Where's Olin?

     Mississippi Map Turtles have experienced a steady decline in numbers over the last few years. This is due to pollution, people, and invasive species .Here, I am going to give you a problem and work through it with you.

  A picture of Olin. My Mississippi Map Turtle.

To figure out the number of  Mississippi Map Turtles along a tributary of the great Mississippi River, herpetologist have begun to record and mark the map turtles they find. Last year herpetologist caught and marked 80 turtles. Then, they caught 30 more within a time frame of 6 months, and it is found that 10 of those are are marked. Estimate how many map turtles are in the river.

First lets set this problem into two ratios that are equal to each other.

         X  turtles      =              80 turtles        
  30 marked turtles             10 marked turtles

Now we simply cross-multiply.

X * 10 = 30 * 80

Now we divide both sides by 10.

X=        30(80)       
                 10

X= 240

The estimated number of Mississippi Map Turtle is 240.

Here's a  Link  if your interested in learning more about simple probability.
 

What will Lucy do?

Hi all. I have a new problem for you. In this experiment I used my little chug (chihuahua and pug mix), Lucy to help me solve a problem. Enjoy!

If the probability of Lucy choosing a rawhide treat is 5%, what are the odds of her choosing a tasty yam treat over the rawhide? ( she has enjoyed many treats)

So, to start this problem we are going to find the probability that Lucy will pick the rawhide treat. There are 5 chances out of 100 that she will pick the rawhide treat. So we would write this as 5/100.Simplify.

P(rawhide)= 1/20

Next, we have to find the probability that she will pick the tasty yam treat instead.

P(yam treat)= 1-1/20= 19/20

The probability that Lucy picks the rawhide treat is 1/20, and the probability that Lucy picks the yam treat instead is 19/20.

The odds against the rawhide are   19/20
                                                      1/20

So the odds against Lucy picking the rawhide vs the yam treat is 19 to 1. She really loves yams!

       Here's a great Link I found on odds

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.

Piece of the pie

Hello everyone.
    So I was going back over some homework and I remembered that I was having issues with one of the problems. At first I was a little confused. I figured out that my initial hangup was not about the problem itself but of the terminology. I had forgotten what the definitions of multiple and factor were ( I know it was foolish of me but keep in mind that I slept a lot this past summer).
  Anyways, I am going to go over the problem and the definitions. Hopefully this might help some of you on the test if you happen to blank out on a problem like this.


First, I will go over the definitions.

Factor- a number that will divide into another number exactly (ex. factors of 10 are 1,2,5)

Multiple- the product of the number and any other whole number (ex. 2,4,6,8 are multiples of 2)

Now let's tackle the problem.

The spinner shown below is spun. (Assume that the size of each sector is the same.) Find each possibility.

a)   P(factor of 15)

b)   P(multiple of 2)

Now let's find the factors of 15. There are 3 numbers that will divide into 15. They are: 1,3,5

Since there is a total of 8 sections in the spinner and we only want to know how many of the 8 sections are factors of 15 we put 3 over 8.

a) P(factor of 15)= 3/8


Now let's find the multiples of 2. There are 4 numbers that are multiples of 2. They are: 2,4,6,8

Since there is a total of 8 sections in the spinner and we only want to know how many of the 8 sections are multiples of  2 we put 4 over 8.

b) P(multiple of 2)= 4/8 = 1/2


The definitions came from http://math.about.com/library/blm.htm and the problem came from MyMathLab.

 link
A link for an example on how to complete a problem similar to this problem.

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.