How to find joint cdf and pmf?

Question

Suppose that you have an urn with 5 white balls and 8 red balls, and you pick 2 balls with replacement.

Let X1 be the random variable that is 1 if the rst ball is white and 0 if not. Let X2 be the random

variable that is 1 if the second ball is white and 0 if not.

(a) Find the joint cdf of X1 and X2.

(b) Find the joint pmf of X1 and X2.

 

Please show your work step by step please so I can learn how to do it.

Answer 1

a)

Total number of balls in urn is 5 white + 8 red = 13 balls .

The Random variable  X₁ is picking first ball from urn .

The successes probability of random variable X₁ is p( X₁=1 ) = 5/13 .

                                                             [Getting white color is the success ]

The failure probability of random variable X₁ is p( X₁=0 ) = 1 - ( 5/13 ) = 8/13 .

The Random variable  X₂ is picking second ball from urn with replacement so the conditions are unchanged for Random variable  X₂ .

The successes probability of random variable X₂ is p( X₂=1 ) = 5/13 .    

The failure probability of random variable X₂  is p( X₂  = 0 ) = 1 - ( 5/13 ) = 8/13 .

The joint commutative distribution function is F( X1 < x₁ , X2 <  x₂) = F(X1 < x₁) * F(X2 <  x₂)  

F( X1 < 1 , X2 <  1) = F(X1 < 1) * F(X2 < 1)  

                            = (5/13)*(5/13) = 25/169 .

So the joint commutative distribution = 25/169 .

 

Answer 2

(b)

Joint probability density functions P(X1 = x , X2 = y) = P(X1 = x) * P(X2 = y)  .

In this situation 4 case are present 

When 2 time the ball is white .

                P(X1 = 1 , X2 = 1) = P(X1 = 1) * P(X2 = 1) = (5/13) * (5/13) = 25/169 

When 1st ball is white and second is red .

             P(X1 = 1 , X2 = 0) = P(X1 = 1) * P(X2 = 0) = (5/13) * (8/13) = 40/169 

When 2 time the ball is red .

                P(X1 = 0 , X2 = 0) = P(X1 = 0) * P(X2 = 0) = (8/13) * (8/13) = 64/169 

When 1st ball is red and second is white .

                P(X1 = 0 , X2 = 1) = P(X1 = 0) * P(X2 = 1) = (8/13) * (5/13) = 40/169