General Multiplication Rule and Probability

Total set: $10$ numbered chips consisting of $5$ red, $2$ blue, and $3$ green.
Probability of selecting an even chip: $P(\text{Even}) = \frac{5}{10} = 50\%$ .
Probability of selecting a blue chip: $P(\text{Blue}) = \frac{2}{10} = 20\%$ .
Probability of selecting a green chip: $P(\text{Green}) = \frac{3}{10} = 30\%$ .

Two events are independent if knowing that one occurred does not change the probability of the other.
Blue and even chips: exactly $1$ chip out of $10$ is both blue and even ( $P(\text{Blue} \cap \text{Even}) = \frac{1}{10} = 10\%$ ).
Multiplication condition for independence:
- $P(\text{Blue}) \times P(\text{Even}) = 0.2 \times 0.5 = 0.1$ ( $10\%$ ).
Consistency across conditional subsets:
- Given an even chip, the probability of it being blue is $P(\text{Blue} \mid \text{Even}) = \frac{1}{5} = 20\%$ .
- Given a blue chip, the probability of it being even is $P(\text{Even} \mid \text{Blue}) = \frac{1}{2} = 50\%$ .

Events are dependent if knowing the outcome of one alters the probability of the other.
Green and even chips: exactly $2$ chips out of $10$ are both green and even ( $P(\text{Green} \cap \text{Even}) = \frac{2}{10} = 20\%$ ).
Failure of the independent multiplication test:
- $P(\text{Even}) \times P(\text{Green}) = \frac{5}{10} \times \frac{3}{10} = \frac{15}{100} = 15\% \neq 20\%$ .
Dependency explanation: out of the $3$ green chips, $2$ are even; knowing a chip is green raises the probability of it being even above $50\%$ .

The universal formula for the joint probability of two events $A$ and $B$ :
- $P(A \cap B) = P(A) \times P(B \mid A)$
Application to even ( $E$ ) and green ( $G$ ) chips:
- Out of $5$ even chips, $2$ are green ( $P(G \mid E) = \frac{2}{5}$ .
- $P(E \cap G) = P(E) \times P(G \mid E) = \frac{5}{10} \times \frac{2}{5} = \frac{2}{10} = 20\%$ .
Universal application: when events are independent, $P(B \mid A) = P(B)$ , simplifying the general formula to $P(A \cap B) = P(A) \times P(B)$ .