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Mathematics 22 Online
OpenStudy (anonymous):

Two marbles are drawn without replacement from a box with 3 white, 2 green, and 1 blue marble. Find the probability of drawing one white marble and one blue marble. please help.

OpenStudy (anonymous):

thank you for taking the time to help.

OpenStudy (anonymous):

There are 6 marbles, drawing 2 means \[_6 C_2=\frac{(6)(5)}{2}=15~ways\]

OpenStudy (anonymous):

1/2 3/56 3/28 1/4 this are my answer choices.

OpenStudy (anonymous):

I tried that.

OpenStudy (anonymous):

3/6 x 1/5 maybe

OpenStudy (anonymous):

i tried 3/5* 1/6=3/10=1/10

OpenStudy (anonymous):

for one white marble we have the this \[_3 C_1=3~ways\]the \[p(white)=\frac{_3 C_1}{_6 C_2}=\frac{3}{15}=\frac{1}{5}\]

OpenStudy (anonymous):

i don't know what else to do.

OpenStudy (anonymous):

i also tried 1/6C1

OpenStudy (anonymous):

got it 1/2

OpenStudy (anonymous):

I think of each white marble as identical. I could be wrong. But I think the probability of picking a white is 3/6. The prob. of picking a blue is 1/5 because there's only one blue marble, and there are only 5 marbles left to choose from because you've already removed one. You multiply the two probabilities: 3/6 times 1/5 = 1/10 Do you think this is correct?

OpenStudy (anonymous):

how did u get that Orion if you don't mind showing me.

OpenStudy (anonymous):

jam333 that was my first answer, but when i looked at the answer choices it was not one of them.

OpenStudy (anonymous):

\[p(1~white~and~1~blue)=\frac{_6 C_2}{_6 C_1 \times _5 C_1}=\frac{15}{(6)(5)}=\frac{15}{30}~or~\frac{1}{2}\]

OpenStudy (anonymous):

the first choice is 1/2...

OpenStudy (anonymous):

i see. you r correct.

OpenStudy (anonymous):

it's really fun solving math problems... :)

OpenStudy (anonymous):

thank you for showing me how to do it. you r a life saver.

OpenStudy (anonymous):

thanks also i recall probability...

OpenStudy (anonymous):

i know. im taking this class so that i can be able to apply to the nursing program, Contemporary math is one of the course required to apply.

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