{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

# Lesson3 - Lesson 3 THE REVISIONS ARE INDICATED BY THE RED...

This preview shows pages 1–5. Sign up to view the full content.

Lesson 3 Atul Roy Please read the chapters 3 and 4 . We are going to review p robability concepts with the use of simple examples. Example 1. If we roll an ordinary six faced die, what is the probability that the face with two dots will show up? A six faced die is a cube with 1,2,3,4,5,6 dots on the faces. In the above phenomenon, the sample space (the set of all possible outcomes) is or in simple terms {1,2,3,4,5,6} Since there are six equally likely outcomes, and 2 dots show up in only one of them, P(the face with two dots will show up) = 1 6 moreover: P(number of dots facing up is even) = 3 6 Example 2. 1

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
A mathematics department has 61 sections of different courses for th is term. Course Number of Sections Finite Mathematics 19 Statistics 15 Precalculus 9 Calculus I 7 Calculus II 6 Calculus III 2 Differential Equations 1 Linear Algebra 1 Abstract Algebra 1 Total=61 If a section is selected at random for evaluation, the probability that the course is precalculus is 9 61 the probability that the course is not statistics = 61 15 61 = 46 61 Example 3. Suppose that we have a group of five people named Alen, Bina, Chris, David and Erin. We would like to pick a simple random sample of three people from this group. Call the members of the group as A,B,C,D,E. a) Let us look at the list of all possible samples of size 3 .Theyare ABC ABD ABE ACD ACE ADE BCD BCE BDE CDE b) W e would like to find the probability that a sample containing Bina and Chris is selected. Note that there are three samples ABC, BCD, BCE containing Bina and Chris. Therefore the probaility that a sample with Bina and Chris is selected is 1 10 . 2
Example 4. If we roll two dice, for convenience, suppose one is green and one is yellow the possible outcomes are or in simple terms, it is (1,1),(1,2),(1,3),(1,4),(1,5),(1,6) (2,1),(2,2),(2,3),(2,4),(2,5),(2,6) (3,1),(3,2),(3,3),(3,4),(3,5),(3,6) (4,1),(4,2),(4,3),(4,4),(4,5),(4,6) (5,1),(5,2),(5,3),(5,4),(5,5),(5,6) (6,1),(6,2),(6,3),(6,4),(6,5),(6,6) Note that there are 36 outcomes. If we have to calculate the probability that the sum of the dots facing up is 5 . note that there are 4 options that give a sum of 5, 3

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document