Adam's logic: Problem understanding a probability theorem On Stanford's Encyclopedia of Philosophy, on the entry on Logic and Probability, section $2.2$, some principles of Adam's logic are defined. On the very beginning of the section, we are told one can easily show' that the probability of a conclusion is the sum of the probabilities of its premises minus one. For example, the probability of a conclusion $q$ with premises $p\vee q$ and $p\to q$ is
$P(q) = P(p\vee q) + P(p\to q) - 1.$
However, this makes no sense to me for two reasons. Firstly, all probabilities are axiomatically constrained to $[0, 1]$. However, if the sum of the probability of each premise is less than $1$, then $P(q)$ would be negative, which makes no sense.
Secondly, it can also be the case that the probabilities add up exactly to 1 (for example, if both premises had a probability of $\frac{1}{2}$). In such a case, $P(q) = 0$, which also doesn't' make sense: both premises would be quite probable, why would that mean the conclusion is necessarily false?
 A: 
For example, the probability of a conclusion $q$ with premises $p\vee q$ and $p\to q$ is $P(q) = P(p\vee q) + P(p\to q) - 1.$

Noting that $P(a∨b)=P(a)+P(b)-P(a∧b),$
$$P(p∨q) + P(p\to q) - 1
\\=P(p∨q) + P(¬p∨q) - 1
\\=P\Big((p∨q) ∨ (¬p∨q)\Big)+P\Big((p∨q) ∧ (¬p∨q)\Big)- 1
\\=P\Big(\top\Big)+P\Big((p∧¬p) ∨ (p∧q) ∨ (q∧¬p) ∨ (q∧q)\Big)- 1
\\=P\Big((p∧¬p) ∨ (p∧q) ∨ (q∧¬p) ∨ (q∧q)\Big)
\\=P\Big( (p∧¬p) ∨ \big(q∧(p∨¬p)\big) ∨ (q∧q) )\Big)
\\=P\Big( \bot ∨ q ∨ q \Big)
\\=P(q).$$

However, if the sum of the probability of each premise is less than $1$, then $P(q)$ would be negative, which makes no sense.

Suppose the former is $0.9;$ then $P(q)=0.9+0.9-1=0.8\not<0.$

On the very beginning of the section, we are told one can easily show' that the probability of a conclusion is the sum of the probabilities of its premises minus one.

That particular sentence in the article is phrased a tad misleadingly: it actually means to refer specifically to the above example.

Secondly, it can also be the case that the probabilities add up exactly to 1 (for example, if both premises had a probability of $\frac{1}{2}$). In such a case, $P(q) = 0$, which also doesn't' make sense

Those two premises $(p∨q)$ and $(¬p∨q)$ are not independent of each other; the sum of their probabilities exceed $1$ precisely because the atomic formula $q$ is not a contradiction.
