A bag contains red, blue, and green balls. What is the minimum number of balls one must draw from the bag (without looking) to guarantee that at least balls of the same color are drawn?
15
14
16
13
Hint 1: Think about the worst-case scenario: drawing as many balls as possible without ever getting 5 of the same color.
Hint 2: How many balls of each color can you draw such that no color has 5 or more balls? That would be 4 red, 4 blue, and 4 green.
Hint 3: Add these numbers together and then add 1 more to guarantee the threshold is met: .
Step 1: To guarantee at least balls of the same color, we consider the "worst-case scenario" (the boundary where we draw as many balls as possible without satisfying the condition).
Step 2: The condition is violated if we draw at most balls of each color:
Maximum red balls drawn without getting 5:
Maximum blue balls drawn without getting 5:
Maximum green balls drawn without getting 5:
Step 3: The maximum total number of balls we can draw without obtaining of any one color is:
Step 4: Drawing one additional ball (the th ball) must result in either a red, blue, or green ball. By the Pigeonhole Principle, this will increase the count of one of the colors to at least .
Thus, the minimum number of balls to draw is .
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