Do you agree with this definition of probability concept? Citing from a book:

First, count the number of possible results that satisfy your desired
condition. Then, count the total number of possible results (including
those that satisfy the condition and those that do not). Divide the
first number by the second and that is the probability that the
condition happens on any roll.

If yes, you agree with this definition, here is my second question. Consider two problems:
Problem 1
If the coin is tossed 3 times what is the probability of getting only one heads?
Problem 2
The same as Problem 1, but coins are biased:
Probability of heads is 85%
Probability of tails is 15%
If the definition above is right, then probably you should agree that these two problems have the same answers?
 A: As mentioned in the comments, the definition mentioned makes sense when there are a finite number of outcomes, each of which is equally likely. This is so in the first case, but not so in the second case. You can recast the second problem as a new experiment where outcomes are equally likely and proceed.
(1) The favorable cases are $TTH,HTT,THT$ out of a totality of eight cases: $HHH,HHT,HTH,THH,TTH,HTT, THT, TTT$. So the probability is $3/8$.
(2) To apply the above definition, let us suppose a single toss is equivalent to drawing a ticket out of a box containing 100 slips, out of which 85 are marked heads (and are further labeled as 1st head, 2nd head etc) and 15 are marked tails (and are further labeled 1st tail, 2nd tail etc). Our experiment consists of drawing a slip, replacing it and then drawing again. This is equivalent to having 3 such boxes and drawing a slip from each. Now everything is equally likely. A counting argument shows that there are a totality of $100^3$ cases out of which $3\cdot 15^2\cdot 85=57375$ are favorable. So the probability, in this case, ought to be $\frac{57375}{100^3}$.

Concerning the comment '"equally likely" is a dangerous word.':
Actually, the underlying philosophy for defining probability followed by your book is the Principle of Insufficient Reason. According to this principle, if we have no evidence of the propensity of any particular outcome of the experiment over others, we should accord an equal measure of certainty to all the outcomes. In other words, if there are $n$ outcomes possible for our experiment and we have no other information, then the probability of each outcome should be $\frac{1}{n}$. As a corollary, if the the set of our favorable cases $A$ has $m$ outcomes in it and there are a total of $n$ outcomes, we should define $P(A)=\frac{m}{n}$.
This principle makes sense in the first case, but not so in the second case where we have some evidence regarding the coin's biasedness that we ought to incorporate. To apply the principle we can recast the second problem as a new experiment where we have no information again.
Note that this principle is not a mathematical law and we are at liberty to accept it or not. It's just that it suits a large number of situations.
A: Already mentioned in the comments but formally, a probability model for an experiment with finitely many outcomes is a tuple: $(S, p)$ where $S$ is the set of all possible outcomes (assumed finite and often called sample space) and $p$ is a function $S \to [0, 1]$ satisfying that $\sum\limits_{o \in S} p(o) = 1.$ When such a tuple is given, then by definition, the probability of a result $R$ (simple or compounded) is $\mathbf{P}(R) = \sum\limits_{o \in R} p(o),$ that is, you sum the individual probabilities assigned to each outcome that is possible in the result. It cannot be stressed enough but the assignation $p$ is what determines the probability and not $S$ (i.e. not the results of the experiment). William Feller in his book An Introduction to Probability Theory and Its Application, vol. 1, discusses situations that arose in physics where much confusion happened. Physicists thought that the results of an experiment where the ones determining $p$ while this is not so and there are physical experiments that have the same underlying space of outcomes ($S$) but different probability distributions ($p$). Anyway, Since $S$ is finite, it has some number $k$ of elements. You can check that $p(o) = \frac{1}{k}$ is a valid assignation (out of many) and this results in the classical definition of probability $$
\mathbf{P}(R) = \sum\limits_{o \in R} \frac{1}{k} = \frac{\text{number of positive outcomes}}{\text{total number of outcomes}}, \quad \text{when} \quad p = \frac{1}{k} \text{ on } S.
$$
Other common assignations widely used in practice are Binomial distribution (more generally, Multinomial distribution), Hypergeometric, truncated versions of Poisson, Geometric and Negative Binomial, etc.
