Question

Two marbles are drawn without replacement from a box with 3 white, 2 green, 2 red and 1 blue marble. Find the probability that one marble is white and one marble is blue.

Answer 1

how many total marbles?

Answer 2

you are drawing w/o replacement... does the order matter in this case?

that is, if you were to select a white marble first and a blue marble second, is that considered the same as drawing a blue forst and a white second? Does the order matter or just the group (1 white and 1 blue)?

Answer 3

8 total marbles

Answer 4

and how many will you select?

Answer 5

the probability of no red marbles _-15/28 probability of at least on red marble_ 1-15/28=13/28

Answer 6

order does not matter, (i guess). i must select only two, one first then the other. need to know what the probability is of selecting one white and one blue marble

Answer 7

all you do is multiply the prob of pulling a white one (3/8)then * that by theprob of pullin a blue(1/7)

Answer 8

right... order doesn't matter. so long as we get 1 white and 1 blue marble, we have the event we want.

so we have \(_8C_2 \) total different groups of 2.

that's going to be our denominator.

we have 3 white marbles, of which we need to choose 1 and 1 blue marble, of which we need to choose 1.

so we get this for our probability:\[P \left( 1 \text{ white and }1\text{ blue} \right)=\frac{ \left(\begin{matrix}3 \\ 1\end{matrix}\right) \left(\begin{matrix}1 \\ 1\end{matrix}\right)}{ \left(\begin{matrix}8 \\ 2\end{matrix}\right) }\]

Answer 9

do you know how to compute this?

Answer 10

@saue7414 that is incorrect because that is the probability of selecting a white 1st and then a blue. there is another sequence which also satisfies the requirement...
blue 1st then a white. the combinations account for this, your method doesn't.

Answer 11

okay thanks for the correction that would be if order mattered right?

Answer 12

correct!

Answer 13

ohhhhhhhh LOL

Answer 14

All i can see from the question is "Find the probability that one marble is white and one marble is blue." Does that say a white marble MUST be chosen first before the blue marble? that I'M not sure of. But with the question worded as such, it probably makes a difference as the outcome. So to err on the side of caution, I'd say that we would have to go with the probability of selecting a white marble first, PRIOR to selecting a blue marble. because if we selected the blue marble first, that would change the outcomes probability

Answer 15

Am I correct in assuming that probability of drawing a white marble first, then a blue marble, would be changed if a blue marble was drawn first, followed by the probability of drawing a white marble? I guess that the order would change the probability? Right?

Answer 16

no... you have to look at the group... the group contains 1 white and 1 blue. the order of selection is irrelevant. so long as 1 white and 1 blue are chosen, you have satisfied the requirement.

you can do this in one of 2 ways... you can select a white first, followed by a blue or you can select a bleu first, followed by a white.

in terms of the probabilities, we have P({W,B}) = P(W 1st and B 2nd) + P(B 1st and W 2nd)
=(3/8)*(1/7) + (1/8)*(3/7) = 2*(3/56) = 3/28

this is the same as\[\frac{ \left(\begin{matrix}3 \\ 1\end{matrix}\right)\left(\begin{matrix}1 \\ 1\end{matrix}\right) }{ \left(\begin{matrix}8 \\ 2\end{matrix}\right) }\]

Answer 17

If what you are saying is correct, and I have no reason to doubt that, wouldn't there be a distinct difference in probability if 1 white marble, (of which there are three in the box) is drawn after 1 blue marble of which there is only one in the box? Wouldn't the order in which they are drawn efect the probability?