Here is another problem for you to take back to your classroom and try with your learners but before you do let's discuss it. Post your solutions, the process that you used to solve it and identify the math standards or mathematical practices.
A bowl contains 75 candies, identical except for color. Twenty are red, 25 are green, and 30 are brown. Without looking, what is the least number of candies you must pick in order to be absolutely certain that three of them are brown?
Have a great week!
Brooke
Comments
Hmmm... my first reaction is 48. Everything else would have to be gone first, so 20 (red) + 25 (green) + 3 (brown)
Since I answered that more quickly than your other problems, there might be faulty logic in there somewhere!
48 its possible but unlikely that the first 45 are the other 2 colors.
Nice problem, Brooke!
Connie, my initial thought is the same as yours! Although not likely, the first 45 picks could possibly all be red or green, so in order to insure that three brown ones are chosen, that would mean picking 48.
What do you say, Brooke? Set us straight if we have an error in our thinking.
Margaret :)
The solution is 48.
You could pick all the red candies, of which there are 20, and all of the green ones, of which there are 25, and still not have a single brown candy in your hand. That's already 45 candies. You'll then need to pick three brown ones, giving you a total of 48.
This is the number you must pick to be absolutely certain that there are three brown candies. However, it's certainly possible that you would have three brown candies if you picked fewer. You just can't be absolutely certain that you'll have three brown ones.
-Brooke