If an event has a probability of, say, 70% of occurring and 30% of not
occurring, how can the probabilities be calculated that this event
will occur none, once, twice, 3 times or all 4 times when attempted,
say, 4 or n times?

Hi!!

This is a typical problem of Binomial Distribution, see for references
the following pages:
"The Binomial Distribution":
http://www.stat.tamu.edu/stat30x/notes/node66.html

"Binomial distribution (1 of 3)":
http://davidmlane.com/hyperstat/A2301.html


Acording to the above pages the formula for the case where the number
of trials is n=4 we have:

P(X=0) = (4!/0!4!) * (0.7)^0 * (0.3)^4 = 
       = 1 * 1 * 0.0081 = 
       = 0.0081 (or 0.81%)

P(X=1) = (4!/1!3!) * (0.7)^1 * (0.3)^3 =
       = 4 * 0.7 * 0.027 =
       = 0.0756 (or 7.56%)

P(X=2) = (4!/2!2!) * (0.7)^2 * (0.3)^2 =
       = 6 * 0.49 * 0.09 =
       = 0.2646 (or 26.46%)

P(X=3) = (4!/3!1!) * (0.7)^3 * (0.3)^1 =
       = 4 * 0.343 * 0.3 =
       = 0.4116 (or 41.16%)

P(X=4) = (4!/4!0!) * (0.7)^4 * (0.3)^0 =
       = 1 * 0.2401 * 1 =
       = 0.2401 (or 24.01%)



For n trials and x succeses just apply the formula:

P(X=x) = (n!/x!(n-x)!) * (0.7)^x * (0.3)^(n-x)



For more visit the following page, which has a Java applet where you
can confirm the results and links to more examples:
"The Binomial Distribution":
http://www.stat.wvu.edu/SRS/Modules/Binomial/binomial.html



Search strategy:
My own knowledge to find the topic and the keyword "binomial
distribution" at Google.com


I hope this helps you. If you find something unclear please do not
hesitate to request for a clarification, I will be glad to give you
further assistance on this question if you need it.


Regards,
livioflores-ga

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Request for Answer Clarification by vaac-ga on 04 Jul 2006 14:22 PDT

Thanks for your answer and please excuse my poor background in
statstics. Should, for istance,your third equation be read:

P(X=3) = (4!/(3!*1!)) * (0.7)^3 * (0.3)^1 =
       = 4 * 0.343 * 0.3 =
       = 0.4116 (or 41.16%)
or
P(X=3) = ((4!/3!)*1!) * (0.7)^3 * (0.3)^1 =
       = 4 * 0.343 * 0.3 =
       = 0.4116 (or 41.16%)

---

Clarification of Answer by livioflores-ga on 05 Jul 2006 06:20 PDT

Hi!!

The correct version is:
P(X=3) = (4!/(3!*1!)) * (0.7)^3 * (0.3)^1 =
       = 4 * 0.343 * 0.3 =
       = 0.4116 (or 41.16%)

Note that the formula to be used to solve problems like this one is:

P(X=x) = n!/[x!*(n-x)!)] * P^x * (1-P)^(n-x)

where:
n is the number of trials
x is the number of successes
P is the probability for a success per trial (in this problem is 0.7)


Hope this clarify the answer for your better understand.

Regards,
livioflores-ga

Very good and to the point answer.Hope I will not have any problems
with it in the future