Probability. Robert P. Dobrow

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      Computing probabilities for equally likely outcomes takes a fairly simple form. Suppose upper A is an event with s elements, with s less-than-or-equal-to k. As upper P left-parenthesis upper A right-parenthesis is the sum of the probabilities of all the outcomes contained in upper A,

upper P left-parenthesis upper A right-parenthesis equals sigma-summation Underscript omega element-of upper A Endscripts upper P left-parenthesis omega right-parenthesis equals sigma-summation Underscript omega element-of upper A Endscripts StartFraction 1 Over k EndFraction equals StartFraction s Over k EndFraction equals StartFraction Number of elements of upper A Over Number of elements of normal upper Omega EndFraction period

      In other words, probability with equally likely outcomes reduces to counting elements in upper A and normal upper Omega.

       Example 1.9 A palindrome is a word that reads the same forward or backward. Examples include mom, civic, and rotator. Pick a three-letter “word” at random choosing from D, O, or G for each letter. What is the probability that the resulting word is a palindrome? (Words in this context do not need to be real words in English, e.g., is a palindrome.)There are 27 possible words (three possibilities for each of the three letters). List and count the palindromes: DDD, OOO, GGG, DOD, DGD, ODO, OGO, GDG, and GOG. The probability of getting a palindrome is

       Example 1.10 A bowl has red balls and blue balls. A ball is drawn randomly from the bowl. What is the probability of selecting a red ball?The sample space consists of balls. The event has elements. Therefore, .

      A model for equally likely outcomes assumes a finite sample space. Interestingly, it is impossible to have a probability model of equally likely outcomes on an infinite sample space. To see why, suppose normal upper Omega equals StartSet omega 1 comma omega 2 comma ellipsis EndSet and upper P left-parenthesis omega Subscript i Baseline right-parenthesis equals c for all i, where c is a nonzero constant. Then summing the probabilities gives

sigma-summation Underscript i equals 1 Overscript infinity Endscripts upper P left-parenthesis omega Subscript i Baseline right-parenthesis equals sigma-summation Underscript i equals 1 Overscript infinity Endscripts c equals infinity not-equals 1 period sigma-summation Underscript i equals 1 Overscript infinity Endscripts upper P left-parenthesis omega Subscript i Baseline right-parenthesis equals sigma-summation Underscript i equals 1 Overscript infinity Endscripts left-parenthesis one half right-parenthesis Superscript i Baseline equals left-parenthesis one half right-parenthesis StartFraction 1 Over 1 minus left-parenthesis 1 slash 2 right-parenthesis EndFraction equals 1 period

      We introduce some basic counting principles in the next two sections because counting plays a fundamental role in probability when outcomes are equally likely.

      Counting sets is sometimes not as easy as 1 comma 2 comma 3 comma ellipsis. But a basic counting principle known as the multiplication principle allows for tackling a wide range of problems.

      Multiplication principle

      If there are m ways for one thing to happen, and n ways for a second thing to happen, there are m times n ways for both things to happen.

      More generally—and more formally—consider an n-element sequence left-parenthesis a 1 comma a 2 comma ellipsis comma a Subscript n Baseline right-parenthesis period If there are k 1 possible values for the first element, k 2 possible values for the second element, ellipsis, and k Subscript n possible values for the nth element, there are k 1 times k 2 times midline-horizontal-ellipsis times k Subscript n possible sequences.

      For instance, in tossing a coin three times, there are 2 times 2 times 2 equals 2 cubed equals 8 possible outcomes. Rolling a die four times gives 6 times 6 times 6 times 6 equals 6 Superscript 4 Baseline equals 1296 possible rolls.

       Example 1.11 License plates in Minnesota are issued with three letters from A to Z followed by three digits from 0 to 9. If each license plate is equally likely, what is the probability that a random license plate starts with G-Z-N?The solution will be equal to the number of license plates that start with G-Z-N divided by the total number of license plates. By the multiplication principle, there are possible license plates.For the number of plates that start with

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