Binomial Random Variable with unknown probability $p$, with the condition $P\left( |X-E(X)|\le 3 \right)=1$. How to find range of $p$

Question

I am struggling with this questions on Binomial Random Variable that requires to find the range of parameter $p$.

Let $X$ be a binomial random variable with parameters $(5, p)$. Find the range of $p$ for which $P\left( |X-E(X)|\le 3 \right)=1$.

Unable to use the relation $P\left( |X-E(X)|\le 3 \right)=1$.

My approach is, to assume $|X-E(X)|\le 3$ to true as probability of the same is given as $1$.

Answer 1

As $X$ ~ $B(5, p)$, $E[X] = 5p$

$ \displaystyle \begin{align} P(|X-E[X]| \leq 3) &= P(|X-5p| \leq 3) \\ &= P(-3 \leq X-5p \leq 3) \\ &= P(5p-3 \leq X \leq 5p+3) \\ &= 1 \end{align} $

It means that every $X$ is in $[5p-3, 5p+3]$

We know that, $X$ varies from $0$ to $5$.

Thus, we get

$ \begin{cases} 5p-3 \leq 0\\ 5p+3 \geq 5 \end{cases} $

which implies $\displaystyle\frac{2}{5} \leq p \leq \frac{3}{5}$

Answer 2

Given that $P\left( |X-E(X)|\,\le 3 \right)=1$, that means it is definite event and $|X-E(X)|=1$

Since $E(X)=np=5p$

$\Rightarrow \,|X-np|\,\le 3$

Since $X$ is a Random Binomial Variable with parameters $(5,\,p)$, the value of $X$varies from $0$ to $5$. And absolute difference of $X$ and $np$ should be less than or equal to $3$. So $np$ can be within the shaded portion of the line shown below so that its absolute difference from $X$never exceeds $3$.

So, $2\le np\le 3$

$\Rightarrow \frac{2}{3}\le p\le \frac{3}{5}$