1 7 +3 4 3 +1 10 +3 4 4 +1 13 +3 4 5 16
- Dependent on kids and colors the money goes up linearly
- If you add the amount of gumballs, the coins will go up dependent on how many children there are also. 2 children result in 1 coin increase if you increase the gumballs
- Also observed that the children and the difference in coins, is the same linear value as the amount of coins, when you add a new gumball
I definitely learned how to expand upon a problem efficiently and correctly because I usually will not expand on a problem. This definitely gave me a sense of challenge in math and gave me confidence in the field of probability and math as a whole. (Sharing in class and seeing multiple classmates present, help give me confidence.) I used the habit starting small because I definitely began to try and find the ratios, fractions etc in the beginning. Once seeing classmates give simple answers to somewhat simple problems, I began to start small and see what else I could build on. The problems were actually very simple and through using this habit, I definitely gained a sense of where to start when even building upon larger problems, that can potentially be challenging in the future. I feel like I deserve a 10/10 because I collaborated with others, and I thoroughly understood the circumstances and what I needed to complete. Overall this problem has given me a sense of confidence in math, and I am able to confidently present and discuss the topic of probability.