Find $\mathbb{P}(X>6\;|\;X>2)$

Question

Let $X$ be a random variable with density:

\begin{equation} f(x) = \begin{cases} \frac{1}{4}e^{-\frac{x}{4}}, & x\geq0\\ 0, & \text{elsewhere}\\ \end{cases} \end{equation}

1. Find $\mathbb{P}(X>3)$.
2. Find $\mathbb{P}(1\geq X<10)$.
3. Find $\mathbb{P}(X>6\;|\;X>2)$.

My attempt:

I know how to do the first two problems but I don't know how to do the third one. Is there a formula I should use to convert the conditional probability into individual ones?

1. $\mathbb{P}(X>3)=\int_3^\infty\frac{1}{4}e^{-\frac{x}{4}}dx=[-e^{-\frac{x}{4}}]^\infty_3=e^{-\frac{3}{4}}$
2. $\mathbb{P}(1\geq X<10)=\int^{10}_1\frac{1}{4}e^{-\frac{x}{4}}=[-e^{-\frac{x}{4}}]^{10}_1=e^{-\frac{1}{4}}-e^{-\frac{5}{2}}$

Answer 1

Hint.

In general, the conditional probability is given by

$$P(A|B)=\frac{P(A\cap B)}{P(B)}\;.$$

For (3), let $A=\{X>6\}$ and $B=\{X>2\}$. Note that $A\cap B=A$.

So you basically calculate $P(X>6)$ and $P(X>2)$ and then divide.

Alternatively, you may use the memorylessness of the exponential distribution so that you only need $P(X>4)$.