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I need to apply Bayes' theorem for a conditional probability which in turn makes use of continuous random variables.

Context: A continuous random variable $X$, expresses in minutes, the duration of the telephone communications. Its density function is as follows: if $x>0$ $\Rightarrow f(x)=(5/6)ℯ^{-(5/6)x}$

  • I have $F_X(x) = 1-ℯ^{-(5/6)x}$
  • I need to calculate $\mathsf P(X>5\mid X>2)$

How can I express $\displaystyle P(A_i|B) = \frac{P(B|A_i) P(A_i)}{\sum_{k=1}^n P(B|A_k) P(A_k)}.... [1]$ through integrals?

If you like to contribute something more about the exercise, it is also welcome. Thank you very much.

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Hint: $$ \Pr(X>5\mid X>2) = \frac{\Pr(X>5,X>2)}{\Pr(X>2)} = \frac{\Pr(X>5)}{\Pr(X>2)} = \frac{1-F_X(5)}{1-F_X(2)}.$$

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    That's because if $A$ the call lasts for more than 5 minutes then $B$ has lasted for more than 2 minutes, so $A→B\Rightarrow A \subseteq B \Rightarrow A\cap B=A$. Or why is it? – 2017-02-14
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    Yes you are right. $\{\omega:X(\omega)>5\} \cap \{\omega:X(\omega)>2\} = \{\omega:X(\omega)>5\}.$ – 2017-02-14