Bernoulli Trials (silly conceptual doubt) Suppose we conduct n independent Bernoulli trials, each with probability of success p. If k is such that the probability of k successes is equal to the probability of k + 1 successes, then


*

*$(n + 1)p = n(1 + p)$

*$np = (n − 1)(1 + p)$

*$np$ is a positive integer

*$(n + 1)p$ is a positive integer


This was a question in my exam. 
To answer this I equated ${^nC_k}~p^k~(1-p)^{n-k}$ to ${^nC_{k+1}}~p^{k+1}~(1-p)^{n-k-1}$.  
Solving this I got $k+1=p(1+n)$, thus I marked answer as part (4), but my teacher says correct answer is part (1). I opened up part(1) to get its value as $[n=p]$, but it did not make sense to me because $n$ can be any natural number but $p$ will be $[0,1]$. Am I making a mistake in my calculation? Please help.  
 A: We have
$$ \frac{n!}{k!(n-k)!} p^k (1-p)^{n-k} = \frac{n!}{(k+1)!(n-k-1)!} p^{k+1}(1-p)^{n-k-1}$$
or
$$(k+1)(1-p)=(n-k)p.$$
Then $k+1-pk-p = pn-pk$, so we have $k+1=pn+p$, or $p=\frac{k+1}{n+1}$.
I agree with your answer, $(n+1)p$ should be a positive integer.
As for claim 1, we do get $n=p$ which is silly.
For claim 2, we get $np=np+n-p-1$, or $n=p+1$, which is also silly. Finally, for claim 3, we have $np$ is a positive integer, but if $p$ is $1/2$, then as was mentioned in the comments, only if $n$ is odd do we have a $k$ such that the probability of $k$ successes is the same as the probability of $k+1$ successes, and then $np$ is definitely not an integer. Therefore none of the other options even make sense. (4) must be correct.
A: A trivial counterexample to demonstrate that option (1) is not true is the random variable $$X \sim \operatorname{Binomial}(n = 3, p = 1/2),$$ with $k = 1$.  Then $$\Pr[X = 1] = \Pr[X = 2] = \frac{3}{8},$$ but $n(1+p) \ne (n+1)p$, and your teacher is busted.
