A contradiction hidden in the definition of the probability of the intersection of events?

Consider a set $C$ composed of three distinguishable kinds of elements $A,B,G$, and let $\alpha,\beta,\gamma>0$ be the numbers of elements of each kind, and $|C|=c=\alpha+\beta+\gamma$.

We define the three events $L_n^A, L_n^B, L_n^G$ as to get, in $n$ trials with replacement, at least one element of kind $A$, at least one element of kind $B$, and at least one element of kind $G$.

The probabilities of these events are $$P(L_n^A)=1-\left(\frac{c-\alpha}{c}\right)^n=1-\left(\frac{\beta+\gamma}{c}\right)^n,$$

$$P(L_n^B)=1-\left(\frac{c-\beta}{c}\right)^n=1-\left(\frac{\alpha+\gamma}{c}\right)^n,$$

$$P(L_n^G)=1-\left(\frac{c-\gamma}{c}\right)^n=1-\left(\frac{\alpha+\beta}{c}\right)^n.$$

We evaluate $P(L_n^A\cap L_n^B)$. By definition of conditional probability and applying the property of the opposite event, we have $$P(L_n^A\cap L_n^B)=P(L_n^A|L_n^B)P(L_n^B)=[1-P(\overline{L_n^A}|L_n^B)]P(L_n^B)=P(L_n^B)-P(\overline{L_n^A}|L_n^B)P(L_n^B).$$

By means of Bayes' theorem, $P(\overline{L_n^A}|L_n^B)P(L_n^B)=P(L_n^B|\overline{L_n^A})P(\overline{L_n^A})$. If we know that the event $L_n^A$ did not take place, all the $n$ extractions are either of kind $B$ or of kind $G$. Therefore, $P(L_n^B|\overline{L_n^A})=1-\left(\frac{\beta+\gamma-\beta}{\beta+\gamma}\right)^n=1-\left(\frac{\gamma}{\beta+\gamma}\right)^n$.

In conclusion, since $P(\overline{L_n^A})=\left(\frac{\beta+\gamma}{c}\right)^n$, we obtain $$P(L_n^A\cap L_n^B)=P(L_n^B)-P(\overline{L_n^A}|L_n^B)P(L_n^B)=P(L_n^B)-P(L_n^B|\overline{L_n^A})P(\overline{L_n^A})=$$ $$=1-\left(\frac{\alpha+\gamma}{c}\right)^n-\left[1-\left(\frac{\gamma}{\beta+\gamma}\right)^n\right]\left(\frac{\beta+\gamma}{c}\right)^n=$$ $$=1-\left(\frac{\alpha+\gamma}{c}\right)^n-\left(\frac{\beta+\gamma}{c}\right)^n+\left(\frac{\gamma}{c}\right)^n.$$

We notice that if $n=0$ (or $n=1$), then we correctly have $P(L_n^A\cap L_n^B)=0$.

We now evaluate $P(L_n^A\cap L_n^B\cap L_n^G)$. As we have done before, we find $$P(L_n^A\cap L_n^B\cap L_n^G)=P(L_n^G|L_n^A\cap L_n^B)P(L_n^A\cap L_n^B),$$ which implies that,

for $n=0$, it must also be $P(L_n^A\cap L_n^B\cap L_n^G)=0$.

We go on with the calculation of $P(L_n^A\cap L_n^B\cap L_n^G)$, and we apply (again) first the definition of opposite event $$P(L_n^A\cap L_n^B\cap L_n^G)=P(L_n^G|L_n^A\cap L_n^B)P(L_n^A\cap L_n^B)=[1-P(\overline{L_n^G}|L_n^A\cap L_n^B)]P(L_n^A\cap L_n^B)=$$ $$=P(L_n^A\cap L_n^B)-P(\overline{L_n^G}|L_n^A\cap L_n^B)P(L_n^A\cap L_n^B),$$ and then the theorem of Bayes on the second term $$P(\overline{L_n^G}|L_n^A\cap L_n^B)P(L_n^A\cap L_n^B)=P(L_n^A\cap L_n^B|\overline{L_n^G})P(\overline{L_n^G}).$$

If we know that event $L_n^G$ does not take place, the probability to get at least one element of kind $A$ and at least one element of kind $B$ is $$P(L_n^A\cap L_n^B|\overline{L_n^G})=1-\left(\frac{\alpha}{\alpha+\beta}\right)^n-\left(\frac{\beta}{\alpha+\beta}\right)^n.$$

Therefore, since $P(\overline{L_n^G})=\left(\frac{\alpha+\beta}{c}\right)^n$,

$$P(L_n^A\cap L_n^B\cap L_n^G)=P(L_n^A\cap L_n^B)-P(\overline{L_n^G}|L_n^A\cap L_n^B)P(L_n^A\cap L_n^B)=P(L_n^A\cap L_n^B)-P(L_n^A\cap L_n^B|\overline{L_n^G})P(\overline{L_n^G})=$$ $$=1-\left(\frac{\alpha+\gamma}{c}\right)^n-\left(\frac{\beta+\gamma}{c}\right)^n+\left(\frac{\gamma}{c}\right)^n-\left(\frac{\alpha+\beta}{c}\right)^n+\left(\frac{\alpha}{c}\right)^n+\left(\frac{\beta}{c}\right)^n.$$

If we now substitute $n=0$ in this expression we obtain $P(L_n^A\cap L_n^B\cap L_n^G)=1$,

which is in contradiction with what we observed before, i.e. that $P(L_n^A\cap L_n^B\cap L_n^G)=0$ with $n=0$.

There is likely a mistake in this reasoning, but I am not able to spot it.

• What’s the Cliffs notes version? Jul 28, 2018 at 18:01
• @Randall Hi Randall, sorry I don't know what are you referring to!
– user559615
Jul 28, 2018 at 18:03

There are a couple of problems that I can see.

1. You're using Bayes' theorem when the event conditioned on has probability $0$, but Bayes' theorem does not necessarily hold in that case.
2. In some cases your formulae are only valid for $n\geq 1$. For example, you calculate the probability that you get at least one each of A and B conditional on having no G as $1$ minus the probability of getting $n$ As, minus the probability of getting $n$ Bs. This assumes that these events are disjoint - that you can't get $n$ As and $n$ Bs - which is only true for $n\geq 1$. (Indeed, your formula gives a negative probability when $n=0$.)
• Thanks! I see very well now!
– user559615
Jul 28, 2018 at 18:28
• But, still I don't get why, with $n=0$, we have a reasonable result for $P(L_n^A\cap L_n^B)$ but not for $P(L_n^A\cap L_n^B\cap L_n^G)$, although we used in both cases Bayes' theorem...
– user559615
Jul 28, 2018 at 18:36
• I think the Bayes' theorem issue may not actually be a problem - when you use it, it looks like both sides of the equation are zero anyway. So the main issue is assuming two events are disjoint, which I think only happens in the triple intersection calculation. Jul 28, 2018 at 18:56
• But then it is however $P(L_0^A\cap L_0^B\cap L_0^G)=P(L_0^A\cap L_0^B)=0$, isn't it? How can I prove it?
– user559615
Jul 28, 2018 at 19:25