The relation of $P(X=x+1)$ and $P(X=x)$ in binomial distribution

Question

If I substitute the values to the binomial probability theory, it appears as such

$${n \choose x+1} p^{x+1} (1-p)^{n-x-1}$$

I don't know how to move on...

What am I doing wrong, or are you expected to approach this some other way?

Answer 1

$$\begin{align}\require{cancel} \Pr[X=x]&={n\choose {x}}p^{x}(1-p)^{n-x}\\ \Pr[X=x+1]&={n\choose {x+1}}p^{x+1}(1-p)^{n-(x+1)}\\ &={n\choose x+1}\left(\dfrac p{1-p}\right) p^x(1-p)^{n-x}\\ &={n\choose x+1}\left(\dfrac p{1-p}\right)\dfrac{\Pr[X=x]}{n\choose x}\\ &=\dfrac {n\choose x+1}{n\choose x}\left(\dfrac p{1-p}\right)\Pr[X=x]\\ &=\dfrac {\left[\dfrac {\cancel{n!}}{(x+1)!(n-x-1)!}\right]}{\left[\dfrac {\cancel{n!}}{x!(n-x)!}\right]} \left(\dfrac p{1-p}\right)\Pr[X=x]\\ &=\dfrac{x!(n-x)!}{(x+1)!(n-x-1)!}\left(\dfrac p{1-p}\right)\Pr[X=x]\\ &=\dfrac{\cancel{x!}\left[(n-x)\cancel{(n-x-1)!}\right]}{\left[(x+1)\cancel{x!}\right]\cancel{(n-x-1)!}}\left(\dfrac p{1-p}\right)\Pr[X=x]\\ &=\left(\dfrac {n-x}{x+1}\right)\left(\dfrac p{1-p}\right)\Pr[X=x] \qquad \blacksquare \end{align}$$

Answer 2

Recall that if $X \sim \mathrm{Binomial}(n,p)$, then $$\Pr[X = x] = \binom{n}{x} p^x (1-p)^{n-x}, \quad x = 0, 1, 2, \ldots, n.$$ Then use this to calculate the ratio $$\frac{\Pr[X = x+1]}{\Pr[X = x]},$$ being careful to cancel like terms.