I'm trying to answer a textbook question that asks: in a binomial distribution with $n$ trials and probability of success $p$, what value of $k$ successes has the maximum probability?

What I tried is the following ratio:

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

Where do I go from here? I'm not sure how to simplify this. Any help is appreciated.

  • $\begingroup$ Can you articulate why you are looking at that ratio? $\endgroup$
    – paw88789
    Commented Oct 10, 2021 at 13:49
  • $\begingroup$ @paw88789 Using the ratio would allow me to find the last term that is greater than 1 before the ratio becomes less than one, which would be the k with maximum probability $\endgroup$
    – fmtcs
    Commented Oct 10, 2021 at 14:19

1 Answer 1


Hint: You can simplify the fractions quite a bit. Powers of $p$ can be simplified top and bottom; as can powers of $(1-p)$. Also if you write out the binomial coefficients using their factorial definitions, there will be a lot of simplifying with them also.

  • $\begingroup$ I've gotten to $k\lt p(n+1)$. Is the maximum then at $k=p(n+1)$? $\endgroup$
    – fmtcs
    Commented Oct 10, 2021 at 14:32
  • $\begingroup$ I set the inequality so that the numerator has to be greater than the denominator, i.e. the ratio must be greater than 1. But I don't know how to get "the last" $k$ before the ratio goes less than 1. $\endgroup$
    – fmtcs
    Commented Oct 10, 2021 at 14:35
  • $\begingroup$ @statematics If $p(n+1)$ is an integer it's interesting to ask about what is going on when $k=p(n+1).$ If $p(n+1)$ is not an integer, what do you do? $\endgroup$
    – David K
    Commented Oct 10, 2021 at 15:10

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .