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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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  • $\begingroup$ P(max(X,Y)≤4) = P(X≤4)P(Y≤4) because independence $\endgroup$
    – Coolwater
    Commented Jan 30, 2014 at 11:25

1 Answer 1

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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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  • $\begingroup$ but isn't this max of one of two? so shouldn't be divided by two since only one of them can occur? $\endgroup$
    – Solix
    Commented Jan 30, 2014 at 11:26
  • $\begingroup$ No. $ $ $ $ $ $ $\endgroup$
    – Did
    Commented Jan 30, 2014 at 11:31
  • $\begingroup$ 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. $\endgroup$
    – Solix
    Commented Jan 30, 2014 at 11:36
  • $\begingroup$ Sorry but I fail to see how this is related to the question. $\endgroup$
    – Did
    Commented Jan 30, 2014 at 18:39

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