im not sure why probability problems give me fits....
ok so here is the problem.
a bag contains 3 red marbles and 3 blue marbles. what is the probability that you draw two red marbles without replacement?
ok so my reasoning is this...
we have the outcome that we want divided by the total possible?
2/6 = 1/3
of course my answer is wrong, im just not sure why
Without replacement is the key here. The chance of getting anything is number of wanted outcomes/total outcomes.Originally Posted by slapmaxwell1
If you're having trouble use a tree diagram, they are excellent for problems like this
On the first draw we have 3 red marbles (wanted) and 6 in total so the chance of a red marble is
On the second draw we now have 2 red marbles from a total of 5 so the chance is
Since our problem is an AND type you need to multiply your chances together: = \dfrac{1}{2} \cdot \dfrac{2}{5} = \dfrac{1}{5})
Without replacement means our first ball is gone from the bag forever in contrast to with replacement when we'd put it back in before the second draw. Unless I've misread the question I've assumed that we only draw twice and we want to know the chance of both being red.
We don't pull three marbles on the first draw, we pull one but there are six marbles in the bag, of which three are red. Thus we have a 3/6 = 1/2 chance of pulling out a red ball.
If we do not draw a red ball then essentially the "game" is over so for the second draw we say that the first one was red. However, we've took a red marble out and because we've not replaced it there are 2 red marbles in and the original three blue for a total of 5.
Thus there is a 2/5 chance on the second time
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