# probability of maximum of two independent random variable

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 :

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!

• P(max(X,Y)≤4) = P(X≤4)P(Y≤4) because independence Commented Jan 30, 2014 at 11:25

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, $P(M\geqslant4)=1-(1-\mathrm e^{-4})^2=2\mathrm e^{-4}-\mathrm e^{-8}$.
• No.   
• 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. Commented Jan 30, 2014 at 11:36