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How would I calculate the probability of a certain outcome of a probabalistic equation after X trials?

Suppose P = (cos x)^2. Success is when P > 0.5 and we assume x is a truly random number between 0 and 90.

If we use s as some target number of successes, what is an equation to tell us the probability Ps of achieving this target after running N trials?

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    I don't know much about statistics, which is why this isn't an answer, but couldn't you calculate the percentage of the image of P that is greater than .5 to get the probability of a single success, call that p, then wouldn't the probability of s successes be $\frac{N!}{(s!)(N-s)!}p^s$? I wrote out the definition of the binomial coefficient because I'm not sure how to use latex to write it out in the standard way – 2017-01-04

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Since $ x_1, ... , x_n $ are random numbers between $ 0 $ and $ 90 $, then $ x_1, ... , x_n $ ~ $ U[0,90] $.

Let $ Y $ be a Bernoulli random variable such that $ Y = 1 $ if $ cos^2(X) > 0.5 $ and $ Y = 0 $ otherwise. Since $ cos^2(x) > 0.5 \rightarrow cos(x) > \sqrt{0.5} \rightarrow x > 45 $, then $ Y = 1 $ when $ x > 45 $.

What is the probability of any $ X $ being greater than 45?

$$ P(X>45) = \int_{45}^{\infty} f_{X}(x) dx = \int_{45}^{90} \frac{1}{90} dx = \frac{1}{90} x \big|_{45}^{90} = \frac{1}{90} (90-45) = \frac{45}{90} = \frac{1}{2} $$

Therefore, $ Y $ is said to be a Bernoulli random variable with $ p = \frac{1}{2} $.

Now let $ S = \sum_{i=1}^{n} Y_i $, i.e. let S be the sum of $ n $ independent Bernoulli trials. Then $ S $ follows a binomial distribution with parameters $ n $ and $ p $, i.e.

$$ f_{S}(S=k) = \bigg(\frac{n!}{k!(n-k)!}\bigg) p^{k} (1-p)^{n-k} = \bigg(\frac{n!}{k!(n-k)!}\bigg) 0.5^{k} 0.5^{n-k} = \bigg(\frac{n!}{k!(n-k)!}\bigg) 0.5^{n} $$

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    Awesome. Some questions: what is the variable *k* in the last formula? What does the horizontal arrow symbol mean? And I notice that there's no "dividing line" between n! and k!(n-k)! below it in the formula. Is there supposed to be? If not then what does it mean? Thanks. – 2017-01-05
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    @CommaToast 1) $ k $ is an integer greater or equal to 0 and less than or equal to $ n $. If you want to know the probability of having $ 2 $ successes in $ n $ trials, then $ S = 2 $... More generally, if you want to know the probability of having $ k $ successes in $ n $ trials, then $ S = k $. 2) You are right about the dividing line! I've updated my answer accordingly. The term is known as a binomial coefficient, and it can be expressed as $\frac{n!}{k!(n-k)!}$ or ${n}\choose{k}$. More info on the binomial distribution: https://en.wikipedia.org/wiki/Binomial_distribution – 2017-01-05
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    Thanks! I found that Wolfram Alpha can take a sentence input and do the trials for me, so I got what I needed (at least in part). Now I just need to master Hilbert spaces and quantum mechanics. – 2017-01-05