# Math Help - Drawing marbles

1. ## Drawing marbles

Box 1 contains 5 red and 3 blue marbles while box 2 contains 2 red, 3 blue, and 1 white marble.

If you select a box at random and select a marble at random from it, then toss that marble into the other box and randomly draw a marble from it, what is the probability that the marble you draw is blue?

1. Approximately 43.55 percent

2. 439/1008

3. 3/7

4. Approximately 10.71 percent

5. 3/6

2. ## Re: Drawing marbles

Originally Posted by jthomp18
Box 1 contains 5 red and 3 blue marbles while box 2 contains 2 red, 3 blue, and 1 white marble.

If you select a box at random and select a marble at random from it, then toss that marble into the other box and randomly draw a marble from it, what is the probability that the marble you draw is blue?

1. Approximately 43.55 percent

2. 439/1008

3. 3/7

4. Approximately 10.71 percent

5. 3/6
What have you tried? Did you draw a probability diagram? Is this a conditional probability? what are your thoughts on the matter?

3. ## Re: Drawing marbles

Neither! I dont know! I was trying to decide the answer

4. ## Re: Drawing marbles

Originally Posted by jthomp18
Box 1 contains 5 red and 3 blue marbles while box 2 contains 2 red, 3 blue, and 1 white marble.
If you select a box at random and select a marble at random from it, then toss that marble into the other box and randomly draw a marble from it, what is the probability that the marble you draw is blue?
The string $1~B_1B_2$ stands for box 1 was chosen then a blue ball and then another blue from box 2.

$P(1~B_1B_2)=\frac{1}{2}\frac{3}{8}\frac{4}{7}= \frac{12}{112}$

Now there are four more such strings. If you add those up the answer is one listed.

5. ## Re: Drawing marbles

10.71 percent! Right?

6. ## Re: Drawing marbles

Originally Posted by jthomp18
10.71 percent! Right?
No indeed!
Did you find out the other four strings and calculate the probabilities?

7. ## Re: Drawing marbles

Yes we did. We went back and looked again. We think now that it would be either 3/7 or 3/6.

8. ## Re: Drawing marbles

Events "tossing marble from Box1 into Box2" and "tossing marble from Box2 into Box1" are equally probable. What matters in both cases is the color of the marble being tossed into the second box, and that depends on the box from which it originates.

$\frac{1}{2}\left(\frac{3}{8}\cdot \frac{4}{7}+\frac{5}{8}\cdot\frac{3}{7}\right)+ \frac{1}{2}\left(\frac{3}{6}\cdot \frac{4}{9}+\frac{3}{6}\cdot\frac{3}{9}\right)= \frac{1317}{3024}=\frac{439}{1008}$

Numbers in the first summand are derived as follows:
(probability that you have the case of Box1->Box2 marble tossing)*((probability that the blue marble is selected from box1 and tossed into box2)*(probability that the blue marble is drawn from box2 now that you have one more blue marble in it)+(probability that the marble from box1 tossed into box2 is of some color other than blue)*(probability that the blue marble is drawn from box2 now that you have one more marble in it of some color other than blue))

Second summand covers the case box2->box1

9. ## Re: Drawing marbles

AHH!! That does match!! 439/1008! thank you MathoMan!

10. ## Re: Drawing marbles

you've got the right answer but do you 'get it'? that is more important because here 'the trip itself worths more than the final destination'.

11. ## Re: Drawing marbles

yes we actually wrote out a tree diagram.

12. ## Re: Drawing marbles

We just did not think to use 1/2!

13. ## Re: Drawing marbles

That is confusing me now?

14. ## Re: Drawing marbles

Scenarios box1->box2 and box2->box1 are equally likely since you make a decision at random. When choosing among n possibilities at random, each possibility is equally probable with probability 1/n. That is what making a random decision means, 'its all the same to you', or just word 'whatever'.

15. ## Re: Drawing marbles

Originally Posted by jthomp18
yes we actually wrote out a tree diagram.
Yay