Equally likely outcomes

Under this assumption only, $P(A) = \frac{\text{number of outcomes in } A}{\text{total number of outcomes}}$, so counting is enough to get a probability. On a fair six-sided die each face has probability $1/6$ and $P(\text{roll above 4}) = 2/6 \approx 0.3333$. Now load the die so a 6 turns up with probability 0.30 and the other five faces share the remaining 0.70 equally at 0.14 each: counting would still say $1/6 \approx 0.1667$ for a 6, which is wrong by a wide margin. Whenever the outcomes are not symmetric you have to add the actual probabilities instead of counting.