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Footballguy
Can someone help me with a math problem?
10 independent events each with an 85% chance of occurring.
What is the probability that it doesn't happen in 9/10 cases? 8/10 cases? 7/10 cases? etc
@caustic This has been very helpful to my original question above--thanks. I'd like to tweak the question and assumptions slightly.So if this is a binomial distribution with P(success) = .85, P(failure) = .15, and 10 trials, then
P(9 failures): 10!/(9!1!) * (0.85)^1 * (.15)^9 = .0000003
P(8 failures): 10!/(8!2!) * (0.85)^2 * (.15)^8 = .000008
P(7 failures): 10!/(7!3!) * (0.85)^3 * (.15)^7 = .0001
etc.
First, based on the above, I calculated the probability of 6 failures to be 0.12% and the probability of 5 failures to be 0.85%.
However what if I tweak the question to read what is the probability of 6 or more failures? And also change the chance of success to 88%. Can you show me the equation? Thanks.
	
		
  
  gains, but I feel like this is only the beginning of the ride up for these 2.