Keep in mind, I've only just begun to delve into probability, and I'm still trying to wrap my head around the methods and the reasons for using certain formulas and what they actually mean. My question is less for the progressing my understanding of the mathematics, and more for progressing my interests in this game.
For any given event, where n-quantity of dice are rolled, a certain number of successes are required to pass. On a single die, a result of 1, 3, or 5, represent 0 successes, while 2 and 4 represent 1 success, and 6 represents 2 successes.
The given outcomes and their probabilities on any single trial are as follows;
- Outcome A: 0 Success 50%
- Outcome B: 1 Success 33%
- Outcome C: 2 Success 17% (Rounded)
Let's say I roll 5 Dice, and need 2 successes at least to pass. If I've done this correctly (Using Binomial Distribution), and assuming I haven't already overthought this entire process, my odds of getting AT LEAST 2 instances of Outcome B is 53.25%, and my odds of getting AT LEAST 1 instance of Outcome C is 60.61%.
How do I calculate my odds of Overall Success? Do I add these percentages (113.86% (Impossible))? Do I add the odds of failure for each (The inverse)((1-53.25%)+(1-60.61%)=86.14%)? or am I missing the process entirely?
I've read several questions similar to mine, but I can't seem to find an answer.
You fail if you have exactly $0$ or exactly $1$ successes. The probability of the first is $\left({1\over2}\right)^5$, and the probability of the second is ${5\choose1}\cdot{1\over3}\cdot\left({1\over2}\right)^4$. Therefore you pass with probability $$p=1-\left({1\over32}+{5\over48}\right)={83\over96}=86.46\%\ .$$