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I have a word problem question where a "man" is waiting for two buses, bus A or bus B.

Let $T_1(T_2)$ be the time till the next bus A(B) arrives.

$T_1$ and $T_2$ are independent continuous variables. They can be defined as follows:

$T_1$~$Exp(\lambda)$ and $T_2$~$Exp(\mu)$

Let $T$ be the time till the first of these two buses arrives.

I must find $P(T>t)$ and thus find the cdf for $T$.

Can anyone help me with this question? I am aware of the different properties of exponential distributions such as $P(T>t)=1-P(T \le t)= e^{-{\lambda}t}$, although the thing I am struggling is, is having the two different continuous variables.

Krish
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Hint: We have $$\Pr(T\gt t)=\Pr((T_1\gt t)\cap (T_2\gt t))=\Pr(T_1\gt t)\Pr(T_2\gt t).$$

André Nicolas
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  • So that would give: $(1-P(T_1 \le t))((1-P(T_2 \le t))=e^{-\lambda t}e^{-\lambda t}=e^{-2\lambda t}$. That doesn't really sound right to me? – Krish Nov 27 '13 at 23:01
  • The two random variables do not necessarily have the same parameter. It is $e^{-(\lambda+\mu)t}$. – André Nicolas Nov 27 '13 at 23:04
  • Of course, I missed that from the original question. So if that is $P(T>t)$ and I am asked to find the cdf for T (which is equal to $P(T<t)$), would this be as simple as writing $F_T(t)=1-e^{-(\lambda+\mu)t}$? – Krish Nov 27 '13 at 23:09
  • Yes. Mild (and irrelevant in the continuous case) it is $\Pr(T\le t)$. Your answer is correct, $T$ has exponential distribution with parameter $\lambda+\mu$. – André Nicolas Nov 27 '13 at 23:12