a parricular video game has 1 winner for every 3 lossers. A second game has 17 winner for every 51 lossers .Which game is better odds winning? if third game has 18 lossers how many winners are needed to keep the same ratio as the first game?
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Step-by-step explanation:
the odds of winning in the first game= 1/4
the odds of winner in the second game = 17/68
now let's find out which game has better odd of winning. lcm of 4and 68=68
1/4* 17/17= 17/68
17/68*1/1=17/68
they have the same odd.
second question.
let x be the number of winners
the odds of winner= x/(x+18)
so to have the same ratio= 17/68=x/(x+18
= 17(x+18)=68x
=17x+308=68x
308=68x-17x
x=308/51
=6
the odds in the final game = 6/6+18=6/24
hope this help
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