Question

A die is thrown twice. Let X1 and X2 denote the outcomes, and define random variable X to be the minimum of X1 and X2. Determine the distribution of X.

Answer 1

Explanation:

Given that a die is thrown twice.

X1, X2 are the outcomes in 2 throws

X1 = minimum of two

X1 can be either 1 or 2...6

Total outcomes are 36.

For x1 =1, fav ourable outcomes are (1,1) (1,2)...(1,6) (6,1)...(2/,1)= 11

P(X1=1) =
`(11)/(36)`\

P(X1=2) =Prob for one die showing two and other die showing 2 to 6

=
`(9)/(36)`

P(x1=3) = Prob for one die showing three and other die showing 3 to 6

=
`(7)/(36)`

thus we find that probability is reducing by 2 in the numerator

P(x1=4) = 5/36 followed by 3/36 for 5 and 1/36 for 6