I'm currently learning discrete binomial distribution in highschool. I'm confused because the following equation is in our textbook:
\begin{equation} \sum _{k=1}^{n}{n \choose k}p^{k}(1-p)^{n-k} = (p+1-p)^{n}=1 \end{equation}
I'm almost certain that the equation is wrong. Shouldn't $k=0$ ? Because if $k$ starts at 1 the probability of 0 successes isn't taking into account. And how is the lefthand side equal to $(p+1-p)^n$ ?
Can anybody explain if and why this equation is either correct or incorrect? Any help is appreciated but don't be mean pls.
By Binomial theorem,
$$ \boxed{(a + b)^n = \sum\limits_{k = 0}^n \ \left( \matrix{ n \cr k \cr } \right) \ a^k b^{n - k} }$$
A binomial random variable $X$ has the probability law given by
$$ P(X = k) = \left( \matrix{ n \cr k \cr } \right) \ p^k q^{n - k}, \ \mbox{where} \ \ p > 0, q > 0, 1 + q = 1.$$
It is common to interpret $p$ as the probability for success and $q$ as the probability for failure in a finite sequence of $n$ Bernoulli trials having only two outcomes. This is how the Binomial Probability law is derived.
We can verify that the total probability is equal to one for the binomial probability distribution.
$$ \sum\limits_{k = 0}^n \ P(X = k) = \sum\limits_{k = 0}^n \ \left( \matrix{ n \cr k \cr } \right) \ p^k q^{n - k} = (p + q)^n = (1)^n = 1 $$
(Note that $p + q = 1$.)