chap3 - 3. Probability Counting Techniques Some probability...

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3. Probability – Counting Techniques Some probability problems can be attacked by specifying a sample space S = { a 1 ,a 2 ,...,a n } in which each simple event has probability 1 n (i.e. is “equally likely"). Thus, if a compound event A consists of r simple events, then P ( A )= r n . To use this approach we need to be able to count the number of events in S and in A , and this can be tricky. We review here some basic ways to count outcomes from “experiments". These approaches should be familiar from high school mathematics. 3.1 General Counting Rules There are two basic rules for counting which can deal with most problems. We phrase the rules in terms of “jobs" which are to be done. 1. The Addition Rule: Suppose we can do job 1 in p ways and job 2 in q ways. Then we can do either job 1 or job 2, but not both, in p + q ways. For example, suppose a class has 30 men and 25 women. There are 30 + 25 = 55 ways the prof. can pick one student to answer a question. 2. The Multiplication Rule: Suppose we can do job 1 in p ways and an unrelated job 2 in q ways. Then we can do both job 1 and job 2 in p × q ways. For example, to ride a bike, you must have the chain on both a front sprocket and a rear sprocket. For a 21 speed bike there are 3 ways to select the front sprocket and 7 ways to select the rear sprocket. This linkage of OR with addition and AND with multiplication will occur throughout the course, so it is helpful to make this association in your mind. The only problem with applying it is that questions do not always have an AND or an OR in them. You often have to play around with re-wording the question for yourself to discover implied AND’s or OR’s. 15
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16 Example: Suppose we pick 2 numbers at random from digits 1, 2, 3, 4, 5 with replacement. (Note: “with replacement” means that after the first number is picked it is “replaced” in the set of numbers, so it could be picked again as the second number.) Let us find the probability that one number is even. This can be reworded as: “The first number is even AND the second is odd, OR, the first is odd AND the second is even.” We can then use the addition and multiplication rules to calculate that there are (2 × 3) + (3 × 2) = 12 ways for this event to occur. Since the first number can be chosen in 5 ways AND the second in 5 ways, S contains 5 × 5=25 points. The phrase “at random” in the first sentence means the numbers are equally likely to be picked. Therefore P ( one number is even )= 12 25 When objects are selected and replaced after each draw, the addition and multiplication rules are gen- erally sufficient to find probabilities. When objects are drawn without being replaced, some special rules may simplify the solution. Problems: 3.1.1 (a) A course has 4 sections with no limit on how many can enrol in each section. 3 students each randomly pick a section. Find the probability: (i) they all end up in the same section (ii) they all end up in different sections (iii) nobody picks section 1.
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This note was uploaded on 12/01/2010 for the course MATH Stat 230 taught by Professor A during the Spring '07 term at Waterloo.

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chap3 - 3. Probability Counting Techniques Some probability...

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