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Unformatted text preview: /k ). When a key is put in a box that already contains a key, we say that a collision occurs. The goal is to nd the expected number of such collisions X . We also let Y be the number of boxes with at least one key (at the end of the allocation process). For instance, if we have 3 boxes and 4 keys, with no keys in the rst box, 4 in the second and 1 in the third, X is 3 (3 keys were added to the rst one in box 2) and Y is 2. The problem is divided in several elementary steps, and you should not be surprised to nd strong similarity with one of the homework problems: 1) What is the probability that a given box will recieve no keys at all? At least one? 2) Using the familiar trick of dening random variables with values 0 or 1, nd the expected value of Y . 3) Find a simple relationship between X and Y , and use it to compute the expected value of X (Hint: 3=52 in our example)...
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This note was uploaded on 10/02/2009 for the course PSTAT 5E taught by Professor Eduardomontoya during the Spring '08 term at UCSB.
 Spring '08
 EduardoMontoya

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