Red, blue and green balls
You have a bag containing n red balls, n blue balls and n green balls, where n > 2 .
If you pick 3 balls out at random (without replacement) the probability of getting 3 of the same color is greater than 10%.
What is the fewest number of balls you can have in the bag altogether?
The answer is 63.
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1 solution
Sorry again for the report. I'm feeling somewhat sheepish right now ...
Anyway, just thought that I'd note that the limiting percentage for which there is a finite value for n is 9 1 0 0 %. If it were 1 1 % then n would be ≈ 2 0 0 , and it were 1 1 . 1 % then n would be ≈ 2 0 0 0 . In general, if the percentage were x 1 0 0 % then we would have
n = 2 ( x − 9 ) ( 3 x − 9 ) + x 2 + 2 6 x + 9 .
For x = 1 0 , as is the case here, we have n ≈ 2 0 . 1 0 5 .
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No problem at all, @Brian Charlesworth ... Thanks for all your feedback! Usually when I get feedback from you, I figure, well if Brian could interpret the problem that way, then I guess anyone could... I better change the wording! :^)
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I always keep an eye out for your problems. Thanks for posting them! :)
The probability they will be the same is given by:
P ( n ) = ( 3 3 n ) 3 ( 3 n )
This is a monotonically increasing function, and P ( 2 0 ) < 0 . 1 , P ( 2 1 ) > 0 . 1 .
Therefore, the minimum number you can have in the bag is 2 1 ∗ 3 = 6 3