A bag contains 10 blue marbles, 8 brown marbles and 8 green marbles. If 4 green marbles are added to the bag, and 3 brown and 2 blue marbles are taken out, what is the probability of choosing at least one blue marble when two marbles are randomly chosen?

Question

anonymous · Answer

one sec. i got this one.

anonymous · Answer

Sure, take your time.  :)

anonymous · Answer

do you add and remove marbles first and then after adding and removing marbles is that when you pick the two marbles out of the bag.

anonymous · Answer

10 blue marbles, 8 brown marbles and 8 green marbles. If 4 green marbles are added to the bag, and 3 brown and 2 blue marbles are taken out

so now we have

8 blue marbles, 5 brown marbles and 12 green marbles.

anonymous · Answer

now we need to figure out

what is the probability of choosing at least one blue marble when two marbles are randomly chosen?

anonymous · Answer

Yeah, that's where we use permutations/combinations, right?

anonymous · Answer

for the probability?

anonymous · Answer

P(choosing at least one blue marble) = 1 - P(choosing no blue marbles)

anonymous · Answer

^ Yes. But the question says 'at least' so it could be
1 blue marble + 1( Brown/ Green Marble)
Or both blue marbles.

anonymous · Answer

8 blue marbles, 5 brown marbles and 12 green marbles.


P(choosing no blue marbles in 2 draws) = P(not choosing a blue marble on the first draw AND not choosing a blue marble on the second draw)

since the two events are independent, we can break the AND (intersection) into a product of two probabilities.

P(not choosing a blue marble on the first draw AND not choosing a blue marble on the second draw) = P(not choosing a blue marble on the first draw) * P(not choosing a blue marble on the second draw)


P(not choosing a blue marble on the first draw ) = (5 + 12)/(5 + 12 + 8)=.68

P(not choosing a blue marble on the second draw) = (5 + 12 -1)/ (5 + 12 + 8 -1)=.666666

in the first case you are dividing the number of non blue marbles by the number of total marbles.

in the second case you are dividing the number of non blue marbles minus one (because you already drew one non blue marble) divided by the total number of marbles (previous total minus one)

.68 * .666666 = .4533333

so we know that the P(choosing no blue marbles in 2 draws) = .4533333

now to find the probability of choosing at least one blue marble you just do 1-.45333333.

anonymous · Answer

I think I kind of, sort of understood it. 
Thank you. :)

anonymous · Answer

you're welcome