Q.3.76 - A First Course in Probability

Question

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

Step-by-Step Solution

Answer

Thus,

$p=\sum _{i\in \mathrm{\mathbb{N}}}P{\left[\right(E\cup F{)}^c]}^{i}P\left(E\right)$

$=P\left(E\right)\sum _{i\in \mathrm{\mathbb{N}}}P{\left[\right(E\cup F{)}^c]}^i$

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

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

$p=P\left(E\right)\frac{1}{P(E\cup F)}$

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

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

Step 1: Given Information

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

Step 2: Explanation

$\text{Events:}E,F,E\cap F=\varnothing$

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

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

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

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

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

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

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.

$p=\sum _{i\in \mathrm{\mathbb{N}}}P{\left[\right(E\cup F{)}^c]}^{i}P\left(E\right)$

Step 3: Explanation

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

$\sum _{n\in \mathrm{\mathbb{N}}}{q}^n=\frac{1}{1-q}$

Thus,

$p=\sum _{i\in \mathrm{\mathbb{N}}}P{\left[\right(E\cup F{)}^c]}^{i}P\left(E\right)$

$=P\left(E\right)\sum _{i\in \mathrm{\mathbb{N}}}P{\left[\right(E\cup F{)}^c]}^i$

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

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

$p=P\left(E\right)\frac{1}{P(E\cup F)}$

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

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

Step 4: Final Answer

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

$p=P\left(E\right)\frac{1}{P(E\cup F)}$

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

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