# Summation of infinite series

1. Feb 3, 2008

### Somefantastik

I'm having trouble picking apart this summation:

$$\sum$$$$^{inf}_{n=1}$$ P(E)*P(1-p)$$^{n-1}$$; where p = P(E) + P(F)

I know I need to use the identity of a geometrical series when |r| < 1 : 1/(1-r)

I'm getting $$P(E)/(1-(P(E)+P(F))$$

But I need to be getting P(E)/((P(E)+P(F));

The entire problem is

Let E & F be mutually exclusive events in the sample space of an experiment. Suppose that the exp is repeated until either event E or F occurs. What does the sample space of hte new super experiment look like? Show that the probability of event E before event F is P(E)/(P(E)+P(F)).

2. Feb 4, 2008

### HallsofIvy

That's a bit confusing. If p= P(E)- P(F), then what is just P by itself? Asuming your terms are just A(1-p)n-1, then I would be inclined to first change the summation index: let i= n-1 so this becomes
[tex\\sum_{i=0}^\infty A(1-p)^i[/tex]
That sum is, of course,
$$\frac{A}{1- (1-p)}= \frac{A}{p}= \frac{P(E)*P}{P(E)+ P(F)}$$
because 1-(1-p)= p. I can't say more because I am still not sure what "P" is.

3. Feb 4, 2008

### Somefantastik

Its a probability question, so P(1-p) is just the probability that p has not occurred yet, where p = P(E U F). Since E,F are mutually exclusive, can say p = P(E) + P(F).

4. Feb 4, 2008

### Somefantastik

But you pretty much answered my question...naturally it was a mistake in arithmetic. Thanks for helping me with that...sometimes one gets so wrapped up in the global outcome of the problem and loses track of the little details.