Question

If the probability of winning a chess game is 1/3more than the twice of probability of losing the match.Find the probability of winning and losing the match.

Answer 1

We know that the maximum probability for a task is 100. So let the probability of winning the game be x and that of losing be y.

So we have two equations:
i) 2y + 1/3 = x ; 6y + 1 = 3x
ii) x + y = 100 ; x = 100 - y 

Now putting the value of x in (ii) in (i), we have,
6y + 1 = 3(100-y)
= 6y + 1 = 300 - 3y
= 9y = 299
= y =299/9 = 33.22% (losing)
Therefore, probability of winning = 100 -  33.22 = 66.78%

Hope that helps!!

Answer 2

losing chance=x
winning chance=[tex] \frac{1}{3}+2x \\ \\ \frac{6x+1}{3} [/tex]

losing chance+winning chance= total chance=1

[tex] \frac{6x+1}{3} +x=1 \\ \\ \frac{6x+1+3x}{3} =1 \\ \\ 9x+1=3 \\ \\ 9x=2 \\ x= \frac{2}{9} [/tex]

losing chance= $\frac{2}{9}$

winning chance= [tex](2* \frac{2}{9} )+ \frac{1}{3} \\ \\ \frac{4}{9} + \frac{1}{3} = \frac{12+9}{27} \\ \\ = \frac{21}{27} [/tex]
winning chance=$\frac{21}{27}$