Quote Originally Posted by braddy
HI,
captain I need your help for some problem I have with the probability.

Problem 1
2 random variables are independent and each has a binomial distribution with success probability of 0.3 and 2 trials.
1- find the joint probability distribution.
Here I tried to compute the binomial but I get 0!
b(1,2,0.3)=0

2- find the probability that the second is greater than the first. ??
RV X and Y are identicaly independent distributed with distribution B(0.3,2)

Hence their joint distribution is the product of their individual distributions:

p(X=x, Y=y) = 2!/[(2-x)! x!] 0.3^x (1-0.3)^(2-x) 2!/[(2-y)! y!] 0.3^y (1-0.3)^(2-y)

x=0, 1 or 2 y=0, 1 or 2.

p(X>Y) = p(X=0, Y=1 or 2) + p(X=1, Y=2) = p(X=0,Y=1) + p(X=0,Y=2) + p(X=1,Y=2)

which can be evaluated from the joint distribution given above.

RonL