Re: MathPuzzle 182: Balls in a shoe box

On Jun 1, 4:06 pm, Patrick Hamlyn <p...@xxxxxxxxxxxxxxxxxxxxxxx>
wrote:
Ted Schuerzinger <f...@xxxxxxxxxxxx> wrote:
Somebody claiming to be Patrick Hamlyn <p...@xxxxxxxxxxxxxxxxxxxxxxx>
wrote at Fri, 01 Jun 2007 01:37:49 GMT:

And my suggestion:
Change
The other balls have other colors, with equal numbers of each
color.
to
There are equal numbers (not necessarily 20) of each other color.

It's still slightly clumsy, but it eliminates the possibility that
when OP said 'equal numbers' he meant 'also equal to 20'.

You can eliminate that possibility: the OP wrote:

Removing all 20 white balls from the box does not change the
probability that two balls selected from the box at random and without
replacement (i.e., a pair is chosen together) have the same color.

If there were 20 of every color, it would not be possible for the
probabilities to remain identical. The probability of removing a pair
with the white balls in the box would be 19/(20x-1), where x is the number
of colors (including white), whereas with the white balls removed that
probability would be 19/(20x-21), where x is the same as before. There is
no integer x which will yield an equality of those two fractions.

Quite right. So the sentence can be simply:
Every other color occurs with equal frequency.
--
Patrick Hamlyn posting from Perth, Western Australia
Windsurfing capital of the Southern Hemisphere
Moderator: polyforms group (polyforms-subscr...@xxxxxxxxxxx)



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t - how many balls?


210 balls:

20 whites + 10 each of 19 different colours
p(pair) = 1/21 with or without the 20 whites





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