Question

If ${\text{X}}$ follow a binomial distribution with parameters ${\text{n = 100}}$ and ${\text{p = }}\dfrac{1}{3}$ then ${\text{P(X = r)}}$ is maximum when $r =$___________

Answer 1

This problem comes under probability in binomial distribution which means the possible outcomes of the trial has two that is success or failure. There must be a fixed number of trails $\left( n \right)$. First we need to find the mode of binomial distribution and complete step by step explanation.

Complete step-by-step answer:
Let $r$ is mode $P(X = r)$
We need to solve in two cases for the mode of binomial distribution because the values may be an integer and not an integer.
Case (i) : The mode of binomial distribution is $(n + 1)p$ is an integer and there are two modal values.
They are $(n + 1)p{\text{ and }}(n + 1)p - 1$
Thus case (i) does not exist because there may be two values.
Case (ii) : The mode of binomial distribution is $(n + 1)p$ is not an integer and there exists only unique modal value.
Like previous case the model values exists same but unique and has only one value and integral part of $(n + 1)p$
Thus case (ii) exists, we get
The given binomial distribution has parameters
$\Rightarrow n = 100$
$\Rightarrow p = \dfrac{1}{3}$
Now, substitute $n$ and $p$ value in integral part of binomial distribution,
$\Rightarrow (n + 1)p$
$\Rightarrow (100 + 1)\dfrac{1}{3}$
Now, add values and the value will be in the mixed fraction and we want to convert in improper fraction, we get
$\Rightarrow \dfrac{{101}}{3}$
Now, diving the fraction we get
$= 33.67$ which is not an integer.
Hence, the unique mode is $33$, by taking an approximate value.
$\therefore P(X = r)$ will be maximum when $r = 3$

Note: The problem has to be in the form of unique integral value, for that we have found in two cases. Like this type of problems need attention on mode and parameters this type comes when there are a fixed number of trials $\left( n \right)$.