what is the probability of drawing two red marbles...

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

Re: what is the probability of drawing two red marbles...

Quote:

Originally Posted by

**slapmaxwell1** 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.

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 $\displaystyle \dfrac{3}{6} = \dfrac{1}{2}$

On the second draw we now have **2** red marbles from a total of **5** so the chance is $\displaystyle \dfrac{2}{5}$

Since our problem is an AND type you need to multiply your chances together: $\displaystyle P(RR) = \dfrac{1}{2} \cdot \dfrac{2}{5} = \dfrac{1}{5}$

Re: what is the probability of drawing two red marbles...

ok i have a question, why is it on our first draw we pull 3 marbles, instead of 2? can you explain the term "without replacement again"

Re: what is the probability of drawing two red marbles...

this problem was taken from dr chung's SAT Math pg 52

Re: what is the probability of drawing two red marbles...

Quote:

Originally Posted by

**slapmaxwell1** ok i have a question, why is it on our first draw we pull 3 marbles, instead of 2? can you explain the term "without replacement again"

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