I came across an interesting problem on probability which is as follows:

Consider that there are two buckets A and B. A has N differently numbered balls {1,2,3,...,N} while B has a subset of balls that are already in A.

e.g. N = 5, A = {#1,#2,#3,#4,#5} and B = {#1,#4,#5}

What is the probability of picking d balls from bucket A such that exactly one ball is new to B

e.g. d = 2, Picked = {#2,#4} and #2 is unique

The answer can I could derive is $$ \text{Prob} = \frac{\binom r{d-1}(N-r)}{\binom N{d}} $$

Now I wanted to generalize the above problem, but failed. The problem is as follows.

Consider that there are N different numbered balls that exist {1,2,...,N}.

Moreover, there are two buckets A having n and B having m uniquely numbered balls. A and B might have balls of the same color.

e.g. N = 10, A = {#1,#2,#3} and B = {#2,#3,#4}

What is the probability P that if we pick d balls from A, exactly 1 ball is absent in B.

e.g. d = 2. So we pick {#1,#2} and #1 is unique to bucket B.

The answer can I could derive is:

Given we know the number of balls common in A and B as k,

$$ \text{Prob} = \frac{\binom k{d-1}(m-k)}{\binom n{d}} $$

Did I miss anything important?

  • $\begingroup$ Where you have $\left(\frac r{d-1}\right)$ etc., it should say $\binom r{d-1}$ etc. You can get this either using \binom r{d-1} or using {r\choose d-1}. Also, if you have text like "probability" in an equation that shouldn't be italicised, you can use e.g. \text{probability}. $\endgroup$ – joriki Jul 4 '16 at 6:00
  • $\begingroup$ Thanks @joriki for correcting my notations. Really appreciate the feedback :D $\endgroup$ – Mobashyr Mohammad Jul 5 '16 at 6:32

The problem isn't really any different than the one you already solved. Removing the balls in $B$ that aren't in $A$ and relabeling the balls transforms it into the first problem; the balls in $B$ that aren't in $A$ aren't playing any substantial role, and neither is the hypothetical limit $N$ on the numbers. So the result is

$$ \frac{\binom k{d-1}\binom{n-k}1}{\binom nd}\;. $$

You can tell that your result can't be right e.g. by adding another ball in $B$ that wasn't in either bucket. Then your result involving $m$ would change, even though this clearly doesn't change the desired probability.

  • $\begingroup$ Thanks @joriki. True, my solution for the second variant was incorrect and I understand your answer totally. Thanks for taking time out :D/ $\endgroup$ – Mobashyr Mohammad Jul 5 '16 at 6:38

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