Question 1160681

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 {{{2/10=1/5}}}


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) : 

{{{3/9=1/3}}}

probability to draw one red and one blue: {{{(1/5)(1/3)=1/15}}}


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 {{{5/10 =1/2 }}}

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 {{{4/9}}}

then, the probability of two green marbles is:

 {{{(1/2 )(4/9) = 4/(2*9)= 2/9}}}


for an answer you have option: 

C. {{{2/9}}}  (in case you want to draw two green marbles)