Making sense

Ok. So I have another problem to show everyone. It goes like this:

A law firm consists of 10 members: 6 women and 4 men. 3 members are selected at random to take on the next big case. The names are drawn from a large bowl. What is the probability that all 3 selected are men?

So for the first picking 10 names are in the bowl. The probability of getting a man's name the first time is 4/10.

Now we go to the second drawing. There are 9 names left in the bowl. 3 of these names are men. So that makes our next fraction 3/9.

For the final selection we have 8 names left in the bowl. Only 1 is a man. This means that our fraction will be 1/8.


Now we are going to take all 3 fractions and multiply them together. This will give us the probability of selecting those 3 men.

4/10 * 3/9 * 1/8 = 12/20

This fraction will reduce leaving us with an answer of 1/60. So there is a 1/60 chance that all three men will be drawn from the bowl.


Here's a Link for further explanation.

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.

Wednesday August 29th class

   Class today was a little tricky. We were learning about Probability with playing cards and pom poms.  It was a fun activity until we reach the pom pom section. First, I'll explain the deck of cards and then I'll get to the ugly stuff.
     The playing card activity was pretty straight forward. I got that the probability of getting a queen in a deck of cards was 4/52 which reduces to 1/13. We also could write this as P(Q) = 1/13. This is because there are only 4 queens out of all the cards in a deck.Most of the problems that Mrs.Klassen gave us on this activity were hard at first, but once you wrapped your head around it, it made sense. It really helped me to envision the deck of cards in my head.
    After this excessive I was pretty hopeful and then (dun dun dun) I lost it. We started out with a box containing 3 white balls and two black balls. This was our problem:
 A ball is drawn at random from the box and not replaced. Then a second ball is drawn from the box. Draw a tree diagram for this experiment and all possible outcomes. Find the probability that the two balls are different colors. (straight from Mrs. Klassen's worksheet)

This is what I wrote down.

    I'm still completely lost but Mrs. Klassen did say she was going to review it next class period.

 Link
Here's a link to a video by Khan Academy. This will show you examples of playing cards and probability.