Our first class today consisted of a pre-test for our upcoming (TOMORROW) test on probability.

There were 5 questions, the last being a two parter. :O!

1) A combination lock requires the owner to choose three numbers from 1-40 in order to open the lock. If the numbers can be repeated, the probability of an individual correctly guessing the combination to this lock is:

a) 1/

_{40}P

_{3} b) 1/

_{40}C

_{3} **c)1/40**^{3 } d) 1/3x40

The answer is C. The numbers may repeat, but they have to be in order, otherwise you're not going to get the locker open.

2) For this one, you're supposed to find the probability that a chosen route will pass through point B, when trying to get from point A to C. This is assuming the shortest routes from A to C are travelled.

a) 2/7 b) 3/7

**c) 4/7** d) 5/7

The answer is C because there are 20 options to get from A to B. Then, the shortest way to get to C from there is to go directly right. That said, all that has to be done is 20/35, simplified that'd be 4/7.

3) Using the word FOOD, the probability that an arrangement of this word will begin with the two O's if all the letters are used, correct to the nearest hundreth, must be:

4!/(4-2)! = 12. You'd do 2/12 because there are two O's, and you can't tell them apart. Simplified that's 1/6, which in decimal form would be 0.17 (rounded).

4) Events A and B have the following probabilities of occuring; 0.2 = P(A), 0.5 = P(B). If these events are mutually exclusive, the value of P(A or B), correct to the nearest tenth, must be:

P(A or B) = P(A) + P(B) - P(A and B) ... And, because these events are mutually exclusive, P(A and B) = 0. Therefore, .2 + .5 - 0 = .7

5) Rupert has either milk or cocoa to drink for breakfast with either oatmeal or pancakes. If he drinks milk, then the probability that he is having pancakes with milk is 2/4. The probability that he drinks cocoa is 1/5. If he drinks cocoa, the probability of him having pancakes is 6/7.

a) List the sample space of probabilities using a tree diagram or any other method of your choice.

MO

MP

CO

CP

b) Find the probability that Rupert will have oatmeal with cocoa tomorrow morning.

P(oatmeal and cocoa)

P(C) = 1/5

P(O) = 1/7

Multiply 1/5 and 1/7 and voila, you have a 1/35 probability that Rupert will have the oatmeal with cocoa.

We got into groups to work on this after we did it ourselves. Is that even a word, ourselves?

Anywho, the afternoon class consisted of going over that pain of a stencil.

We didn't go over all of the answers. I'm wondering if I should go over them. I can't really explain them, but I understand. Does that make any sense? I'm a bit worried that I'll mislead or explain incorrectly. What the hell, here we go.

3) Seven people reach a fork in a road. In how many ways can they continue their walk so that 4 go one way and 3 the other?

You do 7 choose 4. That means that 4 will go one way. Then, 3 choose 3. You have three people to choose from left, and you choose all three of them. You mutiply them and get 35. Then you do 7 choose 3, multplied by 4 choose 4. This all equals 70.

5) A party of eight boys and eight girls are going for a picnic. Six of the party can ride in one car, and four in another. The rest must walk. (assume anyone can drive.)

a) In how many ways can the party be distrubuted for the trip?

16 (the party) choose 6 (car one) x 10 (ten people left) choose 4 (second car) x 6 (people left) choose 6 (the suckers that have to walk) = 1 681 680

b) In how many of these will no girl have to walk?

16 choose 6 (first car/girls) x 2 choose 2 (two girls left, guaranteed spots in the second car) x 8 choose 2 (8 people left, two of the boys get to be in the second car) x 6 choose 6 (six boys left, and they all get to walk.. bwahahahahahaHA)

c) Ditto to the last question, except there has to be a boy in each car driving

1 choose 1 (the driver of the car) x 8 choose 5 (5 of the 8 girls in the first car) x 1 choose 1 (the next driver) x 3 choose 3 (3 girls left, and 3 spots left in the second car) x 6 choose 6 (again, the suckers that get to walk)

7) In how many ways can 4 balls be drawn together from a bag containing 5 red balls, 7 white and 3 green? ... 15 choose 4

a) 2 red and 2 white balls?

5 choose 2 x 7 choose 2 .. That means you choose 2 of the 5 red balls, and 2 of the 7 white balls.

b) Just two white balls?

7 choose 2 x 8 choose 2 ... 2 of the 7 balls, and then any 2 of the remaining balls of the other two colours

c) Only white balls?

7 choose 4 ... It's only white balls, right? So you just choose four of them.

d) Balls of only one colour?

5 choose 4

**+** 7 choose 4 *** It's + because it can be wither 4 of the 5 red balls, or 4 of the 7 white balls. You cannot use the green balls, as there are only 3 of them. Poor green... :( :P

e) No red balls?

10 choose 4 ... You just add the 7 white balls and the 3 green ones

8) Ten points are marked in a plane, no three of them lying in a straight line. How many triangles do they determine?

This one is 10 choose 3.

10) From 7 French books and 9 English books, in how many ways can we select 3 French books and 4 English books and arrange them all on a shelf?

7 choose 3 x 9 choose 4 x *7! ... *Because they're arranged on a shelf .. 7! = 7x6x5.. etc.

After we went through the stencil, we started a review on page 43. We're to do questions 1 through 7.

I hope I explained all of these properly. Tomorrow's scribe is going to be Mr. mohamed (aka Mr. Smartypants)