The definition of conditional probability can be turned around to find the probability that two things both happen, one after the other. For independent events this is plain multiplication; otherwise the second factor must be conditional.
Idea clave
$$P(A \cap B) = P(A) \cdot P(B \mid A).$$ If A and B are independent, $P(B \mid A) = P(B)$ and this becomes $P(A)P(B)$. On a tree diagram: multiply along the branches of a path.
Por qué funciona
Multiply both sides of $P(B \mid A) = P(A \cap B)/P(A)$ by $P(A)$. In words: first A must happen, and then, in the world where A happened, B must happen. Drawing without replacement changes the world after the first draw (one fewer of that color, one fewer in total), so the second probability must be the conditional one.
Ejemplo resuelto: Without replacement
A bag holds 5 red and 3 blue marbles. Draw two without replacement. Find $P(\text{both red})$.
- First draw: $P(\text{red}) = \dfrac{5}{8}$.
- Given a red is gone: 4 red of 7 left, $P(\text{red} \mid \text{red}) = \dfrac{4}{7}$.
- Multiply: $\dfrac{5}{8} \cdot \dfrac{4}{7} = \dfrac{20}{56} = \dfrac{5}{14}$.
Respuesta: $\frac{5}{14}$
Ejemplo resuelto: With replacement
Same bag, but the first marble is put back before the second draw. Find $P(\text{both red})$.
- Replacing the marble restores the bag, so the draws are independent.
- $\dfrac{5}{8} \cdot \dfrac{5}{8} = \dfrac{25}{64}$.
Respuesta: $\frac{25}{64}$
Ejemplo resuelto: At least one
Without replacement, find the probability that at least one of the two marbles is blue.
- "At least one blue" is the complement of "both red."
- $1 - \dfrac{5}{14} = \dfrac{9}{14}$.
- Check with the tree: $\frac{15}{56} + \frac{15}{56} + \frac{6}{56} = \frac{36}{56} = \frac{9}{14}$.
Respuesta: $\frac{9}{14}$
The rule chains: $$P(A \cap B \cap C) = P(A)\,P(B \mid A)\,P(C \mid A \cap B)$$ Three reds in a row from the same bag: $\frac{5}{8} \cdot \frac{4}{7} \cdot \frac{3}{6} = \frac{5}{28}$.
Error común
Multiplying unconditional probabilities when drawing without replacement: $\frac{5}{8} \cdot \frac{5}{8}$ for two reds. After the first red is gone, the bag has changed. Check: write down what is left in the bag before each draw.