Probability Concepts

What is Probability?

Probability - A value between zero and one, inclusive, describing the relative possibility (chance or likelihood) an event will occur.

• Experiment - A process that leads to the occurrence of one (and only one) of several possible observations
• Simple event or Outcome - A particular result of an experiment
• Event - A collection of one or more outcomes of an experiment
• Sample space - All possible simple events.

Approaches to Probability

• Relative Frequency Concept

  

• Classical - Equally likely simple events or outcomes

  

  • Mutually exclusive - The occurrence of any one event meansthat none of the others can occur at the same time

  • Collectively Exhaustive At least one of the events must occur when an experiment is conducted.

• Subjective Probability - The likelihood (probability) of a particular event happening that is assigned by an individual based on whatever information is available. For example everytime an entertainment industry executive decides to go ahead with a TV program because they feel it has a good chance of being a success they have used their judgement to estimate a probability.
  

  Some Rules of Probability

  • Rules of Addition

    • Special Rule of Addition - Events must be mutually exclusive

      P(A or B) = P(A) + P(B)

    • General Rule of Addition

      P(A or B) = P(A) + P(B) - P(A and B)

      Joint Probability - P(A and B) - probability that measures the likelihood that two or more events will happen concurrently.

    Idea of Venn Diagram, Venn diagrams are used to show how events relate to one another. The total area of the figure shown is 1.0 (the probability that something will happen). The area of each event represents its probability of occurence. If events are mutually exclusive their areas do not overlap. If they are not mutually exclusive there will be an area of overlap.

    

    Complement rule:P(A) + P(~A) = 1.0 or P(A) = 1.0 - P(~A)

  • Rules of Multiplication

    • Special Rule of Multiplication - P(A and B) = P(A) P(B)

      Independent - The occurrence of one event has no effect on the probability of the occurrence of the other event.

    • General Rule of Multiplication - P(A and B) = P(A) P(B|A)

      Conditional probability - P(B|A) -The probability of a particular event occurring (event B) given that another event (event A) has already occurred.

  
  

  Tree Diagrams, The tree diagram is a way to depict a probability-based decision process. The first branch represents the probabilities of events A and ~A (Not A). The second branch, coming from the ends of the first branch represents the conditional probabilities of events B or ~B. If events A and B are independent then the second branch just has the probability of B and ~B (left side of the figure below). If P(B) depends on whether A occurred or not then the second branch is P(B|A) or P(B|~A). These are the probability of B given that A or ~A occurred first. This is called a conditional probability. To give this numerical meaning we will work two examples. In case one the probability of B will not depend on A and in case two the probability of B will depend on A.

  

  
  

  First consider the case where you have two jars with balls in them. The first jar has seven amber balls (A) and three that are not amber. So, P(A)=7/10 = 0.7 and P(~A)=3/10=0.3. The second jar has four balls that are blue (B) and six that are not blue (~B). So, P(B) = 0.4 and P(~B)=0.6. If we draw one ball from the first jar and one from the second jar we have the situation depicted in the left hand tree shown below. The draws are indepndent (second jar draw does not depend on the result of our draw from the first jar). The probabilities for the four possible combinations of A,~A, and B, ~B are shown in the figure.

  Now let us consider a case where all of the balls are in one jar. We have seven amber (A) and three not amber (~A) in the one jar. What we draw on the first draw will have an impact on what we draw on the second draw. If we draw an A on the first draw the chance for an A on the second draw will be P(A₁ and A₂)= 7/10 * 6/9. The chance for an A on the first draw and ~A on the second draw will be P(A₁ and ~A₂)= 7/10 * 3/9. The other possibilities are shown in the figure.

  

  Some Principles of Counting

  • The Multiplication Formula, If you have several groups of objects and you are going to select one from each group to make an arrangement the total number of arrangements = (m) (n) (r) (etc...). For example if your new car comes in four exterior colors, three body styles and five interiors the number of possible arrangements is (4)(3)(5)=300
  • The Permutation Formula, If you are going to select x objects from a group of size n, and the order of selection matters, the total number of arrangements is calculated with the Permutation formula (shown below). If you wanted to know how many arrangments you could form from a group of 5 colors where you are selecting 3 at a time (and the order matters) the number of arrangements would be 5!/(5-3)! = (5*4*3*2*1)/(2*1) = 60.
  • The Combination Formula, If you are going to select x objects from a group of size n, and the order does not matter, the total number of arrangements is calculated with the Combination formula (shown below). Suppose you wanted to know how many arrangements you could form from a group of 5 colors where you are selecting 3 at a time (and the order does not matter) the number of arrangements would be 5!/[3!(5-3)1]=(5*4*3*2*1)/[(3*2*1)(2*1)]=10. Note that the number of combinations is much smaller than the number of permutations.

  How do I know if order matters?

  • You are selecting a committee of three from a group of 9. The first person selected is President, the second person is VicePresident and the third person cleans the room up after all meetings. Order matters!, Permutation.
  • You go to the supermarket to buy some lettuce. The scan code for lettude is || || (two regular bars and two dark bars). Lettuce is $1.00 a pound. The scan code for Beluga Caviar is || || (two regular bars and two dark bars, but now the dark bars come first!). Beluga Caviar is $125.00 an ounce. Both have scan codes of two dark bars and two light bars the difference comes from the order of the bars, order matters!, Permutation.
  • You are selecting a committee of 3 from a group of 9. The 3 selected will study a problem given to the committee. Nobody has a special position, the three members are equal, order does not matter, Combination.

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  Some problems from the textbook to use as examples

  • Fundamentals - Finding the Probability of a Home Run -Baseball player Barry Bonds broke a record when he hit 73 home runs in the 2001 season. During that season, he was at bat 476 times. If one of those at "at bats" is randomly selected, find the probability that it is one of the times he hit a home run.

    He hit 73 home runs in 476 "at bats" using the relative frequency concept P(home run)= 73/476 = 0.1534.
    
    

  • Fundamentals - Probability of Brand Recognition - In a study of brand recognition, 831 consumers knew of Campbell's Soup, 18 did not. Use these results to estimate the probability than a randomly selected consumer will recognize Campbell's Soup.

    First you must recognize that the total number of "objects" (in this case people) that were looked at was 831 + 18 = 849. The probability that someone will recognize Campbell's Soup is 831/849=0.979
    
    

  • Addition Rule - Birthday and Complement If someone is randomly selected, find the probability that his or her birthday is not in October. Ignore leap years.

    There are 365 days in a normal year. October has 31 days so 334 days are not in October. The Probability of a selected date not being in October is 334/365.
    
    

  • Multiplication Rule - Quality Control: A production manager for Telektronics claims that her new process for manufacturing CDs is better because the rate of defects is lower than 2%, which had been the rate of defects in the past. To support her claim, she manufactures a batch of 5000 CDs, then randomly selects 15 of them for testing, with the result that there are no defects among the 15 selected CDs. Assuming that the new method has the same 2% defect fate as in the past, find the probability of getting no defects among the 15 CDs.

    We are going to assume that the probability of a good CD is constant at 98% (P(defective) = 2%) and the CDs are independent of one another. The probability of all 15 being good is P(all good) = 0.98¹⁵ = 0.739.
    
    
    

    Calculation of Combinations and Permutations

    The following tools will allow you to calculate cominations and permutations. The inputs to the two functions are n, the number of trials, and x the number of successess. Since order matters for permutations, for given values of n and x the number of permutations will be considerably greater than the number of combinations.

    Calculation of permutations, _nP_x

    For the calculations to have meaning n, the number of trials, must be greater than x, the number of successes. Here are some sample problems that you can try.

    Permutation problems
    1. n = 10, x = 3, result is 720
    2. n = 8, x = 5, result is 6720
    3. You have 7 paint colors that you will use 4 at a time, how many arrangements are possible if order matters? Answer, 840
    4. You are selecting 3 people from a group of 12. First is president, second is Vice President, third is Secretary, so order matters. How many arrangements are possible? Answer, 1320

    Sample size, n
    Number of successes, x
    n P x =

    Calculation of combinations, _nC_x

    For the calculations to have meaning n, the number of trials, must be greater than x, the number of successes. Here are some sample problems that you can try.

    Combination problems
    1. n = 10, x = 3, result is 120
    2. n = 8, x = 5, result is 56
    3. You have 7 paint colors that you will use 4 at a time, how many arrangements are possible if order does not matter? Answer, 35
    4. You are selecting 3 people from a group of 12. They will be members of a cleanup crew, so order does not matter. How many arrangements are possible? Answer, 220

    Sample size, n
    Number of successes, x
    n C x =