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Consider an urn which contains three different kinds of balls A, B and C. We suppose that there is at least one ball of each kind in the urn.

We define the event $A$ as "to get, in $n$ independent trials, at least one ball of kind A" and, similarly, we define the events $B$ and $C$. In each trial, we extract only one ball, and then we put it back in the urn.

These three events are clearly linked to each other through the constraint $P(A\cup B\cup C)=1$.

In which conditions does the constraint $P(A)=P(\overline{B})$ hold?

This constraint can be formulated as follows: "The chance to get at least one ball of kind A, in $n$ trials, is equal to the chance not to get any ball of kind B".

This question is a special case of the problem treated in this post A weird problem of probability! and, more in general, in this other post A problem of conditional probability.

Especially Lime
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1 Answers1

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Let's suppose there are $a,b,c$ balls of each type.

We have $P(A)-P(\overline B)=1-P(\overline A)-P(\overline B)$. Also $P(\overline A)=\big(\frac{b+c}{a+b+c}\big)^n$ and $P(\overline B)=\big(\frac{a+c}{a+b+c}\big)^n.$ Multiplying $1-P(\overline A)-P(\overline B)=0$ by $(a+b+c)^n$ we need $(a+c)^n+(b+c)^n=(a+b+c)^n$, which has no solutions for $n>2$ by a result of Wiles. It is easy to see there are no solutions for $n=1$ either, since the two sides differ by $c>0$.

If $n=2$ we have a Pythagorean triple $x^2+y^2=z^2$. Any such triple gives a solution, setting $c=x+y-z$, $a=z-y$, $b=z-x$. (e.g. for $3,4,5$ we get $a=1,b=2,c=2$.)

Especially Lime
  • 46,692
  • Thanks a lot! Very clear! –  Jun 22 '18 at 07:22
  • So still a solution for $n=2$. I overlooked that and deleted my answer. (+1). – drhab Jun 22 '18 at 07:44
  • @Especially Lime I wrote a post you might be interested in. You may have a look to help me to write it better https://math.stackexchange.com/q/2828661/559615. Thanks for your help. –  Jun 23 '18 at 10:36