Conditional_probability_2_stud

Conditional_probability_2_stud - Conditional_probability...

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Unformatted text preview: Conditional_probability_ 2_stud . doc // This file demonstrates the solutions of // Bayes’ theorem problems by means Feb 22’10 // of tree diagrams and tables. COMBINING THE RULES OF PROBABILITY Many probability problems can be represented by tree diagrams . When this is possible, the addition and multiplication rules can be applied quite easily. To illustrate the use of tree diagrams, let us consider the following problem. Two balls are drawn at random without replacement from a box containing one each of red, blue , and white balls. Find the probability of drawing (in any order) one red and one blue ball. The following outcomes (events) are possible: 1 st drawing: 2 nd drawing: Event R 1 : red ball is selected first . Event R 2 : red ball is selected second . Event B 1 : blue ball is selected first . Event B 2 : blue ball is selected second . Event W 1 : white ball is selected first . Event W 2 : white ball is selected second . The tree diagram representing this experiment (see Fig. 2) shows mutually exclusive outcomes of a first drawing and then outcomes of a second drawing. (M.V.: Each outcome of the whole experiment is characterized not only by the color of two selected balls but also by the order in which the balls have been selected; the order is shown by subscripts). A tree diagram representing all possible outcomes at the first and the second stage of experiment and the corresponding probabilities is given in Fig. 2. Fig. 2 P(B 2 |R 1 ) = ½ P(R 1 ) = ⅓ /---------------------- B 2 (1) P(B 2 ∩R 1 ) = P(R 1 )*P(B 2 |R 1 ) = 1/6 /--------------- R 1 * / \ P(W 2 |R 1 ) = ½ / \-------------------- W 2 (2) P(W 2 ∩R 1 ) = P(R 1 )*P(W 2 |R 1 ) = 1/6 / / / P(R 2 |B 1 ) = ½ / P(B 1 ) = ⅓ /--------------------- R 2 (3) P(R 2 ∩B 1 ) = P(B 1 )*P(R 2 |B 1 ) = 1/6 / / *------------------------ B 1 * \ \ P(W 2 |B 1 ) = ½ \ \-------------------- W 2 (4) P(W 2 ∩B 1 ) = P(B 1 )*P(W 2 |B 1 ) = 1/6 \ \ \ P(R 2 |W 1 ) = ½ \ P(W 1 ) = ⅓ /-------------------- R 2 (5) P(R 2 ∩W 1 ) = P(W 1 )*P(R 2 |W 1 ) = 1/6 \ / \ ------------- W 1 * \ P(B 2 |W 1 ) = ½ \------------------- B 2 (6) P(B 2 ∩W 1 ) = P(W 1 )*P(B 2 |W 1 ) = 1/6 1 After the tree has been drawn and labeled, we need to assign probabilities to each branch of the tree. Let us assume that in the experiment in question it is equally likely that any ball would be drawn at each stage (but of course, the probabilities are different at different stages). Now we can assign a probability to each branch segment of the tree , as shown in Fig. 2. Notice that a set of branches that initiate from a single point (*) have a total probability of 1 . In this diagram there are four such sets of branch segments. The tree diagram shows six possible distinct outcomes. Reading down : branch ( 1 ) shows the sequence of outcomes (R 1 , B 2 ), branch ( 2 ) shows (R 1 , W 2 ), and so on. (Note : Each outcome for the experiment is represented by a branch that begins at the common starting point and ends at the terminal points at the right.starting point and ends at the terminal points at the right....
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Conditional_probability_2_stud - Conditional_probability...

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