## Tuesday, May 27, 2008

### Combinations of Permutations and Probability and Combinations O_o

Why hello there!
Benofschool here. I never scribed for a while, I was beginning to think that you guys forgot about me =(. Well anyways time for a scribe. Mr.K was later than usual today but he wasn't late for school which was good. Today's class was a workshop. We were broke up into groups like every other workshop class. Unfortunately the slides aren't up. Questions were put up and we were off...

Question 1
We had to find the probability of getting an ace or a diamond from a deck of cards. Fairly simple. Francis went up to answer it. To get the answer we just add the probabilities of getting an Ace and a Diamond and subtract the probability of getting the Ace of Diamonds. We add because it is an "or" question and we subtract that single card because it was counted twice when we calculated the probability of getting an Ace or Diamond. So the answer was 4/13.

Question 2
The second question involved a venn-diagram and venn skills from grade 11 logic in pre-calculus. The question was what is the probability of picking a person who doesn't like Dr. Pepper or the diet version if 7 people liked Dr.Pepper, 11 people liked the diet version, and 3 people liked both. So first we have to create a two circle venn-diagram. Rule is start with the inside and work outwards. So we put in the middle which means that 3 people prefer both drinks. Then lets work with on side. But remember from grade 11 we have to subtract the number of objects in the center from the separate values. If we look at the first question, we didn't count the Ace of Diamonds because the probability of getting an Ace or a Diamond is not mutually exclusive as well as the Dr. Pepper situation. If one thing occurs, the other probability can still happen. So that means we have to subtract the number of people that liked both drinks from the separate preferences. We get the sum of those numbers (4 +3+8=15) and subtract from the total number of students in the class (25-15=10). That will be the numerator in the probability and the sample space would be the total number of students. So the answer is 2/5.

Question 3
The 3rd question is about picking dresses. There are 15 dresses: 6 green, 5 blue and 4 yellow dresses. We wanted to know what is the probability of getting exactly 2 green dresses from picking 6. This question involves both combinations and probability. So 6 choose 2 because they are indistinguishable. multiplied by 9 choose 4 for the remainder of the dresses divided by 15 choose 6 which is the ways that we can choose 6 dresses out of 15. So we get an answer of 37.8%.
Question 4
Another combination question. A couple has 4 children that are about to be born. What is the probability of getting at least 2 girls. So what we do is find the probability of getting 2 girls first then 3 girls and finally all girls. Then we just find the sum of those values and we get the answer. The answer is:

That were all of the questions we did in today's class. Remember DEV due dates are arriving in a future now. Check the calendar on the right side bar to check for your due dates. Tomorrow's double will be used as a DEV work day. So bring the stuff you want to bring to work on your DEVs. That is all I have to say about today's class. The next scribe will be kristina. Good Night and see you all tomorrow!!!