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Suppose $X$ and $Y$ are two independant random variable with exponential distribution with paramet $\lambda=1$ and $M=$max{$X$,$Y$}. Then $P(M \ge 4)$ is equal to :

Answer: 0.036

how do i come to this solution?

I tried finding CDF for each X and Y ,since they are both exponential that will give me $F(x)=1-e^{-x}$ & $F(y)=1-e^{-y}$ so $F(m)=1-P(M \le 4)$ so $F(m)=1-P(max{X,Y} \leq 4)$ , and I am stuck right about here!

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P(max(X,Y)≤4) = P(X≤4)P(Y≤4) because independence – Coolwater Jan 30 '14 at 11:25
up vote 3 down vote accepted

Note that $P(M\geqslant4)=1-P(M\lt4)$ and $[M\lt4]=[X\lt4]\cap[Y\lt4]$ hence $P(M\lt4)=P(X\lt4)\cdot P(Y\lt4)$ and $P(M\geqslant4)=1-P(X\lt4)\cdot P(Y\lt4)$. This uses only the independence of $X$ and $Y$.

In your case, $P(M\geqslant4)=1-(1-\mathrm e^{-4})^2=2\mathrm e^{-4}-\mathrm e^{-8}$.

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but isn't this max of one of two? so shouldn't be divided by two since only one of them can occur? – Solix Jan 30 '14 at 11:26
No. $ $ $ $ $ $ – Did Jan 30 '14 at 11:31
its definitely correct your answer but I am curious to know why? for example if you have two bag and wanna draw a marble from it with some colors say 2 red,2 blue for each, then the probability of getting blue would be $\frac{P(blue from bag 1)+P(blue from bag 2)}{2 bags}$. I can be totally lost in probability since it is a very new subject for me. would appreciate if you could elaborate more on that. – Solix Jan 30 '14 at 11:36
Sorry but I fail to see how this is related to the question. – Did Jan 30 '14 at 18:39

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