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