I cannot figure this out. The problem goes A boy has a bag filled with balls:

  • 1 red
  • 2 green
  • 4 Blue
  • 7 White

A child plays a game in which he withdraws one ball, writes down its color, and then replaces the ball. In order to win the game he must write down at least one of each color of ball. What is the probability that this will happen on, or before the N-th trial???

I initially assumed this problem involved a geometric probability distribution function, based on the 4 probabilities multiplied together, for N trials. But I dont know. Please help!


This is $1$ minus the probability that after $N$ trials, one or more of the colours is missing.

So let us find the probability that some colour is missing. We will use the Method of Inclusion/Exclusion.

The probability red is missing is $\left(\frac{14}{15}\right)^N$.

We can write down similar expressions for the probability green is missing, blue is missing, white is missing.

If we add up these $4$ probabilities, we will have double-counted the situations in which $2$ of the colours are missing. So we must subtract the probability that red and green is missing, together with $5$ other similar expressions. The probability that red and green is missing is, for example, $\left(\frac{12}{15}\right)^N$.

But we have subtracted too much, for we have subtracted once too many times the situations in which $3$ colours are missing. So we must add back $4$ terms that represent these probabilities.

  • $\begingroup$ Thank you so much! what would happen though if there were an odd number of balls in the bad? would you do the same and then just subtract the time when 4 colors are missing? $\endgroup$ – user2778291 Nov 3 '13 at 19:10
  • $\begingroup$ One can use the same idea for any number. though it rapidly gets unpleasant unless there are symmetries among the numbers. $\endgroup$ – André Nicolas Nov 3 '13 at 19:34

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