SOLUTION: An opaque bag contains 5 green marbles, 3 blue marbles, and 2 red marbles. Two marbles are randomly drawn without replacement. A. 1/5 B. 3/10 C. 2/9 D. 1/3 The que

Algebra ->  Sequences-and-series -> SOLUTION: An opaque bag contains 5 green marbles, 3 blue marbles, and 2 red marbles. Two marbles are randomly drawn without replacement. A. 1/5 B. 3/10 C. 2/9 D. 1/3 The que      Log On


   

Question 1160681: An opaque bag contains 5 green marbles, 3 blue marbles, and 2 red marbles. Two marbles are randomly drawn without replacement.
A. 1/5
B. 3/10
C. 2/9
D. 1/3

The question and the answer choices in the link here: https://i.imgur.com/CdhrAdS.png
Answer by MathLover1(20850) About Me  (Show Source):
You can put this solution on YOUR website!

there are 5 green marbles, 3+blue marbles, and 2 red marbles-> total 10
as I can see on a link, you need to draw twice
in first draw you need 1 red:
if total number of marbles is+10 and you have 2 red marbles:
probability to draw one is 2%2F10=1%2F5

second draw you want 1 blue:
we have 3 blue marbles and total number of 9 marbles (remember one red is gone,there is no replacement) :
3%2F9=1%2F3
probability to draw one red and one blue: %281%2F5%29%281%2F3%29=1%2F15

since there is no such a choice in given options, I will try to see what is the probability of choosing two green marbles

there are 5 green marbles, 3+blue marbles, and 2 red marbles-> total 10
to draw, for example, one green marble, probability is 5%2F10+=1%2F2+
since without replacement, now total number of marbles is 9 and total number of green marbles is 4
and, to draw one more green marble without replacement , probability is 4%2F9
then, the probability of two green marbles is:
%281%2F2+%29%284%2F9%29+=+4%2F%282%2A9%29=+2%2F9

for an answer you have option:
C. 2%2F9 (in case you want to draw two green marbles)