Find $P(X\ge1|X\le6)$ from the cdf

Question

$$\begin{array}{c|c} X\in&(-\infty,0)&[0,1)&[1,2)&[2,3)&[3,4)&[4,5)&[5,6)&[6,\infty)\\\hline F_X=&0&0.06&0.19&0.39&0.67&0.92&0.97&1\end{array}$$ Find $P(X\ge1|X\le6)$.

This is the Question but I'm confused because most my friends do this $$P=\frac{P(1\le X\le6)}{P(X\le6)}=0.94$$ and I do this

Whom one is doing it right and why?

Answer 1

$1\leq X\leq 6$ includes the point on the boundary ($X=1$), so you must exclude only $X<1$, rather than $X\leq 1$.

Thus your denominator should be:

$$\begin{align}\mathsf P(1\leq X\leq 6)&=\mathsf P(X\leq 6)-\mathsf P(X<1)\\&= F_X(6)-F_X(1^-)&&F_X(x^-):=\lim\limits_{a\nearrow x}F_X(a)\\&=1-0.06\\&=0.94\end{align}$$

Or, if you prefer: $$\begin{align}\mathsf P(X{\,\in\,}[1,6])&=\mathsf P(X{\,\in\,}(-\infty,6]{\,\smallsetminus\,}(-\infty,1))\\&=1-0.06\\&=0.94\end{align}$$

Answer 2

You both use the same correct formula for the conditional probability, but you calculated $F(1)$ incorrectly so you have to fix that.