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A box contains 100 balls of different colours: 28 red 17 blue 21 green 10 white 12 yellow and 12 black.Smallest no of balls drawn from box so that at least 15 balls are of same colour is

Well , I had been trying to solve this question the other way round. I am assuming first that I have taken largest no. of ball that is 100 from the box. Then I had gradually been decreasing the number of balls I had taken out from 100 to 99. Like if I keep one ball back in the box then the probability of at least 15 balls out of the box would be hundred percent. (That is at least 15 balls of same colour must be out of the box). Extending this approach I think the answer should be 100-(28+21+2) [all red, all green and 2 blue still in box.] But the answer comes out to be 77. I am not able to understand what went wrong

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    $\begingroup$ Consider the worst case scenario without having $15$ of a color. That would correspond to having drawn $14$ of each except for those who have less than $14$ in which case you draw all of what is available. Once you draw one more then it must be the fifteenth of one of your colors. So, worst case you have $14+14+14+10+12+12=76$ balls taken without $15$ of a color. The $77$'th ball will force that you have $15$ red or blue or green balls. $\endgroup$ – JMoravitz Aug 25 '17 at 17:52
  • $\begingroup$ Ok now I understand it. Thanks... $\endgroup$ – Mridul Kumar Rai Aug 25 '17 at 17:54
  • $\begingroup$ Fixing your approach, the point is you don't need all red to still be in the box, only some of the reds... $\endgroup$ – JMoravitz Aug 25 '17 at 17:55
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The minimum is (1+) 14 red + 14 blue + 14 green + 10 white + 12 yellow + 12 black, hence 77.

First three by pigeonhole. Last three because in total they are $<15$.

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