91Ó°ÊÓ

Show that if independent trials of this experiment are performed, then E will occur before F with probability P(E)/[P(E) + P(F)].

$\text{Events:}E,F,Eâˆ\copyright F=\mathrm{âˆ\dots }$

$\text{Probabilities:}P\left(E\right),P\left(F\right)$

$\text{Calculate}p\text{, the probability that}E\text{occurs before}F\}$

$\text{If}E\text{occurs first,}E\text{has to occur after}i=0,1,2,…\text{occurrences of}(E∪F{)}^c$

$\text{Probability that}i\text{experiments will end in}(E∪F{)}^c\text{, and then}E\text{is (because of independence):}$

$P{\left[\right(E∪F{)}^c]}^{i}P\left(E\right)$

$\text{In the special case,}P\left[\right(E∪F{)}^c]=0⇒p=P\left(E\right)$

The events in which E occurs first after i repetitions are mutually exclusive, thus the probability of their union which is the wanted event is the sum of their probabilities.

$p=\underset{i∈\mathrm{ℕ}}{∑}P{\left[\right(E∪F{)}^c]}^{i}P\left(E\right)$

This is computed by applying the formula for the sum of a geometric sequence.

$\underset{n∈\mathrm{ℕ}}{∑}{q}^n=\frac{1}{1−q}$

Thus,

$p=\underset{i∈\mathrm{ℕ}}{∑}P{\left[\right(E∪F{)}^c]}^{i}P\left(E\right)$

$=P\left(E\right)\underset{i∈\mathrm{ℕ}}{∑}P{\left[\right(E∪F{)}^c]}^i$

$=P\left(E\right)\frac{1}{1−P\left[\right(E∪F{)}^c]}$

The formula for the probability of complement then P(EF)=0 gives:

$p=P\left(E\right)\frac{1}{P(E∪F)}$

$=P\left(E\right)\frac{1}{P\left(E\right)+P\left(F\right)−P\left(EF\right)}$

$=\frac{P\left(E\right)}{P\left(E\right)+P\left(F\right)}$

The formula for the probability of complement then P(EF)=0 gives:

$p=P\left(E\right)\frac{1}{P(E∪F)}$

$=P\left(E\right)\frac{1}{P\left(E\right)+P\left(F\right)−P\left(EF\right)}$

$=\frac{P\left(E\right)}{P\left(E\right)+P\left(F\right)}$