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Unit-2 Probability:- Probability is “How likely something is to happen” or “chance of occurring of an event”. Many events can't be predicted with total certainty. The best we can say is how likely they are to happen, using the idea of probability. Tossing a Coin when a coin is tossed, there are two possible outcomes: heads (H) or tails (T). We say that the probability of the coin landing H is ½. And the probability of the coin landing T is ½. Throwing Dice When a single die is thrown, there are six possible outcomes: 1, 2, 3, 4, 5, 6. The probability of any one of them is 1/6. Probability In general: Probability of an eventhappening = Number of wa yscanhappen Total numberof outcomes Example: The chances of rolling a "4" with a die. Number of ways it can happen: 1 (there is only 1 face with a "4" on it) Total number of outcomes: 6 (there are 6 faces altogether) So the probability ¿ 1 6

Unit 2 Probability

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  1. 1. Unit-2 Probability:- Probability is How likely something is to happen or chance of occurring of an event. Many events can't be predicted with total certainty. The bestwe can say is how likely they are to happen, using the idea of probability. Tossing a Coin when a coin is tossed, there are two possible outcomes: heads (H) or tails (T). We say that the probability of the coin landing H is . And the probability of the coin landing T is . Throwing Dice When a single die is thrown, there are six possible outcomes: 1, 2, 3, 4, 5, 6. The probability of any one of them is 1/6. Probability In general: = Example:The chances of rolling a "4" with a die. Number of ways it can happen: 1 (there is only 1 face with a "4" on it) Total number of outcomes: 6 (there are 6 faces altogether) So the probability = 1 6 Example:There are 5 marbles in a bag: 4 are blue, and 1 is red. Whatis the probability that a blue marble gets picked? Number of ways it can happen: 4 (there are 4 blues) Total number of outcomes: 5 (there are 5 marbles in total) So the probability = 4 5 = 0.8 Note:- Probability is always between 0 and 1
  2. 2. Probability is Just a Guide (Probability does not tell us exactly whatwill happen, it is justa guide) Example:Toss a coin 100 times, how many Heads will come up? Probability says that heads have a chance, so we can expect 50 Heads. But when we actually try it we might get 48 heads, or 55 heads ... or anything really, but in most cases it will be a number near 50. Some words havespecial meaning in Probability: Experiment or Trial: an action where the result is uncertain. Tossing a coin, throwing dice, seeing whatpizza people chooseare all examples of experiments. Sample Space: all the possibleoutcomes of an experiment Example:choosing a card froma deck There are 52 cards in a deck (not including Jokers) So the Sample Space is all 52 possiblecards: {Ace of Hearts, 2 of Hearts, etc... } The Sample Space is made up of Sample Points: Sample Point: justone of the possible outcomes Example:Deck of Cards, the 5 of Clubs is a sample point the King of Hearts is a sample point "King" is not a sample point. As there are 4 Kings, that is 4 different sample points. Event:a single result of an experiment Example Events: Getting a Tail when tossing a coin is an event Rolling a "5" is an event. An event can include one or more possibleoutcomes: Choosing a "King" froma deck of cards (any of the 4 Kings) is an event Rolling an "even number" (2, 4 or 6) is also an event
  3. 3. The Sample Space is all possibleoutcomes. A Sample Point is justone possibleoutcome. And an Event can be one or more of the possibleoutcomes. Hey, let's usethose words, so we get used to them: Example:Alex wants to see how many times a "double" comes up when throwing 2 dice. Each time Alex throws the 2 dice is an Experiment. Itis an Experiment becausethe resultis uncertain. The Event Alex is looking for is a "double", whereboth dice havethe same number. Itis made up of these 6 Sample Points: {1,1} {2,2} {3,3} {4,4} {5,5} and {6,6} The Sample Space is all possibleoutcomes (36 Sample Points): {1,1} {1,2} {1,3} {1,4} ... {6,3} {6,4} {6,5} {6,6} These are Alex's Results: Experiment Is it a Double? {3,4} No {5,1} No {2,2} Yes {6,3} No ... ...
  4. 4. After 100 Experiments, Alex has 19 "double" Events ... is that close to what you would expect? Dependent Events:- Event A and B in the sample spaceS are said to be independent if occurrenceof one of them does not influence the occurrenceof the other. i.e. B is dependent of A if P(B/A)=P(B) Now, P(AB)=P(A).P(B|A)=P(A)P(B) Thus events A and B are independent, if P(AB)=P(A).P(B) Conditional Probabiliry:- Given an event B such that P(B)>0 and any other event A, we define a conditional probability of A is given by ( | ) = ( ) () Example: Let a balanced die is tossed once. Use the definition to find the probability of getting 1, given that an odd number was obtained. Solution:- Probability of A given that the event B has occurred, ( | ) = () () But ( ) = 1 2 , ( | ) = () () = 1 6 1 2 = 1 3 . Law of Total Probability Let 1, 2, be mutually exclusive and let an event A occur only if anyoneof occurs, Then ( ) = (|)() =1 Bayes Theorem Let 1, 2, be mutually exclusive event such that ( ) > 0 for each i. Then for any event 1 such that ( ) > 0, we have ( |) = (|)() (|)() =1 = 1,2, Example:- Three urns A, B, C have 1 white , 2 black, 3 red balls, 2 white 1 black, 1 red balls and 4 white Example:- Let P be the probability that a man aged y years will meet with an accident in a year. What is the probability that a man among n men all aged y years will meet with an accident ? Solution:-
  5. 5. Probability that a man aged y years will not meet with an accident =1-p P (none meet with an accident) =(1 )(1 ) = (1 ) p (atleast one man meets with an accident) = 1 (1 ) So p( atleast one man meets with an accident, a person is chosen) = 1 [1 (1 ) ] Example:- A box contains 4 white, 3 blue and 5 green balls. Four balls are chosen. What is the probability that all three colors are represented ? Solution:- The total number of balls in the box is 12. Hence the total number of ways in which 4 balls can be chosen = 12 4 = 12 11 10 9 4 3 2 1 = 495 Each color will be represented in the following mutually exclusive ways: White Blue Green (i) 2 1 1 (ii) 1 2 1 (iii) 1 1 2 Hence the number of ways of drawing four balls in the abovefashion = 4 2 3 1 5 1 + 4 1 3 2 5 1 + 4 1 3 1 5 2 = 90 + 60 + 120 = 270 So Required Probability = 270 495 . Random variable:- A variable whosevalueis determined by the outcome of randomexperiment is called randomvariable. The value of the randomvariable will vary fromtrial to trial as the experiment is repeated. Random variable is also called chance variableor stochastic variable. There are two types of random variable - discrete and continuous. A random variable has either an associated probability distribution (discrete random variable) or probability density function (continuous random variable). Discrete randomvariable:- If the random variable takes on the integer values such as 0,1,2, then it is called discrete random variable.
  6. 6. Example:- 1. The number in printing mistake in a book, the number of telephone calls received by the phone operator on a farm areexample of discrete random variable 2. A coin is tossed ten times. The randomvariable X is the number of tails that are noted. X can only take the values 0, 1, ..., 10, so X is a discrete random variable. Continuous random variable:- If the randomvariable takes all values, with in a certain interval, then the random is called continuous randomvariable. Example:- 1. The amount of rainfall on a rainy day or in a rainy season, the height and weight of individuals are example of continuous randomvariable. 2. A light bulb is burned until it burns out. The randomvariable Y is its lifetime in hours. Y can take any positive real value, so Y is a continuous random variable. Probability Distribution The probability distribution of a discrete randomvariable is a list of probabilities associated with each of its possiblevalues. Itis also sometimes called the probability function or the probability mass function. More formally, In terms of symbols if a variableX can assumediscrete set of values 1, 2, with respective probabilities 1, 2, where 1 + 2 + + = 1, we say that a discrete probability distribution for has been defined. The function () which has the respectivevalues 1 , 2 , for = 1, 2, is called the probability function or frequency function of . Note:-
  7. 7. a. 0 () 1 b. ( ) = 1 ExpectedValue:- The expected value (or population mean) of a randomvariable indicates its averageor central value. Itis a useful summary value (a number) of the variable's distribution. Stating the expected value gives a general impression of the behavior of some randomvariable without giving full details of its probability distribution (if it is discrete) or its probability density function (if it is continuous). Two randomvariables with the same expected value can havevery different distributions. There are other usefuldescriptivemeasures which affect the shape of the distribution, for example variance. The expected value of a randomvariable X is symbolized by E(X) or . If is a discrete randomvariable with possible values 1, 2, with respective probabilities 1 , 2 , where 1 + 2 + + =1the mathematical expectation of is defined as: = ( ) = ()() Where the elements are summed over all values of the randomvariable X. If X is a continuous randomvariablewith probability density function f(x), then the expected value of X is defined by: = ( ) = ( ) Example
  8. 8. Discrete case: When a die is thrown, each of the possiblefaces 1, 2, 3, 4, 5, 6 (the 's) has a probability of 1/6 (the ()'s) of showing. Theexpected value of the face showing is therefore: = ( ) = (1 1 6 ) + (2 1 6 ) + (3 1 6 ) + (4 1 6 ) + (5 1 6 ) + (6 1 6 ) = 3.5 Notice that, in this case, () is 3.5, which is not a possiblevalue of . Variance of = ( ) = 2 = ( 2) [ ()]2 Standard Deviationof = Example:- Find expectation of the number of points when a fair die is rolled. Solution:- Let be the randomvariable showing number of points. Then = 1,2,3,4,5,6 ( = ) = () Product 1 1 6 1 6 2 1 6 2 6 3 1 6 3 6 4 1 6 4 6 5 1 6 5 6 6 1 6 6 6 ________________ ( ) = 21 6 = 7 2 Expectation = 7 2 .
  9. 9. Normal Distribution In probability theory, the normal (or Gaussian) distribution is a very commonly occurring continuous probability distribution . Strictly, a Normalrandomvariable should be capable of assuming any valueon the real line, though this requirement is often waived in practice. For example, height at a given age for a given gender in a given racial group is adequately described by a Normal randomvariableeven though heights mustbe positive. A continuous randomvariable X is said to follow a Normal distribution with parameters and if it has probability density function ( ) = (, , ) = 1 2 ( )2 22 , > 0, < < In such situation we write ~( , 2) The distribution involves two parameters and 2 . Properties:- a) Distribution is symmetrical. b) = , = 2 .
  10. 10. c) Mean , median and mode coincide. d) () 0 for all . e) ( ) = 1 i.e. Total area under the curve = () bounded by the axix of is 1. : a) Curvey-f(x), called normal curveis a bell shaped curve. b) Itis symmetric about = . c) The two tails on the left and right sides of the mean extend to infinity. d) Put = . Then is called a standard normalvariable and its probability density function is given by ( ) = 1 2 2 2 , . e) Mean of the standard Normaldistributions is 0 and varianceis 1. Write ~(0,1) The simplest case of a normaldistribution is known as the standard normaldistribution. Areaunder the standard normal curve:- a) = 1 = 1 0.6827(Since total area under the standard normal curveis 1) . . (1 < < 1) = 0.6827 b) = 2 = 2 0.9595 . . (2 < < 2) = 0.9595 c) = 3 = 3 0.9973 . . (3 < > 3) = 0.9973 In other words,
  11. 11. EXAMPLE: If a randomvariable has the normaldistribution with = 82.0 and = 4.8, Find the probabilities that it will take on a value (a) less than 89.2 (b) greater than 78.4 (c) between 83.2 and 88.0 (d) between 73.6 and 90.4 Solution: (a) We have
  12. 12. z = 89.2 82 4.8 = 1.5 therefore the probability is 0.4332 + 0.5 = 0.9332. (b) We have = 78.4 82 4.8 = 0.75 therefore the probability is 0.2734 + 0.5 = 0.7734 . (c) We have 1 = 83.282 4.8 = 0.25and 2 = 8882 4.8 = 1.25 therefore the probability is 0.3944 0.0987 = 0.2957. (d) We have 1 = 73.682 4.8 = 1.75and 2 = 90.482 4.8 = 1.75 hereforethe probability is 0.4599 + 0.4599 = 0.9198. Applications of the Normal Distribution EXAMPLE: Intelligence quotients (IQs) measured on the Stanford Revision of the Binet-Simon Intelligence Scale are normally distributed with a mean of 100 and a standard deviation of 16. Determine the percentage of people who haveIQs between 115 and 140. Solution: We have 1 = 115100 16 = 0.9375 and 2 = 140100 16 = 2.5 therefore the probability is 0.4938 0.3264 = 0.1674. Itfollows that 16.74% of allpeople have IQs between 115 and 140. Equivalently, the probability is 0.1674 thata randomly selected person will havean IQ between 115 and 140. Many distribution tend to a normaldistribution in the limit. When a variable is not normal, it can be made normal using using suitable transformation. When the sample size is large distrubution of simple mean, simple variance etc. approach normality. Thus distribution forms a basis for test of significance. Normal distribution is also called distribution of errors.