# Definition of a multiple within a probability problem.

1. Feb 16, 2015

### tempneff

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

$S= \{0,1,2,3,4,5,6,7,8,9\}$

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

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

$P(A \cup B) =\frac{7}{10} \hspace{20pt} P(A \cup B \cup C) = \frac{9}{10}$

2. Feb 16, 2015

### SammyS

Staff Emeritus