# Derivation of equalities from odds ratio

Could someone give me the consecutive steps to derive the following equations:

$$P(Y=1) = \frac{\mathrm{OR}}{1+\mathrm{OR}}$$

and

$$P(Y=0) = \frac{1}{1+\mathrm{OR}}$$

from

$$\mathrm{OR}=\frac{P(Y=1)}{P(Y=0)},$$

where $P(Y=1)$ is the probability of success, $P(Y=0)$ is the probability of failure, and $\mathrm{OR}$ is the Odds Ratio.

It should have something to do with the fact that $P(Y=1) + P(Y=0)=1$, but I can't work it out.

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1. substitute $OR = P(Y=1)/P(Y=0)$ in any of the equation that you have to derive $\phantom{x}\\$ 2. when deriving the equation for $P(Y=1)$ substitute $P(Y=0)=1-P(Y=1)$ $\phantom{x}\\$ 3. simplify – Ilya Jan 11 '13 at 10:01
Thanks. But isn't that using the solution to get to the solution? I mean, could I do it without making use of the derivations yet in step 1. – Marloes Jan 11 '13 at 10:11

This is just two equations with two unknowns $$c=\frac{a}{b},\qquad a+b=1,$$ where $a,b\in (0,1)$ and $c\in (0,\infty)$.
Let us solve for $b$. The first equation yields $a=bc$, and substituting this into the second equation yields $$1=bc+b=b(c+1)\quad\Rightarrow \quad b=\frac{1}{c+1}.$$ Now use that $a=1-b$ to derive an expression for $a$ depending only on $c$, and realize that your example has $a=P(Y=0)$, $b=P(Y=1)$ and $c=\mathrm{OR}$.
Thanks, So that last step would be $a=1-(1/c+1)$ How does that give $a=c/1+c$ – Marloes Jan 11 '13 at 10:34
Try to write $1$ as $\frac{c+1}{c+1}$ and put the two fractions together. – Stefan Hansen Jan 11 '13 at 10:36