A video game told me that I had a 90% of stealing an item. I attempted to steal it 3 times, and each time, I failed. I wanted to find what the probability of that happening was. I found a formula for calculating this on mathwarehouse.com:
I plugged my numbers in, but then I realized that the first part of the formula is something I have never seen before in my life. Two exponents? And one of them is before the variable (C)? I have never seen something like that, so my question is what is that called, and how do I solve it?
In the example, 8C2 is simplified to 28. My formula would be 3C3, right? Since I tried 3 times and had 3 outcomes. So how would that simplify and how do you solve it?
In the formula $_n C_k$, $n$ and $k$ are not exponents. This represents the number of ways to choose $k$ objects from $n$ objects. ($n$ and $k$ must be non-negative integers).
$_n C_k$ is commonly notated as $n\choose k$ and is evaluated as:
$\binom{n}{k} = \frac{n!}{(n-k)!k!}$
It is always true that $_n C_n =\binom{n}{n} =1$ and that $\binom{n}{k}=\binom{n}{n-k}$
The formula in your image is the binomial distribution which determines probability of $k$ successes in $n$ independent Bernoulli trials each with probability of success equal to $p$.
In your case the probability of three failures in three attempts is $(0.1)^3$
(suggesting that a life of crime may not be for you)
An example of using the formula where both the combinatorial formula and both the probabilities of success and failure are needed would be the probability of 3 successful steals in 5 attempts:
$\binom{5}{3} (0.9)^3(0.1)^2$
This is used as well in coin toss problems where $p=0.5$
This link should help:
https://en.wikipedia.org/wiki/Binomial_distribution
Added: just noticed that in the original image, $n$ is used where I used $k$ which is more common and that total number of trials has no variable (is commonly $n$).