In a game of rock-paper-scissors, you have a 1/3 chance of winning, a 1/3 chance of losing, and a 1/3 chance of tying in any given round. What is the probability that you will win at least twice in 3 rounds, given that there aren't any tied rounds in this particular match

1/5

Step-by-step explanation:

given data;

chances of winning = 1/3

chances of losing = 1/3

chances of tying in a given round = 1/3

solution:

probability that you would win atleast 2 in any 3 matches without a tied match is

1/3 /  ( 2 - 1/3 )

= 1/3 / 5/3

= 1/5

the probability of winning 2 of 3 games without a tie is 1/5

here is the

mark brainliest pl

1.3

step-by-step explanation:

3x2=6   9x+6

bring down the 6 on both sides

6+6=12

12= 9x

divide by 9

12/9 =

1.3 =x