{ a } { a, b } { a, b, c } { a, c } { b } 8 subsets. { b, c } { c } Zero elements. One element. Two elements. Three elements. { a } { a, b } { a, b, c } { a, c } { b } 8 subsets. { b, c } { c }

Finding the number of subsets and the cardinal number of a set. Cardinal number is a non-negative integer defining how many elements are in a set. It is denoted as n( ). Example, n( O ) = 0 Let A = { a, b }. How many subsets are there. 4 subsets. Zero elements. One element. Two elements. { a } { b } { a, b } Let n be the cardinal number of set A. Set A has 2n subsets. What would the formula for the number of Proper Subsets? Define Union and Intersection of sets. Intersection of sets are the elements that are the same in both sets. In A and B. Union of sets are all the elements that are in each set. In A or B. Counting Principle (Addition).

How many were registered in College Algebra or English? In a survey of 100 college students, 35 were registered in College Algebra, 52 were registered in English, and 18 were registered in both courses. union How many were registered in College Algebra or English? How many were registered in neither class? Make a tree diagram for two appetizers, 4 entrees, and 2 desserts. 1 2 3 4 5 soup 6 7 8 9 salad 10 11 12 13 14 15 16

Counting Principle (Multiplication). If a task consists of a sequence of individual events, then multiply the number of each event to find total possibilities. The Student Union is having a lunch special value meal. You get to choose one of 4 sandwiches, one of 5 bags of chips, one of 7 drinks and one of 2 desserts. How many different lunch specials can you make?

How many letters in the alphabet? ____ ____ ____

____ ____ ____

men women

4 I’s, 4 S’s and 2 P’s There are 11 letters. n = 11 This is the repeating letters. There are 11 letters. n = 11 4 I’s, 4 S’s and 2 P’s

1’s must be on the corners of the triangle. How to use the graphing calculator to find a row of Pascal’s triangle! Find the 5th row.

( ) ( ) 1( )4( )0 + 4( )3( )1 + 6( )2( )2 + 4( )1( )3 + 1( )0( )4 2x Decreasing power Increasing power ( ) ( ) Need the 4th row of Pascal’s Triangle. 1, 4, 6, 4, 1 1( )4( )0 + 4( )3( )1 + 6( )2( )2 + 4( )1( )3 + 1( )0( )4 2x – 3y 2x – 3y 2x – 3y 2x – 3y 2x – 3y Notice the powers on both variables add up to 4, the power on the binomial.

Remember from the last example that the powers on the x-term and y-term have to add up to be . . . Therefore, 10 – 7 = 3 = r. The y-term is always to the r power. n = 10 Remember that the Combination formula starts counting with zero for the r! r = 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10 r = 6

6 * 6 = 36 total 4 * 13 = 52 total Sample Space (Total = 8 = 23 ) H H 1st flip T H T T H 2nd flip T T T T 3rd flip Sample Space (Total = 8 = 23 ) 1 2 3 4 5 6 1 2 3 4 5 6 7 2 3 4 5 6 7 8 3 4 5 6 7 8 9 2 3 4 5 6 7 8 9 10 J Q K A 4 5 6 7 8 9 10 Hearts 5 6 7 8 9 10 11 Diamonds 6 7 8 9 10 11 12 Clubs Spades 6 * 6 = 36 total 4 * 13 = 52 total

Always find total number of outcomes 1st! Sample Space H H H H H T H T H H T T T H H T H T T T H T T T 1 2 3 4 5 6 1 2 3 4 5 6 7 2 3 4 5 6 7 8 3 4 5 6 7 8 9 2 3 4 5 6 7 8 9 10 J Q K A 4 5 6 7 8 9 10 Hearts 5 6 7 8 9 10 11 Diamonds 6 7 8 9 10 11 12 Clubs Spades

NO WAY! What is the opposite of at least 1 tails? No Tails or All Heads. All Heads can only happen one way.

10 probabilities would have a denominator of 365. The complement is P( all different birthdays ), 1 – P( all different B-days ). There are 365 days in the year and every day is likely to occur. Find the size of the sample space… 10 probabilities would have a denominator of 365. The 1st person has all 365 days available. The numerators are an arrangement where order matters, therefore, we can use Permutation notation. n = 365, r = 10 The 2nd person has 364 days available. The 3rd person has 363 days available. 365 364 363 362 361 360 359 358 357 356 P(different B-days) = 1 – P( different B-days )

P( 2 girls in 5 trials ) Binomial Probability. This probability has only two outcomes, success or failure. Let n = the number of trials. Let r = the number of successes. Let n – r = the number of failures. Let p = the probability of success in one trial. Let q = the probability of failure. 1 – p = q. Binomial Probability Formula P( r successes in n trials ) Always need two categories. Other common categories are Good or Bad, Boys or Girls, True or False, etc. The sum of the two probabilities is 1. The spirit club is 75% girls. What is the probability of randomly picking 5 people, where there are 3 boys and 2 girls? P( 2 girls in 5 trials )

A manufacture has determined that a bulb machine will produce 1 bad bulb for every 2000 bulbs it produces. What is the probability that an order of 200 bulbs are all perfect? There is at least one bad bulb? We need the new Binomial Probability formula. There is one bad bulb? There are two bad bulbs?

Total chips = 10 1st pick 2nd pick 3rd pick 1st pick 2nd pick 3rd pick Notice that the denominator decreases by one on each pick because there is no replacement. Same as first, but order doesn’t matter. 1st pick 2nd pick 3rd pick We have to consider all the different ways the 3 colors can be picked. RWB, RBW, WRB, WBR, BRW, and BWR. 3! = 6 1st pick 2nd pick 3rd pick Same colors, but order doesn’t matter. We have to consider all the different ways the 3 colors can be picked. RWB, RBW, WRB, WBR, BRW, and BWR. 3! = 6

Total chips = 10 Another approach we can use with Probabilities that have NO REPLACEMENT, is to use Permutations or Combinations. Since order matters, this is a Permutation. Since order doesn’t matter, this is a Combination.