204 5.5 counting perm%2c comb

# 204 5.5 counting perm%2c comb - 5.5 Counting Permutations...

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5.5 Counting Permutations and Combinations In calculating classical probabilities we will need to be able to count both the number of possible outcomes of an experiment and the number of outcomes that will be considered success. For example, if we are tossing a six sided die the number of possible outcomes is 6. If we are interested in determining the number of ways to get an even number of dots then there are 3 ways to get a success (2,4, & 6). If the die is fai r (i.e. any side is as likely to appear as any other side) the probability of getting an even number of dots is just 3/6 = 1/2. In other words we would expect that over the long run about 1/2 of the rolls will have an even number of dots showing. Thus, when all outcomes of an experiment are equally likely we can calculate the probability of success by taking the ratio of the number of ways success can be achieved to the total number of ways the experiment can turn out (number of outcomes for success/number of possible outcomes for the experiment). As an example of counting the possible outcomes let’s consider a student filling up a two hour block of their schedule. For the first hour the can take either History, English, or Chemistry. For the second hour the student can take either Mathematics or a Foreign Language. The number of ways to complete the schedule can be calculated by what is known as a tree diagram. 1st hour History English Chemistry 2nd hour Math or Lang Math or Lang Math or Lang There are six (3x2) possible ways to fill the two hour block (h,m) (h,fl) (e,m) (e,fl) (c,m) (c,fl) In general if there are x ways to choose a first step in a two step process and y ways to choose the second step then there are x*y ways to do the two step process. In a k step process with n

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## This note was uploaded on 12/05/2011 for the course MATH 2040 taught by Professor Raysievers during the Fall '10 term at Utah Valley University.

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204 5.5 counting perm%2c comb - 5.5 Counting Permutations...

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