Consider the experiment where a pair of fair dice is thrownS

Consider the experiment where a pair of fair dice is thrown.

all possible outcomes of thwrow of 2 dice are--

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)} so n(S)=36

P[X=1] means that the maximum number appearing on either of the die is 1.

hence only possible outcome out of the 36 possible outcomes is (1,1)

so P[X=1]=1/36

P[X=2] means that the maximum number appearing on either of the die is 2. so possible outcomes are (1,2),(2,1),(2,2)

so P[X=2]=3/36

P[X=3] means that the maximum number appearing on either of the die is 3.

possible outcomes are (3,1),(1,3),(3,2),(2,3),(3,3)

so P[X=3]=5/36

P[X=4] means that the maximum number appearing on either of the die is 4.

possible outcomes are (4,1),(1,4),(4,2),(2,4),(4,3),(3,4),(4,4)

so P[X=4]=7/36

P[X=5] means that the maximum number appearing on either of the die is 5.

possible outcomes are (5,1),(1,5),(5,2),(2,5),(5,3),(3,5),(5,4),(4,5),(5,5)

so P[X=5]=9/36

P[X=6] means that the maximum number appearing on either of the die is 6.

possible outcomes are (6,1),(1,6),(6,2),(2,6),(6,3),(3,6),(6,4),(4,6),(6,5),(5,6),(6,6)

so P[X=4]=11/36

further P[X is divisible by 3]=P[X=3 or X=6]=P[X=3]+P[X=6]=5/36+11/36=16/36 [answer]