# Logit function and probability

I was reading a paper regarding using a logistic regression to predict an outcome of interest. In the paper the logit is given as

logit(p) = log[p/1-p] = b0 + b1*x1 + .... + bn*xn => Which is fine

but then p is given as

p = 1/exp[-(b0 + b1*x1 + .... bn*xn)] => Which I don't understand.

It should be p = 1/(1 + exp[-(b0 + b1*x1 + .... + bn*xn)]), then the probability will be between 0 and 1, right?

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If $\ell = \log \frac{p}{1-p}$ then $\ell = \log\left(-1 + \frac{1}{1-p}\right)$. The advantage of this last way of writing it is that $p$ only appears once. So just invert step-by-step: \begin{align} e^\ell & = -1 + \frac{1}{1-p} \\ 1 + e^\ell & = \frac{1}{1-p} \\ \frac{1}{1+e^\ell} & = 1-p \\ p & = 1 - \frac{1}{1+e^\ell} = \frac{(1+e^\ell) - 1}{1+e^\ell} = \frac{e^\ell}{1+e^\ell} \end{align} Multiplying the numerator and denominator both by $e^{-\ell}$ yields: $$p = \frac{1}{e^{-\ell}+1}.$$ Now put the appropriate expression in place of $\ell$.