Question

A bag contains 5 red marbles and 8 yellow marbles. You are asked to draw 3 marbles from the bag without replacement. In how many ways can you draw 2 reds and 1 yellow?

Answer 1

how many ways, or probability?

Answer 2

how many ways: you have 5 red marbles and so there are \(_5C_2=10\) ways of selecting two of them
then there are \(_8C_1=8\) ways of selecting the one yellow one out of the 8
multiply to get your answer

Answer 3

i think @Nurali made a typo, and meant \(_8C_1=8\) as opposed to \(_6C_1=6\) for the yellow one

Answer 4

5C2*8C1 = 5*4/(2*1)*8 = 80 ways.

Answer 5

its how many ways.
possible answers:
160
40
17
4,838,400
200

Answer 6

This question Is not a valid question with correctly valid answers

Answer 7

maybe we are to assume that order counts in this problem, in which case the answer would be different

Answer 8

Even if order is to matter because that is the way I presumed it there will only be 3 ways

Answer 9

kind of strange though, since it is written in such a way that would make you think order does not count, as you cannot tell the two reds apart

Answer 10

Basically this is why I'm confused. I didn't know if it was just me or...

Answer 11

but if order counts we can do that too

Answer 12

That is what I did and resulted with only 3 possible ways if both reds are the same.

Answer 13

And if it was two different reds then their is still only 6 possible way to draw the marbles for the bag.

Answer 14

but i don't get any of the above answers
if order counts it could be
red red yellow
red yellow red
yellow red red

for each the number of ways would be \(5\times 4\times 8=160\)

Answer 15

so in this case the answer should be (i think) \(3\times 160=480\)

Answer 16

Actualy 160 is a answer choice

Answer 17

yeah, but that is not the answer if order counts

Answer 18

Yes true but this is probably a pulication error or a mistake from the her part

Answer 19

if order counts, then
red red yellow
is different from
red yellow red
is different from
yellow red red

either order counts, or it doesn't
if it doesn't the answer is \(80\)