Monday, March 10, 2008


okay so today we worked on a pre-test since our unit test is tomorrow...

1. A combination lock requires the owner to choose three numbers from 1-40 in order to open the lock. if numbers can be repeated, the probability of an individual correctly guessing the combination to this lock is:
C) 1/ 40^3

this is the correct answer since in a combination lock you are able to use a number more than once. For example a combination could be 12,25,12. And because of this you would have to count all 40 numbers 3 times. 

2. Using the diagram to the right, given the shortest routes from A to C are travelled, what is the probability that a chosen route will pass through point B, in attempting to get from A to C?
C) 4/7

i think that its pretty much self explanatory as long as you use pascals triangle u can get the answer easily.

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

# of ways to start with O. 
1   1  2  1 = 2
-- -- -- --
o  o 

No restrictions 

4! / 2! = 12
the 4! represents the four letters in the word and all the ways they can be arranged and the 2! represents the two O's


p = 2 / 12
   = 1 /6

The answer is 1 / 6

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

P(AorB) = P(A) + P(B)
= 0.2 + 0.5
= 0.7

Since it says A or B, you can understand from the word or that the two must be added together.

Therefore the answer is:

5. Including Peter and Deanna, a particular school council has 10 committees, each containing four members, are chosen from this council. Find the probability that:

a) Peter and Deanna are both chosen for the committee. 

2C2 * 8C2
 ----------------  = 2/15

well the 2C2 stands for both Peter and Deanna and the two ways that they can be together and we multiply that by 8C2 because out of the 8 people that can be in the the two committees of 4 we want Peter and Deanna together and its divided by 10C4 since that is all the possibilities that out of the 10 people the 4 to be chosen for the committee.

b)Either Peter or Deanna is chosen for the committee.

2C1 * 8C3
    ----------------  = 8/15

we use the 2C1 because out of Peter and deanna we only want one of them for the committee, and then it is multiplied by 8C3 because there are 8 people left not including Peter and Deanna and we only want 3 of those people to be in the committee with one of them. It is then divided again by 10C4 for the same reasons.

c) Neither Peter nor Deanna is chosen for the committee.

    -----------  = 1/3

well since we dont want neither of them we just multiply by the rest of the people left over for the committee of four and divide it by all the possible combinations.

okay well i hope i did a good way of going over the pre-test and described everything well for everyone. And the next scribe post will be done by:

SEZY.... luckily you get to wait until after the test tomorrow =)

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