I got it!!!

     So a few days ago I was completely dumbfounded with the Pom Pom activity that I did in class. This activity required us to find possible outcomes regarding 3 white pom poms and 2 colored pom poms. Luckily Mrs. Klassen reviewed a similar problem  in simpler terms using a coin toss as an example of possible outcomes.
    Once I completed the homework which covered problems like this, I finally feel that I understand it. I will give you one of the examples from our homework lesson and try to explain it as best as I can. Below I have posted the questions from the homework and worked the problem out.

A box contains 5 blue balls and 3 green balls. Two balls are drawn at random from the box.

a) If the first ball is drawn and not replaced, the probability that two balls will be drawn of different colors is?

b) If the first ball is drawn and then put back, the probability that two balls will be drawn of different colors is?

Here's the key for the letters I used in the picture of a Tree Diagram below. B= blue balls  G= green balls

 Tree Diagram of the problem:

 a)  
     For the B part, to get the 5/8 you take the number of blue balls over the total amount of balls. Getting the 4/7 is a little trickier. If you remember from the problem above one ball is drawn but not replaced. So that means  if we picked a blue ball from the first drawing and we picked another blue ball from the second draw we would do 5 minus the 1 equaling 4.Then, we take the 4 and put it over the total amount of balls left which would be 7. To get the 3/8 we would use the same pattern just assume that we pick a green ball instead of the blue ball on the first draw. Since we have never drawn a green before, this would just be the original amount of 3. We then take the 3 and put it over the new total of balls from the second draw. We would then use the same steps for drawing a green ball first by adjusting the numbers accordingly. The fractions that I got were 3/8, 4/7, and 2/7.
   Now to answer the first question. To get the probability that two balls will be drawn of different colors, we simply multiply 5/8*3/7. This will equal 15/56. Now we add 15/56 + 15/56 together to get our answer.

5/8*3/7= 15/56         15/56 + 15/56 = 15/28

So our answer for problem (a) would be 15/28

b) For problem (b) we will use the same picture of the Tree Diagram seen above. Now we have to find the probability that two balls will be drawn of different colors, if the ball is first drawn and then put back. For this we would take the 5/8 from the first draw of a blue ball and the 3/8 from the first draw probability of getting a green ball and then multiply them.

5/8*3/8 = 15/64

Then, we take 15/64 and add another 15/64.

15/64 + 15/64 = 15/32

So our answer for problem (b) would be 15/64

Here's a Link for further explanation on the colored ball probability.