A box 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)
Accepted Solution
A:
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%)