We are given the following equality:
$B(k;n,p)=B(k;n+1,p) + pb(k;n,p)$
where $B$ is the binomial cdf, $b$ is the binomial pdf, $n$ is the number of trials and $p$ is the probability of success.
How can we show this equality using a probabilistic interpretation, without using any algebra?
I am going to answer my own question, I think I figured it out but I would like to have some feedback.
Define the events $A$="Having at most $k$ successes in $n$ trials") and $B$="Having at most $k$ successes in $n+1$ trials". If you have at most $k$ successes in $n+1$ trials, then you have at most $k$ successes in n trials, so $B\subseteq A$. However, if you have at most $k$ successes in n trials, this does not imply that you have at most $k$ successes in $n+1$ trials, because you could have exactly $k$ successes in the first $n$ trials and succeed in the last trial, which would give you $k+1$ successes. So $A \nsubseteq B$, which implies $B⊂A$. We know that when $B⊂A$, $P\left( A\right) =P\left( B\right) +P\left( A-B\right)$.
Note that $P(A)=B(k;n,p)$ and $P(B)=B(k;n+1,p)$. $A-B$ are the events we described before that are contained in $A$ but not in $B$: having $k$ successes in $n$ trials and success in the $n+1$th trial. The probability of such events is $pb(k;n,p)$, so $B(k;n,p)=B(k;n+1,p)+pb(k;n,p)$
