Question
Sun September 06, 2009 By:

probability

Expert Reply
Mon September 07, 2009

If  X denote the sum of numbers  obtained when two fair dice are rolled then X can take values

2 ,3,4,5,6,7,8,9,10,11,12

Sample space of two dice rolled =

S = { (1,1),(1,2),(1,3),(1,4),(1,5),(1,6)
         (2,1),(2,2),(2,3),(2,4),(2,5),(2,6)
         (3,1),(3,2),(3,3),(3,4),(3,5),(3,6)
         (4,1),(4,2),(4,3),(4,4),(4,5),(4,6)
         (5,1),(5,2),(5,3),(5,4),(5,5),(5,6)
         (6,1),(6,2),(6,3),(6,4),(6,5),(6,6) }

There are 36 outcomes so

P(X=2)= 1/36

P(X=3)=2/36 p(X=4)=3/36 P(X=5)=4/36  P(X=6)= 5/36 P(X=7)=6/36  P(X=8)=5/36  P(X=9)=4/36 P(X=10)=3/36 P(X=11)=2/36  P(x=12)= 1/36

Now calculate E(x)= xP(X=x) =7

Var X= E(X2) -E(X)2

Stand deviation = sqrt of Var (X)

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