1

GOAL = AT LEAST one black ball AND AT LEAST ONE red ball

Given an urn with 60 balls, 4 red, 4 black, the others white, randomly taking 7 balls without replacement, the probability of getting AT LEAST one black and AT LEAST ONE red is ~14.54%. (Calculation)

What I can not seem to figure out, if I now take one more ball out of the urn (let's call this a draw step) with the remaining 53 balls inside, what is the probability of achieving GOAL? How can I keep computing the probability for each further draw step, where each step draws exactly one ball?

I'm so hung up on this I must miss the very obvious, probably simple solution. Please enlighten and relieve me of my mental block?

Jürgen
  • 121
  • 1
    It is not clear what you want to compute. There is a more efficient way to find the probability of at least one of each in $7$ draws. But it sounds to me as if your new question asks for for probability that the goal is first met at $k$ draws. – André Nicolas Oct 09 '15 at 17:45
  • @AndréNicolas Hi, it is not 7 draws, but rather I take out 7 balls at once, and the probability for getting at least one red and at least one black ball if I take out 7 balls at once, is 14.54%. Could you please elaborate how this can be computed more efficiently? What I am asking in my original question is: How can I keep computing the probability, if after taking out those 7 balls, I take out another ball, how high is the probability of having at least one red and black? What if I take out one more ball? And so forth. I just want to know how to best compute that. – Jürgen Oct 09 '15 at 18:17
  • What I am currently pondering, is taking out 7 balls, and than one more ball, not the same as taking out 8 balls at once? If so, I guess I just found a solution to my answer, but I am not sure about this. I might be overcomplicating matters here... – Jürgen Oct 09 '15 at 18:23
  • 1
    We describe another way to find the probability of at least one red and at least one black when we remove $7$ balls. The probability of none black is $\frac{\binom{56}{7}}{\binom{60}{7}}$. The probability of none red is the same. The probability of all white is $\frac{\binom{56}{7}}{\binom{60}{7}}$. So the probability of at least one black and at least one red is $1-2\frac{\binom{56}{7}}{\binom{60}{7}}+\frac{\binom{52}{7}}{\binom{60}{7}}$. – André Nicolas Oct 09 '15 at 18:48
  • 1
    Yes, taking out $7$ and then $1$ is taking out $8$. But what are you asking about $8$? The probability it is the first time we get one of each, so $7$ was not enough but $8$ is? Or simply the probability $8$ does the job? Neither question is hard to solve, the first is a little easier. But the question (what probability are we asking for) needs to be made clear. – André Nicolas Oct 09 '15 at 18:52
  • @AndréNicolas Thank you! Well, I just wanted to see how the probabilty would increase for each additional ball that would be taken out of the urn. Looks like I completely overcomplicated matters, and on top of all things, I am sitting here, wondering why I implemented such a complex formula, when all it takes is computing the probability of it NOT happening. Quite the eye opener for me! – Jürgen Oct 09 '15 at 22:07
  • You are welcome. When one has computing power a short reach away, one can often just calculate. I come from a time when computers were behemoths with less computing capacity than an MP3 player. – André Nicolas Oct 09 '15 at 22:12
  • If anybody else is reading this: For another take on how to calculate this, also see: http://math.stackexchange.com/questions/1472307/can-this-probability-be-computed-more-efficiently Though I dare say the explanation given by @AndréNicolas is is the most straight forward. – Jürgen Oct 09 '15 at 22:17

0 Answers0