Latest Elitmus Aptitude Question SOLUTION: A & B throws a pair of dice and winner is one who gets a sum of 9 first.if A starts throwing first Then what is the ratio of A winning over B?

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(#M40157786) ELITMUS Numerical Ability question Keep an EYE

A & B throws a pair of dice and winner is one who gets a sum of 9 first.if A starts throwing first
Then what is the ratio of A winning over B?

Asked In Elitmus Sniper (7 years ago)

Unsolved Read Solution (4)

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AMIT KUMAR  (7 years ago)

scince two dice are thrown together
therefor total no of ways = 6*6 = 36
science they wants 9 thatsby that may be (3,6) or (6,3) or (5,4) or (4,5)
total no of ways for winning = 4
probablity of winning of A = 4/36 = 1/9
science A starts the game then B will only win when A will lose
probablity of lossing of A = 32/36 = 8/9
therefor probablity of winning of B
=prbablity of lossing A*probablity of winning of B when B starts the game
= 32/36*4/36
= 8/81

Required ratio = (1/9)/(8/81) = 9 : 8

Is this solution Helpfull? Yes (18) | No (1)  

pp  (7 years ago)

probability of a person throwing first and winning is given by 1/(2-p)
1/(2-(1/9))=9/17 prob of a
prob of b is 8/17
prob (a/b) 9/8

Is this solution Helpfull? Yes (3) | No (9)  

neha  (7 years ago)

total no of ways =36
probability of A to win =
there are 4 cases if he gets (3,6) (6,3) (5,4) (4,5)
probability of winning of A =4/36
=1/9
probability of losing=1-(1/9)
=8/9 =probability of winning of B
REquired ratio=1/9():(8/9)
=1:8

Is this solution Helpfull? Yes | No (4)  

Pritesh Saha  (4 years ago)

1/9 + ( 8/9 x 8/9 x 1/9 ) + ( 8/9 x 8/9 x 8/9 x 8/9 x 1/9 ) + ......
1/9 [ 1 + (8/9)^2 + (8/9)^4 + (8/9)^6 +....infinite G.P series ]
1/9 x [ 1/(1 - (8/9)^2)]
9/17

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