So to find the P A B we can either Break up the set into 3 mutually exclusive

# So to find the p a b we can either break up the set

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So, to find the P ( A B ) we can either • Break up the set into 3 mutually exclusive sets 1. A B c = { 1 , 3 , 5 } 2. A B = { 7 , 9 } 3. A c B = { 6 , 8 , 10 } and by using the axioms P ( A B ) = P ( A B c ) + P ( A B ) + P ( A c B ) • Add P ( A ) + P ( B ) but we need to subtract the probability of the intersection as that probability was double counted since it is included within A and within B , leading to P ( A B ) = P ( A ) + P ( B ) - P ( A B ) . Example 2.8 A sample of 1000 aluminum rods is taken and a quality check was per- formed on each rod’s diameter and length. Diameter Length Too Thin OK Too Thick Sum Too Short 10 3 5 18 OK 38 900 4 942 Too Long 2 25 13 40 Sum 50 928 22 1000 P (Diameter OK) = 928 / 1000 = 0 . 928 P (Length OK) = 942 / 1000 = 0 . 942 P (Diameter OK Length OK) = 900 / 1000 = 0 . 9 P (Diameter OK Length OK) = 0 . 928 + 0 . 942 - 0 . 9 = 0 . 97 2.3 Counting Methods Lemma 2.1 (Product Rule) If the first task of an experiment can result in n 1 possi- ble outcomes and for each such outcome, the second task can result in n 2 possible outcomes, and so forth up to k tasks, then the total number of ways to perform the sequence of k tasks is Q k i =1 n i 22 2.3. Counting Methods Example 2.9 When buying a certain type and brand of car a buyer has the following number of choices: (2) Engine (5) Color (4) Interior Then, the total number of car choices, i.e. number of unique cars, is 2 × 5 × 4 = 40 . Notation: In mathematics, the factorial of a non-negative integer n , denoted by n !, is the product of all positive integers less than or equal to n , with 0! = 1 . For example, 5! = 5 × 4 × 3 × 2 × 1 = 120 In R : factorial(5) 2.3.1 Permutations Definition 2.4 (Permutation) Permutation is the number of ordered arrangements, or permutations, of r objects selected from n distinct objects ( r n ) without replacement . It is given by, P n r = n ! ( n - r )! . The number of permutations of n objects is n ! . Example 2.10 Take the letters A,B,C . Then there are 3! = 6 possible permutations. These are: ABC,ACB,BAC,BCA,CAB,CBA In e ff ect we have three “slots” in which to place the letters. There are 3 options for the first slot, 2 for the second and 1 for the third. Hence we have 3 × 2 × 1 = 6 = P 3 3 . Example 2.11 Let n = 10 with A,B,C,D,E,F,G,H,I,J and we wish to select 5 letters at random, where order is important. Then there are 5 slots with 10 choices for the first slot, 9 for the second, 8 for the third, 7 for the fourth and 6 for the fifth. That is, 10 × 9 × 8 × 7 × 6 = 10! 5! = P 10 5 = 30240 . Example 2.12 An election is held for a board consisting of 3 individuals out of a pool of 40 candidates. There will be 3 positions of: 1. President 2. Vice-President 3. Secretary Chapter 2. Probability and Random Variables  • • • 