A discrete random variable X takes values x = 1, 2, 3, with probabilities given by P(X=x) = k x(x+1) , where k is a constant. The conditional probability P(X 3 X 2) is equal to

Step-by-step solution

We know that the sum of all probabilities must be 1 . _ x=1 ^ P(X=x) = 1 _ x=1 ^ k x(x+1) = 1 Using partial fractions, we can write this as a telescoping series: k _ x=1 ^ ( 1 x - 1 x+1 ) = 1 k [ (1 - 1 2 ) + ( 1 2 - 1 3 ) + ] = 1 k(1) = 1 k = 1 Now, we find P(X 2) : P(X 2) = 1 - P(X=1) = 1 - 1 1(2) = 1 - 1 2 = 1 2 Next, we find P(X 3) : P(X 3) = 1 - P(X=1) - P(X=2) = 1 - 1 2 - 1 2(3) = 1 - 1 2 - 1 6 = 1 3 The required conditional probability is: P(X 3 X 2) = P(X 3 X 2) P(X 2) Since X 3 is a subset of X 2 , the int