Answer:
Step-by-step 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) = [tex]\frac{11}{36}[/tex]\
P(X1=2) =Prob for one die showing two and other die showing 2 to 6
= [tex]\frac{9}{36}[/tex]
P(x1=3) = Prob for one die showing three and other die showing 3 to 6
=[tex]\frac{7}{36}[/tex]
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
