A particular video game has 1 winner for every three losers. a second game has 17 winners for every 51 losers.which game has better odds of winning?if a third game has 18 losers , how many winners are needed to keep the same ratio as the first game?
Answers
Answer:
The first game and second game have the same odds of winning, i.e. 0.25
The number of winners required is 6
Step-by-step explanation:
The probability of winning in a game if given as the number of winners divided by total number of players in the game.
In the first game;
Number of winners= 1
Number of players= 1+3=4
Probability of winning is;
In the second game;
Number of winners= 17
Number of players= 17+51= 68
Probability of winning is;
The first game and second game have the same odds of winning, i.e. 0.25
In the third game;
Number of winners= x
Number of players= x+18
Probability of winning= 0.25
The number of winners required is 6
Step-by-step explanation:
The probability of winning in a game if given as the number of winners divided by total number of players in the game.
In the first game;
Number of winners= 1
Number of players= 1+3=4
Probability of winning is;
Prob=\frac{1}{4}= 0.25Prob=
4
1
=0.25
In the second game;
Number of winners= 17
Number of players= 17+51= 68
Probability of winning is;
Prob=\frac{17}{68} =0.25Prob=
68
17
=0.25
The first game and second game have the same odds of winning, i.e. 0.25
In the third game;
Number of winners= x
Number of players= x+18
Probability of winning= 0.25
\begin{gathered}\frac{x}{x+18}=0.25\\ x=0.25*(x+18)\\x=0.25x+4.5\\x-0.25x=4.5\\0.75x=4.5\\x=\frac{4.5}{0.75}\\ x=6\end{gathered}
x+18
x
=0.25
x=0.25∗(x+18)
x=0.25x+4.5
x−0.25x=4.5
0.75x=4.5
x=
0.75
4.5
x=6
The number of winners required is 6