Calculating P(X>6|X<8) from pdf

Question

I have a pdf $\frac{k}{x^2}$ for $4<x<\infty$ ($0$ otherwise) and need to work out the $P(X>6 | X<8)$ I already worked out k to equal 4.

To work out this probability I did $\frac{P(X>6)}{P(X<8)}$ but I get a weird answer/

For $P(X>6)$ i did $\int_{6}^{\infty} \frac{4}{x^2}dx = \frac{2}{3} $

and for $P(X<8)$ i did $\int_{4}^{8} \frac{4}{x^2}dx = \frac{1}{2} $

My answer of $\frac{3}{4} $ is obviously wrong, but I am not sure where I went wrong.

Answer 1

You should compute \begin{align}\frac{P(6<X<8)}{P(X<8)}&=\frac{P(X<8)-P(X \le 6)}{P(X<8)}\\&=1-\frac{P(X \le 6)}{P(X < 8)} \\&=1-\frac{1-P(X > 6)}{P(X < 8)}\\ \end{align}

Answer 2

$$P(X>6 | X < 8) = \frac{P((6<X)\cap (X<8))}{P(X<8)} = \frac{P(6 < X<8)}{P(X<8)} = \frac{k\int_6^8\frac{dx}{x^2}}{k\int_4^8\frac{dx}{x^2}} = \frac{\int_6^8\frac{dx}{x^2}}{\int_4^8\frac{dx}{x^2}}$$

Note that the probability density function lives on $(4,\infty$). That's why the lower integral starts at 4.