Probability - Odds This is the question from my textbook:

The odds that an event will occur are given by the ratio of the probability that the event will occur to the probability that it will not occur, provided neither probability is zero. Odds are usually quoted in terms of positive integers having no common factor. Show that if the odds are $A$ to $B$ that an event will occur, its probability is $p=\frac A{A+B}$

I'm not sure where to start. I know if you had numbers here, to get the odds you would divide the probabilities - for example, if the probability that the last digit of a car’s license plate is $2,3,4,5,6,$ or $7$ is $\frac 6{10}$, then you would divide it by $\frac 4{10}$ (the remaining $0,1,8,9$ as the last digit by the total possibilities of $0$-$9$) to get $3:2$ or $3$ to $2$ odds. I am having trouble thinking about this in the reverse but also algebraically.
 A: Probability $P$ is by definition the number of possible successful outcomes ($A$) divided by the number of all possible outcomes ($A+B$) (given equal probability of all outcomes).
A: If the odds of something is $A$ to $B$ then it will happen $A$ times and it won't happen $B$ times.  The total number of times we measure whether anything will happen are the $A$ times it does happen and the $B$ times it doesn't and that $A + B$ times something happens.
BY DEFINITION if an event occurs $A$ out of $A+B$ times the the probability is $\frac {A}{A+B}$.
Now, if you want to solve in algebraically.....
If the probability of the event is $P$.  Then the probability of the event not happening is $1-A$. and the odds are BY DEFINITION defined to by $\frac P{1-P} = \frac AB$ and we solve for $P$.
$\frac P{1-P} = \frac AB$
$PB = A(1-P)$
$PB = A - PA$
$PA + PB = A$
$P(A+B) = A$
$P = \frac A{A+B}$
To get an intuitive sense this work:
$\frac P{1-P} = \frac AB$
$\frac {\frac {number\ of\ success}{total\ of\ successes\ and\ failures}}{\frac {number\ of\ failures}{total\ of\ successes\ and\ failures}}=\frac AB$
$\frac {number\ of\ successes}{number\ of\ failures}=\frac {\frac A{A+B}}{\frac B{A+B}}$
So the number of successes must be in ratio to $\frac A{A+B}$ and the number of failures must be in relation to $\frac B{A+B}$.  But as $\frac A{A+B} + \frac B{A+B} =\frac {A+B}{A+B} = 1$ those must actually be the probabilities of success and failure respectively.
A: Your quotation is slightly strange (you can have irrational probability and odds) but points to the answer:

*

*if you can write the odds as $o = \dfrac{p}{1-p}$ or as $o=\dfrac{A}{B}$

*then you can write the probability as $p=\dfrac{o}{o+1}$ by solving $o = \dfrac{p}{1-p}$ for $p$,

*and so $p=\dfrac{\frac{A}{B}}{\frac{A}{B}+1} =\dfrac{A}{A+B}$ by substitution

A: The definition of "probability" is the number of desired outcomes over the number of total outcomes. We already know that the number of favorable possibilities is $A$, and the number of other possibilities is $B$. Therefore, the total number of possibilities is $A+B$. Hence, using our definition, the probability of the event occurring is $\frac A{A+B}$.
The way the textbook stated the definition, in my opinion, is slightly confusing. I wouldn't blame you for being puzzled if you are new to probability, or just math in general.
