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A box contains 28 red balls, 20 green balls,19 yellow balls, 13 blue balls, 11 white balls, and 9 black balls. What is the minimum number of balls that must be drawn from the box without replacement to guarantee that at least 15 balls of a single color will be drawn?

 Sep 21, 2019
 #1
avatar+2361 
+3

Alright, it says to GUARUNTEE

 

So we are going to find the worst case possible.

 

Any color balls under 15 will be drawn

 

So 13 + 11 + 9 = 33

 

Now we have 28 red, 20 green, 19, yellow.

 

The worst case possible is 14 from each

 

14 red, 14 green, and 14 yellow.

 

 

So we drew  14 red, 14 green, 14 yellow, and 13 blue, 11 white, 9 black balls.

 

The next ball we draw NOT MATTER WHAT will make 15 balls of a single color drawn.

 

So we drawn 14 + 14 + 14 + 13 + 11 + 9 + 1

 

76 balls should be the answer.

 Sep 21, 2019
 #2
avatar+196 
+2

Thank you for your answer!!

 Sep 21, 2019
 #3
avatar+2361 
+2

Yes! This problem utilizes the Pigeon Hole Principle.

CalculatorUser  Sep 21, 2019

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