If a first event A has an impact on the probability of a second event B, how do we apply the addition rule for probability?
Problem: Find the probability of drawing a club from a full deck of cards in two draws, without replacement. Use the addition rule $P(\text{A} \cup \text{B}) = P(\text{A}) + P(\text{B}) - P(\text{A} \cap \text{B})$.
Question: What to plug in for $P(\text{B})$, in this particular situation?
Let $P(\text{A})$ = probability of getting a club on the first draw, and $P(\text{B})$ = probability of getting a club on the second draw.
$P(\text{A})$ is straightforward: $\frac {13}{52}$ = $\frac 14$ = 0.25.
$P(\text{B})$ is theoretically the same. But in fact, the first draw reduced the deck by one card, and the new conditional probability depends on whether the first draw was a club. So it is either equivalent to $P(\text{B | A})$ = $\frac {13 - 1}{51}$ = $\frac {12}{51}$ $\approx$ 0.2353, if a club, or, if not, it is equivalent to $P(\text{B | A}^c)$ = $\frac {13}{51}$ $\approx$ 0.2549. Which of these three — the ideal/"true" probability, or either of the real probabilities — becomes $P(\text{B})$ in the standard formula?
If, instead of using the addition rule, we build up our answer by analyzing the sample space outcome by outcome, we reach the following result.
Probabilities to be used:
$P(\text {club on 1st draw})$ = $P(\text{A})$ = 0.25
$P(\text {non-club on 1st draw})$ = $P(\text{A}^c)$ = 0.75
$P(\text {club on 2nd draw, given club on 1st})$ = $P(\text{B | A})$ = $\frac {13 - 1}{51}$ = $\frac {12}{51}$ $\approx$ 0.2353
$P(\text {club on 2nd draw, given non-club on 1st})$ = $P(\text{B | A}^c)$ = $\frac {13}{51}$ $\approx$ 0.2549
$P(\text {non-club on 2nd draw, given club on 1st})$ = $P(\text{B}^c | {A})$ = $\frac {51 - 12}{51}$ = $\frac {39}{51}$ $\approx$ 0.7647
$P(\text {non-club on 2nd draw, given non-club on 1st})$ = $P(\text{B}^c | {A}^c)$ = $\frac {51 - 13}{51}$ = $\frac {38}{51}$ $\approx$ 0.7451
Four possible outcomes of two draws from the deck:
A. Non-club 1st draw, non-club 2nd draw: prob. $P(\text{A}^c)P(\text{B}^c | {A}^c)$ = $\frac 34 \times \frac {38}{51}$ $\approx$ 0.5588
B. Non-club 1st draw, club 2nd draw: prob. $P(\text{A}^c)P(\text{B} | {A}^c)$ = $\frac 34 \times \frac {13}{51}$ $\approx$ 0.1912
C. Club 1st draw, non-club 2nd draw: prob. $P(\text{A})P(\text{B}^c | {A})$ = $\frac 14 \times \frac {39}{51}$ $\approx$ 0.1912
D. Club 1st draw, club 2nd draw: prob. $P(\text{A})P(\text{B} | {A})$ = $\frac 14 \times \frac {12}{51}$ $\approx$ 0.0588
These probabilities add up to 1.000, and so comprise the entire sample space.
Adding together the three possibilities for drawing a club, we have 0.1912 + 0.1912 + 0.0588 = 0.4412, or a probability of 44.12% of getting at least one club in two draws.
Going back to the addition rule, the only way to obtain the right answer seems to be to use the ideal/"true" value for $P(\text {B})$.
We can rewrite the addition rule
$P(\text{A} \cup \text{B}) = P(\text{A}) + P(\text{B}) - P(\text{A} \cap \text{B})$
by substituting the multiplication rule for the subtrahend, as follows:
$P(\text{A} \cup \text{B}) = P(\text{A}) + P(\text{B}) - P(\text{A})P(\text{B} | A)$. Then, putting in the ideal/"true" value for $P(\text B)$, we have
$P(\text{A} \cup \text{B}) = \frac 14 + \frac 14 - \frac 14 \times \frac {12}{51}$ = 0.25 + 0.25 - (0.25)(0.2353) = 0.5 - 0.0588 = 0.4412.
But why would this be?