n(S)=36 The probability of throwing 9 with two dice {(3,6),(4,5),(5,4),(6,3)}=
4
36
=
1
9
∴ probability for not throwing 9=1−
1
9
=
8
9
∴ If A is to win, he should throw dice alternately (i.e., 1st,3rd and so on ) to get 9 and obviously for B is to win. He should throw dice in 2 nd, 4 th and so on. ∴ Probability that ' B ' wins =(