MATH 1780 Lecture Notes Chapter 2 Section 5 and 6

# MATH 1780 Lecture Notes Chapter 2 Section 5 and 6 - To...

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To begin this section, we will be looking at unemployment data from 1989. This will lead us naturally into Conditional Probability.

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Education Employed Unemploy ed Total Elementary School 5,299 406 5,765 High School, 1-3 years 8,144 705 8,149 High School, 4 years 38,171 1763 39,934 College, 1-3 years 19,991 671 20,662 College, 4+ years 26,015 570 26,585 Total 97,620 4,115 101,735 Civilian Labor Force in the USA, 1989 (Figures in Thousands) A common summary of these data is the “unemployment rate,” which is the Percentage of unemployed workers given by: But this figure does not take into account the educational level. We would need to narrow the focus to each row of the above chat. This is called conditioning. ( 29 %. 0 . 4 100 000 , 735 , 101 000 , 115 , 4 =
Here, we see the unemployment rates conditioned to education. Note the increase from 7% to 8% from Elementary School to High-School, 1-3 year. Why do you think this increase occurred? If you were to poll a person, what is the probability that person is unemployed given that he has graduated from High School, but has had Education Employed Unemployed Elementary School 93% 7% High School, 1-3 Years 92% 8% High School, 4 years 96% 4% College, 1-3 years 97% 3% College, 4+ years 98% 2% Unemployment rates by Education

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Suppose we look at the following Venn diagram: Suppose event A represents students who receive an A, and B represents Business Majors. Note that P(A) = 0.1, but what is the probability that a business major gets an A? There are only 20 business majors, and 3 of them got A’s. So the probability that a business major gets an A is 0.15.
If A and B are any two events, then the conditional probability of A given B, denoted by P(A|B), is provided P(B) > 0. The definition of conditional probability basically restricts the sample space down to the event B, because we already know B has happened. Thus we only care about the amount of the event A which is in B. P(B) P(AB) B) | P(A =

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Example 1: From among 5 motors, of which one is defected, two are to be selected at random for use on a particular day. Find the probability that the second motor is non-defective, given that first was non- defective. Solution: Let N i denote that i th motor selected is non-defective. We want to find P(N 2 |N 1 ). From the definition of conditional probability we have First, we can use permutations to figure out how many possible outcomes there are. ( 29 ( 29 ( 29 . N P N N P N | N P 1 1 2 1 2 =
Solution (cont): To see how many outcomes are in N 1 N 2 , i.e. both motors selected are good, we can use the multiplication principle. There are 4 good motors to select from first, but then there are only 3 good motors to select from second. Thus there are 4*3 = 12 outcomes in N 1 N 2 . We can use the multiplication principle

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## This note was uploaded on 09/30/2008 for the course MATH 1780 taught by Professor Snyder during the Fall '08 term at North Texas.

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MATH 1780 Lecture Notes Chapter 2 Section 5 and 6 - To...

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