Combination - Probability of at least 5 of 9 independent variables

E

efoc

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Hi all,

Hoping you can help me with a problem, or point me in the right direction. I am trying to calculate the probability of at least 5 of 9 independent variables occurring. The twist here is that each variable has a different probability. My specific probabilities are as follows:

1608217562764.png

However, I'm also hoping to gain a more general understanding of this. I'm aware of combinations, but not sure how to apply that to discrete events with different probabilities. Thanks so much for your help!
 
Your probabilities sum to greater than 1.
I only count 7 probabilities there, you mention 9 independent variables.
 
Romsek said:
Your probabilities sum to greater than 1.
I only count 7 probabilities there, you mention 9 independent variables.
Click to expand...

Right, sorry see below for an updated data set, but no need to solve for this exact instance.

These are each independent Bernoulli variables with outcome either 0 or 1. So A has Pa probability of being 0 and (1-Pa) probability of being 1, for example.


1608227862764.png
 
I do not see a problem with the sum being greater than 1. If you have two biased coins with p(1 coin = H) = .8 and p( 2nd coin = H) = .7 where is the problem with the sum >1
 
efoc said:
Right, sorry see below for an updated data set, but no need to solve for this exact instance.

These are each independent Bernoulli variables with outcome either 0 or 1. So A has Pa probability of being 0 and (1-Pa) probability of being 1, for example.


View attachment 23830
Click to expand...
What is T1?
 
Jomo said:
Pick any 5 out of the 9. Can you tell me that probability?
Click to expand...

right, yes :

nCr = n! / r!(n-r)!

where n is 9 and r is 5.

If Pa = Pb = Pc = ... = Pi then the answer would be Pa^5 * (1-Pa)^4 * nCr.

My issue is in applying this combination formula where Pa =/= Pb
 
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Ok. Given the varying individual probabilities you don't really have any choice but brute force.

You have to take each possible combo of 5, take the product of the elements probability of occurrence, and then sum this
over all the combinations.

Do you have some decent math software? This is a few lines of code in Mathematica.
 
Thanks Romsek - I ended up using an excel formula:

=MOD(QUOTIENT(ROW()-1,2^(COLUMN()-1)),2)

I dragged this across 9 columns and 512 (or 2^9) rows. I filtered for more than 5 or more "ones", and will sum the probability of each possible event.

I did some research and the best solution is through a Poisson Binomial Distribution. However, since excel doesn't have any formulas for this, and since my coding knowledge is weak, I went for the brute force approach.

Thanks for your help.
 
Romsek said:
yeah.

View attachment 23833
Click to expand...
So I'm no expert on Mathematica (I've never used it). But I just ran this and despite some syntax issues I'm picking up that this is not calculating at least 5 out 9, and is only calculating for 5 out of 9 precisely. Am I wrong here?

I double checked my Matrix, and I don't have any duplicate scenarios, they all have at least 5 "True" and i have the appropriate 256 scenarios mapped.
 
efoc said:
right, yes :

nCr = n! / r!(n-r)!

where n is 9 and r is 5.

If Pa = Pb = Pc = ... = Pi then the answer would be Pa^5 * (1-Pa)^4 * nCr.

My issue is in applying this combination formula where Pa =/= Pb
Click to expand...
As you know, Pa = Pb = Pc = ... = Pi is not true. Literally pick 5 out of the 9, say a, c, d, g and h. What is the probability that you pick does exact 5? You need to do this with each way of picking 5 and in the end add up these results.
 
efoc said:
So I'm no expert on Mathematica (I've never used it). But I just ran this and despite some syntax issues I'm picking up that this is not calculating at least 5 out 9, and is only calculating for 5 out of 9 precisely. Am I wrong here?

I double checked my Matrix, and I don't have any duplicate scenarios, they all have at least 5 "True" and i have the appropriate 256 scenarios mapped.
Click to expand...

sigh... at least... bloody ****... righto let me fix it. Yep we get the same answer.

Clipboard01.jpg
 
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Ouch you want at least 5. Possibly, based on Romsek's earlier post, he may have assumed exactly 5, not at least 5.
 
Romsek said:
sigh... at least... bloody ****... righto let me fix it
Click to expand...
No problem - I'm pretty confident in my answer here. This matrix isn't adaptable but I'm aiming to vary the variables through other inputs that I've modelled and determine the impact on the final outcome. Since I'm not changing the number of variables or the threshold of 5, this should work for me.
 
efoc said:
No problem - I'm pretty confident in my answer here. This matrix isn't adaptable but I'm aiming to vary the variables through other inputs that I've modelled and determine the impact on the final outcome. Since I'm not changing the number of variables or the threshold of 5, this should work for me.
Click to expand...

I updated my sheet and our answers agree.
 
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