Sunday, March 2, 2008

Hey, hey, hey! So last Friday we did were suppose to move on and do something new, but unfortunately most of us ( me included) didn't do the homework assigned. So now we wasted a whole class doing work. Heres the work and how it was solved, and oh, I suck at blogging so bear with me.

We were given the option every week of either $10 or picking two bills out of a bag, which contained two $1 bills, two $5 bills, and a $10 bill. So in the long run which option was better.
-1st we make our tree diagram to show all possibilities of picking 2 bills out of the bag.

So then, once we find the probability of the 1st bill being pulled out, we find the probability of the 2nd bill being pulled out of the bag. To find the probability of 2 bills being pulled out, Multiply the probability of the 1st bill(should be put into fraction) by the probability of the 2nd bill.
ex: first bill is $1, the probability of pulling one dollar is 2 out of 5, because we got 5 bills and 2 of those bills are $1 bills. Lets pretend the 2nd bill is a $1 bill. The probability of pulling another $1 bill after we already pulled out one $1 bill is 1 out of 4. So we multiply the two fractions 2/5 *1/4 = 2/20.

Now if we have 2 numbers that are the same, we just add the fractions together(the denominator does not change).

Now we just find the % of each number possible of pulling out, by dividing the top number by the denominator and Multiply that by 100.

- Now in the long run, which is better? $10 a week or picking 2 bills out of the bag?
For this example we used 100 weeks to make it nice and easy to read.

So basically, there is 10% that you will pull 2 dollars out of the bag. 10% of 100 is 10. take that 10 and multiply it by 2 to get the possibility of getting $20 out of 100 weeks, by pulling two $1 bills. $10 a week is better opposed to only a possibility of $880 in a 100 weeks.


SO the question asked us, how many different 4 digit numbers are there in which all the digits are different. Okay well there are 9 different digits plus 0 that make up the whole number chart.
The question asked us to make 4 digits.

_ _ _ _ = _
1st 2nd 3rd 4th
So in the 1st slot we have 9 different possibilities because we can't use 0. Now on the 2nd slot we have to minus 1 because we used 9, but now we can use 0 so we still have 9 different ways. the 3rd slot we gotta take out the 0 which leaves us with only 8 possibilities. Then we take out 1 more number that we used which leaves us with 7 ways on slot 4. To find the total number of ways of arranging 4 different digits, we multiply each slot by the others.


So now the question asked us how many of those numbers are odd?
-well odd numbers will end in odd. The only odd numbers there is are: 1, 3, 5, 7, and 9.
5 odd numbers. stick 5 at slot 4, so we know it will end with an odd number.
Since we used 1 number at the end, for the 1st slot we minus 1 way which leaves us with 8 ways of arranged a number in slot 1. In slot 2 we can use 0 so that gives us 8 again because we had to abide by the 1st rule ( 4 different digits). then the 3rd slot we have 7 ways. Multiply the slots together and we get our answer.


Now we had to find how many of those numbers are divisible by 5. well with the number ends with 0 or 5 it can be divided by 5.

We had to show 2 diagrams. one with 0 at the end and one with 5 at the end. the concept is the same, but this time once we find our answer for each diagram we add them together to find the real answer.

I didn't really understand this question, so I don't think I can explain it in case I mislead people, but I understand its the same concept as previous questions, with more thinking.

Well hopefully I can understand that question to the point where I can teach it.
--The next scribe is.....................cowmilk

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