A & B THROW A DICE ALTERNATIVELY TILL ONE OF THEM GETS A '6' AND WINS THE GAME. WHAT WILL BE A PROBABILITY?  Choices.
A) 4/11  B) 5/11  C) 6/11  D) 7/11  E) 8/11
Whoever starts first will have a probability of 6/11 and the other has a probability of 5/11. So, if A starts, as has to be logically inferred from the question, pbbl that A wins = 6/11 and that B wins = 5/11.   Pbbl that the starting person wins = (1/6) + (5/6)*(5/6)*(1/6) + (5/6)*(5/6)*(5/6)*(5/6)*(1/6) + ... = G.P. with first term (1/6) and common ratio (5/6)*(5/6) = (1/6)/[1-(5/6)*(5/6)] = 6/11   ANSWER:6/11
<result>C</result>
A & B THROW A DICE ALTERNATIVELY TILL ONE OF THEM GETS A '6' AND WINS THE GAME. WHAT WILL BE A PROBABILITY?