Let X be a discrete random variable with possible values {0,1,2,...} and the following probability mass function:P(X=0) = 4/5 and for k belongs to {1,2,3,...},P(X=k) = 1/10 * (2/3)^k. (a) Verify that the above is a probability mass function (b) For k belongs tp {1,2,...}, find P(X>=k | X >=1)

Q5 Let = {} 1 f(x) = x = 1, 2, 3, 4, 5, 6, otherwise. be the probability mass function of a discrete random variable X. Show that the probability mass function of the smallest observation Y₁ of a random sample of size 5 from this distribution is 5 91 (y₁) = = (²-")" - (0 =³) "₁ 6 6 y = 1, 2, 3, 4, 5, 6. and zero otherwise. Note: you already know how to write down the cumulative distribution function of Y₁, the smallest of X₁, X₂, X3, X₁, and X5; remembering that we are sampling from a discrete distribution, derive the probability mass function of Y₁.

3) For known k ≥ 2 show that the negative binomial distribution with probability mass function, f(y, k, μ) y = 0, 1, 2, .. = - 1 k k k (³ + k − ¹) (¹ − µ 4 k)" (1² k) " 1 Y + 1 belongs to the exponential family of distributions. Find the natural parameter for this distribution.

The joint PDF of two random variables X,Y is given byfXY(x, y) = {k e^(-y-x/4) , x > 0, y > 0                    0,                  otherwisea)  What is the value of k ?b)  Find the probability P( X<Y ).c)  Find P( 0<X<2 ).