Probability of multiple events - Not sure how to articulate it

Hi,

So I've got a data set with 6 events, all with a 65% of occurring. I don't know how to form the equation to solve this, much less answer the probability of 2, 3, etc occurring.

For 1 event occurring I rationalized it should be A, B, C, D, E, F = 65%; P(1) = [((1-A) * (1-B) ... * (1-F))-1]*-1

But for 2 events I'm at a conceptual wall. Do I halve the results of P(1)? Do I use once less data point? Both approaches seem unnaturally inelegant and the results don't seem correct either.

Lexzl said:

So I've got a data set with 6 events, all with a 65% of occurring. I don't know how to form the equation to solve this, much less answer the probability of 2, 3, etc occurring.

Click to expand...

This is a simple binomial probability. The probability of success is 0.65 & failure is 0.35.
That is true for any of the elementary events. The probability of exactly two events occurring is \(\displaystyle \dbinom{6}{2}(.65)^2(.35)^4\).

pka said:

This is a simple binomial probability. The probability of success is 0.65 & failure is 0.45.
That is true for any of the elementary events. The probability of exactly two events occurring is \(\displaystyle \dbinom{6}{2}(.65)^2(.45)^4\).

Click to expand...

Whoopsie!

Probability of failure is 0.35 (=1 - 0.65)

Thanks both of you! That worked like a charm.

Now (and I hope I'm saying this correctly) I'm trying to optimize the population coefficient for a 4 success or better being at 70% or better. It looks like I have to just do trial and error, though, because I'm learning the hard way how hard it is to factor a factorial and logarithms with the same unknown base. I'm going to attempt to write out my work, but I'm just borrowing the code you guys have already made available to attempt to write it out. Oh also I was wrong about my initial probability, it turns out they have a 50% chance which thankfully makes things a bit easier to conceptualize. EDIT: So I realized I shoehorned myself into a max 6 situation because I can't infinitely insert binomial probabilities into the equation. Well, maybe with some calculus but I don't have that skill set. Still I do enjoy mulling over a good problem, and I'm not 100% sure I worked this out correctly below. If anyone has any suggestions on how to better solve this, or correct my work, please let me know.

\(\displaystyle [\dbinom{x}{4}(.5)^4(.5)^{x-4}] + [\dbinom{x}{5}(.5)^5(.5)^{x-5}] + [\dbinom{x}{6}(.5)^4(.5)^{x-6}] = 0.7\);

Expand
\(\displaystyle [(\dfrac{x!}{4!(x-4)!})(.5)^4(.5)^{x-4}] \,+\, [(\dfrac{x!}{5!(x-5)!})(.5)^5(.5)^{x-5}] \,+\, [(\dfrac{x!}{6!(x-6)!})(.5)^6(.5)^{x-6}] = 0.7\)

Expand
\(\displaystyle [(\dfrac{x!}{4!(x-4)!})(0.0625)(.5)^{x-4}] \,+\, [(\dfrac{x!}{5!(x-5)!})(0.03125)(.5)^{x-5}] \,+\, [(\dfrac{x!}{6!(x-6)!})(0.015625)(.5)^{x-6}] = 0.7\)

Multiplied out the 4!(x-4)!
\(\displaystyle [(x!)(0.0625)(.5)^{x-4}] \,+\, [(\dfrac{x!}{5(x-5)})(0.03125)(.5)^{x-5}] \,+\, [(\dfrac{x!}{30x^2-360x+1080})(0.015625)(.5)^{x-6}] = 16.8(x-4)!)\)

Divided out 0.5^x-4 and (possibly erroneously) got 0.5^-1 and 0.5^-2 which I tried to simplify as positive exponents which gave whole integers.
\(\displaystyle [(x!)(0.0625)] \,+\, [(\dfrac{x!}{5(x-5)})(0.03125)(2)] \,+\, [(\dfrac{x!}{30x^2-360x+1080})(0.015625)(4)] = \dfrac{16.8(x-4)!}{0.5^{x-4}})\)

Combine like terms
\(\displaystyle [(x!)(0.0625)] \,+\, [(\dfrac{x!}{5(x-5)})(0.0625)] \,+\, [(\dfrac{x!}{30x^2-360x+1080})(0.0625)] = \dfrac{16.8(x-4)!}{0.5^{x-4}})\)

Divide everything by 0.0625. Finally broke down those parenthesis.
\(\displaystyle x! \,+\, \dfrac{x!}{5(x-5)} \,+\, \dfrac{x!}{30x^2-360x+1080} = \dfrac{1.05(x-4)!}{0.5^{x-4}})\)

From here on out I'm even less comfortable than I was before. Logs. Not even sure how to type them. Multiplied 0.5^x-4 back over.
\(\displaystyle 0.5^{x-4}(x! \,+\, \dfrac{x!}{5(x-5)} \,+\, \dfrac{x!}{30x^2-360x+1080}) = 1.05(x-4)!\)

Not sure how or if I can even proceed from here.