3
$\begingroup$

While reading the first pages of text book in chapter 2 came up with my own example since we were only introduced to Events, Unions, Intersections of events, and containment within events, one example involved a die and it got me thinking, and I made up my own example using a relation, my question was, what is the probability that if two dice are rolled that the sum of their sides will be odd?

Our sample space, $S =\{a,b \in (\Bbb{N}, \Bbb{N}) \mid a\in \{1,2,3,4,5,6\} \wedge b\in\{1,2,3,4,5,6\}\}$

Then S our sample space is:

$$\{(1,1),(1,2),(1,3),(1,4),(1,5),(1,6),(2,1),(2,2),(2,3),(2,4),(2,5),(2,6),(3,1),(3,2),(3,3),(3,4),(3,5),(3,6),(4,1),(4,2),(4,3),(4,4),(4,5),(4,6),(5,1),(5,2),(5,3),(5,4),(5,5),(5,6),(6,1),(6,2),(6,3),(6,4),(6,5),(6,6)\}$$

Lets define a Relation, $R: A \to B$ to be defined as $R=a+b \vert a+b=2k+1$ for some $k \in Z$

Then our relation is:

$$R=\{(1,2),(1,3),(1,5),(2,1),(2,3),(2,5),(3,1),(3,3),(3,5),(4,1),(4,3),(4,5), (5,1),(5,3),(5,5),(6,1),(6,3),(6,5)\}$$

R is not reflexive because $(1,1),(2,2)....(6,6)$ $\notin R$

R is symmetric because $((1,2),(2,1)),((1,3),(3,1)).......((5,6),(6,5))$ $\in R$

This was sort of an unnecessary thing I added in with the relation, in fact $R$ is our event,, $E$, $E(i+j = 2k+1)$ i.e $i+j$ is odd.

As we can see $\vert E \vert=18$

As we can also see $\vert S \vert=36$

From this example it only made sense to me that the probability of two sides of dice adding up to an odd number would be:

$$P(E)=\frac{\vert E \vert}{\vert S \vert} = \frac{18}{36} = \frac{1}{2}$$

Now I am not that far in the book and sort of just made up my own example for fun in fact I am only four pages in out of about 40 pages, so please offer additional insight, like was this in-fact true and is this something that always holds?

$\endgroup$
  • 1
    $\begingroup$ what? the probability is $1/2$ because no matter what the outcome of the first dice is, half the outcomes will make the sum odd. $\endgroup$ – Jorge Fernández Hidalgo Dec 27 '16 at 22:29
  • 2
    $\begingroup$ Yes, the probability that the sum is odd is indeed $\frac{1}{2}$. Yes, your solution is correct. No, you did not need to refer to any relation or even the sample space for this problem. A much cleaner solution is to approach via conditional probability and recognize that regardless what the outcome of the first die is, there is a $\frac{1}{2}$ chance that the second die will be a different parity than the first, implying the result. $\endgroup$ – JMoravitz Dec 27 '16 at 22:30
  • 1
    $\begingroup$ Use \{ and \} to get set braces in MathJax. $\endgroup$ – Noble Mushtak Dec 27 '16 at 22:35
2
$\begingroup$

In general if you have a finite sample space $\Omega$ and assign equal probability to each outcome in the sample space and an event $E\subset \Omega$, then indeed, $$ P(E)=\frac{\lvert E \rvert}{\lvert \Omega \rvert} $$ where $\lvert\cdot\rvert$ is cardinality (size). In your case $\Omega$ is the set of all possible dice pairs and $E$ is the set of all pairs which have odd sum. If you directly count as you did you get $1/2$. There are often more efficient methods to solve these problems though.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.