Abox contains 3 red marbles, 6 blue marbles, and 1 white marble. the marbles are selected at random, one at a time, and are not replaced. find the probability. p(red and red and red)

Let's work this step by step. There are a total of 10 marbles in the box. The chance of getting a red marble on the first pick can be calculated like this.

Total number of red marvels / Total number of marble s
3/10 = 0.333

Multiply it by 100 to put the answer in percent

0.333*100 = 33.3%

33.3% chance of getting red marble at the first try.

Next since we can't replace the marvel back into the box now we have a total number of 9 marvels and 2 red marbles ( 2 red marbles , 6 blue marble s and 1 white marble )
Total number of red marbles / Total number of marbles
2/9 = 0.22
then multiply by 100
0.22*100 = 22.2%

22.2% chance of getting red marble at the second try.

Now there is only one red marble and a total of 8 marbles(1 red marble 6 blue marbles and 1 white marble)
Total number of red marbles / Total number of marbles
1/8 = 0.125
then multiply by 100
0.125*100 = 12.5%

12.5% chance of picking red marble at the third try.
all together p(33.3% , 22.2% , 12.5%)

Igot 108 hope that’s right !