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

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(#M40000339) SELF question Probability Keep an EYE

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

akash nagpal (14 years ago)

Solved C.T.Narayanan Read Solution (5)

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C.T.Narayanan  (14 years ago)

if A starts to play first , his/her prob will be 6/11 and B's prob is 5/11

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Tijo Jose  (14 years ago)

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

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avinash reddy  (14 years ago)

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

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abhijeet nand kumar  (14 years ago)

one who starts probability will be 6/11
so the pro. of other will be =1-6/11=5/11

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devender singh saini   (14 years ago)

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

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