Let x1, . . , xn be independent exponential random variables having a common parameter λ. determine the distribution of min(x1, . . , xn)

Given fx (x) = λe^λx

Fx (x) = 1 – e^-λx      x…0

To find distribution of Min (X1,….Xn)

By applying the equation

fx1 (x) = [n! / (n – j)! (j – 1)!][F(x)]^j-1[1-F(x)]^n-j f(x)

For minimum j = 1

[Min (X1,…Xn)] = [n!/(n-1)!0!][F(x)]^0[1-(1-e^-λx)]^n-1λe^-λx

= ne^[(n-1) λx] λe^(λx)

= nλe^(-λx[1+n-1])

= nλe^(-nλx)

i can try and you the very best that i can.

yo answer should be 6 if im not mistaken

step-by-step explanation: