Question

the probability of winning a game is x/12. If the probability of losing a game is 1/3. Find the value of x.​

Answer 1

Probability of happening a event + probability of not happening a event = 1


X/12+1/3=1

X/12=1-1/3

X/12=(3-1)/3

X/12=2/3

X=(2*12)/3

X=24/3

X=8

Answer 2

The value of x is 8.

Explanation:

We know that in probability , for any event E :

P( E ) +P( not E) =1

Let E be the event of winning the game.

Then, as per given , we have

P(E) = $\dfrac{x}{12}$

P( not E) = $\dfrac{1}{3}$

∵ P( E ) +P( not E) =1

$\dfrac{x}{12}+\dfrac{1}{3}=1\\\\ \dfrac{x}{12}=1-\dfrac{1}{3}=\dfrac{2}{3}\\\\ x=\dfrac{2}{3}\times12=2\times4=8$

Hence, the value of x is 8.

# learn more:

If the probability of winning a game is twice the probability of losing the game find the probability of winning and losing the game​

https://brainly.in/question/8358304