1.Firstly, I think in the first part of the question, you make a mistake on calculation, see image 1, hopefully my solution is the correct one.2.Secondly, why distribution function of U is P(X<=x, y<=x)? could u explain it in more detail please?3.Could u please correct ur solution of part 3 which requires to calculate the joint distribution function based on correct part 1 answer please?thanks a lot :) Step 2: explanation part 2>2. To compute the distribution function and probability density of U, which is the maximum of twoexponential random variables X and Y, we can use a similar approach.a. Distribution function of U: The distribution function F_U(x) for U is the probability that U is less than orequal to x:?F_U(x) = P(U ≤ x) = P(X ≤ x₂ Y ≤ x),ASince X and Y are independent, we can write this as:F_U(x) = P(X ≤ x)*P(Y ≤ x)For exponential random variables with rate parameter λ, the distribution function is given by:F_X(x) = 1- e^(-^x)So:F_U(x) = (1-e^(-A^x))^2 1. If X≤Y₁ l =x: (F_M(x) = 1 - e^(-λx)) XSo, for M = X:F_M(x) = (1 - e^(-λx)) * (1 - P(Y ≤ x))Since Y is also an exponential random variable with rate parameter X, we have:F_Y(x) = 1 - e^(-x)Therefore:F_M(x) = (1-e^(-x))*(1- (1 - e^(-x)))F₁(x) = P(MEX) = P(XEX, Y>x) = PI(XEX) P (Y>X) by independence = P(X{X) (1-P (YEX))P(X≤x,As - Fx(x) = 1 - e²^* = Fycy), we have:--2λxF₁(x) = (1 - e²^*^) (1 - (1 - e²^^ ) ) = (1 - e²^^) e²^^ = e =^^ - e²еX>0so f√(x) =Sf₁(x) = (2x F₁(x) = -de¯^× + 2de¯²^×{0, x < 0

1.Firstly, I think in the first part of the question, you make a mistake on calculation, see image 1, hopefully my solution is the correct one. 2.Secondly, why distribution function of U is P(X<=x, y<=x)? could u explain it in more detail please? 3.Could u please correct ur solution of part 3 which requires to calculate the joint distribution function based on correct part 1 answer please? thanks a lot :)