# Binomial distribution p=1

This may be a silly question, but what if you set p=1 in the binomial distribution probability mass function, doesn't the PMF go to zero? Does this mean the probability of k successes in n trials is zero despite the probability of success in one trial is 100%?

• Can you give the relevant formula which you are referring to? Commented Sep 8, 2017 at 17:24

## 3 Answers

There's 2 cases to look at here.

If $k < n$, we have that the probability of $k$ successes in $n$ trials will be $0$. This is because the binomial distribution parameter, $p$ refers to the probability of success for each individual trial. In the case that $p = 1$, we have that every single trial will be a success. This means that in $n$ trials we will have $n$ successes and so $k$ being less than $n$ guarantees that the probability of having $k$ successes in $n$ trials is $0$.

On the other hand, if $k = n$, using the logic from above, we can find that the probability of $k$ successes in $n$ trials is 1.

Therefore, the PMF is not $0$, as the entire mass of the function is contained at $k = n$.

However, as stated in Rohan's answer, having $p = 1$ doesn't really agree with the idea of a binomial distribution.

• Hi. It's just confusing. Because on Wikipedia's page: en.wikipedia.org/wiki/Binomial_distribution, they say, $p \in [0,1]$. Commented Sep 8, 2017 at 22:01
• Since p is a probability, it's restricted to be between 0 and 1 inclusive. Intuitively though, when you think about a binomial distribution, each trial is representing a success or a failure, with p representing the probability of success. When p is 1 (or 0), it's just a trivial case, because every single trial will be a success (or a failure). Commented Sep 8, 2017 at 22:15

Note that a binomial distribution has two possible outcomes: success or failure. As a result, the random variable $X$ in a binomial distribution denotes the no. of successes.

If the probability of success, $p=1$, then it is no longer a binomial distribution!!

Let us make sense of this in PGF of Binomial Distribution:

$$(p + (1-p))^{n} = \sum_{i=0}^{n} {n \choose i}p^{i} (1-p)^{n-i}.$$

You see that $$p=0$$ is also problematic. Because you will have $$0^{0}$$ which is indeterminate and conveniently equal to 1 and the others are zeroes (that is why the PMF is still 1). Since you will sum all terms anyway because they are indistinguishable, you can just use $$1= (p + (1-p))^{n}$$ for any problem and forget about individual terms because one is arbitrary and the others are $$0$$.