Question

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?

Answer 1

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;

$Prob=\frac{1}{4}= 0.25$

In the second game;

Number of winners= 17

Number of players= 17+51= 68

Probability of winning is;

$Prob=\frac{17}{68} =0.25$

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

$\frac{x}{x+18}=0.25\\ x=0.25*(x+18)\\x=0.25x+4.5\\x-0.25x=4.5\\0.75x=4.5\\x=\frac{4.5}{0.75}\\ x=6$

The number of winners required is 6

Answer 2

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