Infinite sums and probability

perusal

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A and B are two mutually exclusive events in the same sample space. I am asked to prove that to the probability of event A happening before event B is equal to A/A+B. I assume you can calculate the probability like so...

[MATH] sum_{n=1}^{\infty} A*(1-(A+B)^{n-1}) [/MATH]
However, I am unsure how to equate this infinite sum with the equation A/A+B.

Can anyone help?
 
A and B are two mutually exclusive events in the same sample space. I am asked to prove that to the probability of event A happening before event B is equal to A/A+B. I assume you can calculate the probability like so...
[MATH]sum_{n=1}^{\infty} A*(1-(A+B)^{n-1})[/MATH]However, I am unsure how to equate this infinite sum with the equation A/A+B.
@perusal, once again you have used notation that is non-standard to western eyes. You have used A & B in two different ways.
I think you need the events to be independent.
Here is an example: toss a 12 sided die numbered 1 to 12. Then let \(\displaystyle A=\{2,3,5,7,11\}\) a prime & \(\displaystyle B=\{4,8,12\}\) multiples of four.
\(\displaystyle \mathscr{P}r(A)=\frac{5}{12}~\&~\mathscr{P}r(B)=\frac{3}{12}\). Note tha it in important the the sum is less than the whole.
Calculate the probability that A occurs before B.
 
However, I am unsure how to equate this infinite sum with the equation A/A+B.
I just want you to note that A/A + B = 1 + B > 1 which can not be the correct answer. Please write what you meant to write
 
@perusal, once again you have used notation that is non-standard to western eyes. You have used A & B in two different ways.
I think you need the events to be independent.
Here is an example: toss a 12 sided die numbered 1 to 12. Then let \(\displaystyle A=\{2,3,5,7,11\}\) a prime & \(\displaystyle B=\{4,8,12\}\) multiples of four.
\(\displaystyle \mathscr{P}r(A)=\frac{5}{12}~\&~\mathscr{P}r(B)=\frac{3}{12}\). Note tha it in important the the sum is less than the whole.
Calculate the probability that A occurs before B.
If I take A and B from your example, I assume I can caluclulate the probability that A occurs before B like so:

[MATH] {P}r(A) \sum_{n=1}^{\infty} {P}r(A) * (1-({P}r(A) + {P}r(B)))^{n-1} [/MATH]
But I am unsure as to how that equates to the following

[MATH] \frac{{P}r(A)}{{P}r(A) + {P}r(B)} [/MATH]
 
A and B are two mutually exclusive events in the same sample space. I am asked to prove that to the probability of event A happening before event B is equal to a/(a+b), where Pr(A) = a and Pr(B) = b. I assume you can calculate the probability like so...

[MATH]sum_{n=1}^{\infty} a*(1-a+b)^{n-1}[/MATH]
I've changed a couple details to make this clearer, and corrected your series to agree with you latest version, by moving some parentheses.

I think this sum is right; but let's make it easier to see by letting c = 1 - a - b, the probability that neither happens. Then your answer is

[MATH]sum_{n=1}^{\infty} (a*c^{n-1})[/MATH]
It should be clear that this is a geometric series. What is its sum?
 
If I take A and B from your example, I assume I can caluclulate the probability that A occurs before B like so:
[MATH]{P}r(A)\sum_{n=1}^{\infty} {P}r(A) * (1-({P}r(A) + {P}r(B)))^{n-1}[/MATH]
But I am unsure as to how that equates to the following[MATH]\frac{{P}r(A)}{{P}r(A) + {P}r(B)}[/MATH]
Maybe your mathematical maturity is not up to this.
The series \(\displaystyle \sum\limits_{n = J}^\infty {a{r^n}} = \frac{{a{r^J}}}{{1 - r}},\;|r| < 1\) is a geometric series that has a limit.
In this case \(\displaystyle J=0,~r=(1-(\mathscr{Pr}(A)+\mathscr{Pr}(B)))~\&~A=\mathscr{Pr}(A)\)
 
I've changed a couple details to make this clearer, and corrected your series to agree with you latest version, by moving some parentheses.

I think this sum is right; but let's make it easier to see by letting c = 1 - a - b, the probability that neither happens. Then your answer is

[MATH]sum_{n=1}^{\infty} (a*c^{n-1})[/MATH]
It should be clear that this is a geometric series. What is its sum?

Thanks. I think I understand now.

[MATH] \sum_{n=0}^{\infty} a*c^n = a * \frac{1-c^\infty}{1-c} [/MATH]
[MATH] c^\infty = 0 [/MATH]
[MATH] 1 - c = 1-(1-(a+b)) = a+b [/MATH]
then

[MATH] \frac{a}{a+b} [/MATH]
Sorry I am still learning notation
 
On a side note. How did they work out the formulas for the first n terms of the geometric sequences?
 
On a side note. How did they work out the formulas for the first n terms of the geometric sequences?
\(\displaystyle
\begin{gathered}
{S_n} = a + ar + a{r^2} + \cdots + a{r^n} \\
r{S_n} = ar + a{r^2} + a{r^3} + \cdots + a{r^n} + a{r^{n + 1}} \\
{S_n} - r{S_n} = a - a{r^{n + 1}} \\
{S_n} = \frac{{a(1 - {r^{n + 1}})}}{{(1 - r)}} \\
\end{gathered}
\)
 
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Just to clarify that I am understanding this right I attempted the following question.

The probability of winning on a single toss of a dice is p. A starts, and if he fails, he passed the dice to B, who then attempts to win on her toss. They continue you tossing the dice back and forth until one of them wins. What are their respective probabilities of winning

Here's what I did

r = common ratio

common ratio = [MATH](1-p)^2[/MATH]
j = 0

[MATH] \sum_{n=j}^\infty pr^n = \frac{pr^j}{1-r} [/MATH]
which is equal to

[MATH] \frac{p}{1-(p-1)^2} [/MATH]
Which is the probability of A winning first and then you just subtract that probability from 1 to find the changes of B winning first.
 
Thank you pka and Dr peterson. I think I learn a lot today. I am beginning to see how geometric sequences are extremely useful. I am sorry my notation is lousy and my understanding patchy. I am learning by myself from books, so I often unintentionally learn things in an inadequate order.
 
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