1. OKAN UNIVERSITY FACULTY OF ENGINEERING AND ARCHITECTURE Yrd. Doç. Dr. Didem Kivanc Tureli didemk@ieee.org didem.kivanc@okan.edu.tr 18/27/2015Lecture 2 MATH 265 Probability and Random Processes 02Axioms of Probability Fall 2011

2. Multinomial Theorem The multinomial theorem states that: 8/27/2015Lecture 22 Recall that the multinomial coefficients are: But the binomial coefficients are: So

3. 8/27/2015Lecture 23

4. How many ways … How many ways are there to pick k numbers such that the sum of those numbers is N and none of those numbers are zero? 8/27/2015Lecture 24 How many ways are there to pick k numbers such that the sum of those numbers is N and some of those numbers may be zero?

5. Distribution of distinguishable (or unique) balls in urns How many ways are there that we can put n balls in k urns if the balls are distinguishable (that is, you can tell them apart). Ball 1 can go into any one of k urns Ball 2 can go into any one of k urns … Ball n can go into any one of k urns 8/27/2015Lecture 25

6. Distribution of indistinguishable balls in urns How many ways are there that we can put k balls in n urns if the balls are indistinguishable (that is, you can NOT tell them apart). If you can’t tell the balls apart, then the previous method will not work. For example, in the previous question the following two arrangements are different: – Arrangement 1 : if ball 1 went into urn one and all the rest of the balls went into urn 2, – Arrangement 2: if ball 2 went into urn 1 and all the rest of the balls went into urn 2 But if the balls are exactly the same, then the two arrangements above are exactly the same. 8/27/2015Lecture 26

7. Distribution of indistinguishable balls in urns The outcome of this experiment can be summarized by a vector, with every element of the vector tells us how many balls there are in the corresponding urn: 8/27/2015Lecture 27 Then the question becomes: how many vectors of k numbers are there such that the sum of those k numbers is n ? But we know the answer to this question (slide 4), it is:

8. First study the easy case: don’t allow any empty urns. Draw as many circles as there are balls. Now divide those balls into k urns using k-1 bars. E.g. divide 8 balls into 3 urns. You need 8 circles and 2 bars. The figure below represents having 1 ball in urn 1, 3 balls in urn 2 and 4 balls in urn 3. 8/27/2015Lecture 28 Notice that you can’t tell the balls apart, but you can tell the urns apart, so the following arrangement is different and is counted separately

9. There are n-1 spaces for the first bar, and n-2 remaining spaces for the second bar, giving (n-1)(n-2) possible arrangements. However there is one complication. You can end up with this arrangement in two ways: 8/27/2015Lecture 29 Or

10. So in the number of arrangements is: More generally we can repeat this argument to show that the number of ways that n indistinguishable balls can be put into n urns is: 8/27/2015Lecture 210

11. What if we allow empty urns? 8/27/2015Lecture 211