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Definition of a multiple within a probability problem.

  1. Feb 16, 2015 #1
    Below is a test problem I recently had in Probability class. I missed points on this problem (event B) because I counted 0 as a multiple of 3. But...0 is a multiple of 3 right?

    I approached my professor with this concern and he told me that 0 is definetly not a multiple of 3...
    If I am right, can I have some advice on how to approach him again, or if I should at all. I lost 10% of the test grade which I could not afford..

    Here's what I turned in...

    1. The problem statement, all variables and given/known data

    An urn contains 10 identical balls numbered 0,1,...9. A random experiment involves selecting a ball from the urn and noting the number of the ball. The events A,B,C are defined as follows:
    A "number of the ball selected is odd."
    B "number of the ball selected is a multiple of 3,"
    C "number of the ball selected is less than 5,"

    Find ##P(A),P(B),P(C),P(A\cup B), \text{ and } P(A\cup B \cup C).##

    2. Relevant equations
    Let ##\mathbb{Z}## denote the integers.
    Say ##d## divides ##m##, equivalently, that ##m## is a multiple of ##d##, if there exists an integer ##q## such that ##m = qd##.

    Let ##m=q=0 \text{ and } d=3##

    then ##m=qd \Longrightarrow 0=0\times 3##

    ##\therefore 0 \text{ is a multiple of }3##

    3. The attempt at a solution

    [itex]S= \{0,1,2,3,4,5,6,7,8,9\}[/itex]

    ##A= \{1,3,5,7,9\} \hspace{20pt} B= \{ 0,3,6,9\} \hspace{20pt}C=\{0,1,2,3,4\}##

    ##P(A)=\frac{5}{10}\hspace{20pt}P(B)=\frac{4}{10}\hspace{20pt}P(C)=\frac{5}{10}##

    ##P(A \cup B) =\frac{7}{10} \hspace{20pt} P(A \cup B \cup C) = \frac{9}{10}##
     
  2. jcsd
  3. Feb 16, 2015 #2

    SammyS

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