A B THROW A DICE ALTERNATIVELY TILL ONE OF THEM GETS A 6 AND WINS THE GAME WHAT WILL BE

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

Pbbl that the other wins = 1 - Pbbl that starting person wins (as either one of them has to win ultimately) = 1-6/11 = 5/11
15 years agoHelpfull: Yes(1) No(0)
hey u must say ........who starts the trow 1st ................
if A starts 1st den the probability of A wining .....is 1/6+(5/6)^2*1/6+(5/6)^4*1/6-----------infinity....so =6/11...........similarly appying 4r B we the probability of B win the game is 5/11
15 years agoHelpfull: Yes(1) No(1)
one who starts probability will be 6/11
so the pro. of other will be =1-6/11=5/11
15 years agoHelpfull: Yes(0) No(0)
if A starts p(A)=1/6+1/6*(5/6)*(5/6)+1/6*(5/6)*(5/6)*(5/6)*(5/6)+........
=6/11
P(B)=1-6/11=5/11
if B starts then prob will interchange
15 years agoHelpfull: Yes(0) No(0)

A & B THROW A DICE ALTERNATIVELY TILL ONE OF THEM GETS A '6' AND WINS THE GAME. WHAT WILL BE THEIR RESPECTIVE PROBABILITIES?