Unions with Dependent Events

Finding “A or B” When Events Are Dependent

In Union Probability for Independent Events, you used the multiplication rule for independent events to find their overlap. But many events are dependent: learning that one event occurred changes the probability of the other. Card draws without replacement provide a clear example because each drawn card changes what remains in the deck.

The general addition rule still applies whether events are independent or dependent. The challenge is finding $P(A\cap B)$, the probability that both events occur. For dependent events, use a conditional probability in the general multiplication rule, then subtract the overlap when applying the addition rule.

Formula: If $P(A)>0$, the general multiplication rule and addition rule give $$ P(A\cap B)=P(A)P(B\mid A) $$ and $$ P(A\cup B)=P(A)+P(B)-P(A\cap B) =P(A)+P(B)-P(A)P(B\mid A). $$ The event $A\cup B$ means that $A$ occurs, $B$ occurs, or both occur.

The order in the conditional probability matters: $P(B\mid A)$ means the probability of $B$ among outcomes where $A$ has occurred. You could instead use $P(A\cap B)=P(B)P(A\mid B)$, provided $P(B)>0$. Choose the conditional probability that is easiest to find from the situation.

This is a useful combination of ideas from earlier tutorials, including The General Multiplication Rule and Common Errors with the Addition Rule. The general addition rule does not change for dependent events. What changes is how you find the overlap: do not replace $P(A\cap B)$ with $P(A)P(B)$ unless independence has been established.

A Reliable Plan for Dependent Events

For a card-draw question, first define the events precisely. “A is a heart” could refer to the first card, the second card, or either card, so say which draw each event describes. Next, find the marginal probabilities $P(A)$ and $P(B)$, then use the information about the first event to find an appropriate conditional probability.

A standard deck has 52 cards, with 13 cards in each suit and 26 cards of each color. When a card is drawn and not replaced, the next draw is from the remaining 51 cards. The number of cards of interest in that remaining deck may depend on what happened on the first draw.

Define the events.
State exactly what happens on each draw and identify the union being requested.

Find the individual probabilities.
Calculate $P(A)$ and $P(B)$ for the full chance process.

Find the overlap.
Use $P(A\cap B)=P(A)P(B\mid A)$, updating the deck for the condition $A$.

Apply the addition rule.
Calculate $P(A)+P(B)-P(A\cap B)$, then interpret the result in context.

Keep the marginal probability distinct from the conditional probability. For example, the probability that the second card is a heart is $13/52$ before the first draw is known. But if the first card is known to be a heart, the probability that the second is a heart is $12/51$. These answer different questions.

Worked Examples

Worked Example: Hearts on Two Draws

Two cards are drawn one after the other from a standard deck without replacement. Let $A$ be the event that the first card is a heart, and let $B$ be the event that the second card is a heart. Find the probability that the first card or the second card (or both) is a heart.

State. The target is $P(A\cup B)$, the probability that at least one of the two drawn cards is a heart.

Plan. The draws are without replacement, so the events are dependent: if the first card is a heart, only 12 hearts remain among 51 cards. Use the general multiplication rule to find the overlap, then the general addition rule to find the union. Both individual event probabilities are $13/52$, since each position in a random draw is equally likely to contain any card in the deck.

Do. The conditional probability of a heart on the second draw, given a heart on the first draw, is $12/51=4/17$. Therefore,

P(A\cap B)=P(A)P(B\mid A) =\frac{13}{52}\cdot\frac{12}{51} =\frac{1}{4}\cdot\frac{4}{17} =\frac{1}{17}.

Now apply the addition rule. The fraction calculation is $1/4+1/4-1/17=1/2-1/17=15/34$. Thus,

P(A\cup B)=P(A)+P(B)-P(A\cap B) =\frac{13}{52}+\frac{13}{52}-\frac{1}{17} =\frac{1}{2}-\frac{1}{17} =\frac{15}{34} \approx 0.4412.

As a check, the complement of “at least one heart” is “neither card is a heart.” Its probability is $(39/52)(38/51)=19/34$, so the union probability is $1-19/34=15/34$, matching the addition-rule result.

Conclude. The probability that at least one of the two cards is a heart is $15/34$, or approximately $0.4412$.

Worked Example: A Queen on Either Draw

Two cards are drawn without replacement from a standard deck. Let $A$ be the event that the first card is a queen, and let $B$ be the event that the second card is a queen. Find the probability of drawing a queen on the first draw, the second draw, or both.

State. We want $P(A\cup B)$, the probability that at least one of the two cards is a queen.

Plan. Each draw position has probability $4/52=1/13$ of containing a queen. Given that the first card is a queen, three queens remain among 51 cards. Use this conditional probability to find $P(A\cap B)$, then subtract the overlap once.

Do. The general multiplication rule gives

P(A\cap B)=P(A)P(B\mid A) =\frac{1}{13}\cdot\frac{3}{51} =\frac{1}{13}\cdot\frac{1}{17} =\frac{1}{221}.

The addition rule then gives $2/13-1/221$. Since $2/13=34/221$, this is $33/221$, approximately $0.1493$:

P(A\cup B) =\frac{1}{13}+\frac{1}{13}-\frac{1}{221} =\frac{33}{221} \approx 0.1493.

A complement check gives the same result. The probability of no queen on either draw is $(48/52)(47/51)=564/663$, so the probability of at least one queen is $1-564/663=99/663=33/221$.

Conclude. The probability of drawing at least one queen in the two draws is $33/221$, or approximately $0.1493$.

Worked Example: A Black Card Followed by a Red Card

Two cards are drawn without replacement. Let $A$ be the event that the first card is black, and let $B$ be the event that the second card is red. Find the probability that the first card is black or the second card is red (or both).

State. The target is $P(A\cup B)$, the probability that at least one of these events occurs.

Plan. There are 26 black and 26 red cards, so each marginal probability is $1/2$. Given that the first card is black, all 26 red cards remain among the 51 remaining cards. Use $P(A)P(B\mid A)$ for the overlap, and then apply the general addition rule.

Do. The conditional probability of a red second card given a black first card is $26/51$. Thus,

P(A\cap B)=P(A)P(B\mid A) =\frac{1}{2}\cdot\frac{26}{51} =\frac{13}{51}.

Subtract the overlap from the sum of the marginal probabilities:

P(A\cup B) =\frac{1}{2}+\frac{1}{2}-\frac{13}{51} =1-\frac{13}{51} =\frac{38}{51} \approx 0.7451.

For another check, the only way neither event occurs is for the first card to be red and the second card to be black. The probability of that path is $(26/52)(26/51)=13/51$. Its complement is $1-13/51=38/51$, as calculated.

Conclude. The probability that the first card is black or the second card is red (or both) is $38/51$, or approximately $0.7451$.

Why the Independent-Events Shortcut Does Not Apply

For independent events, the earlier tutorial Union Probability for Independent Events showed that the overlap can be found by multiplying the marginal probabilities. In the heart example, that shortcut would give $(1/4)(1/4)=1/16$ for the probability of two hearts. But the correct overlap is $1/17$, because after a heart is drawn, the remaining deck has 12 hearts among 51 cards.

You can see the change directly: $P(B\mid A)=12/51=4/17$, while $P(B)=1/4$. These are not equal, so knowing that the first card is a heart changes the probability that the second is a heart. The events are dependent. Apply the general multiplication rule using the conditional probability, not the independent-events shortcut.

Key distinction: The general multiplication rule uses $P(A\cap B)=P(A)P(B\mid A)$. Only when $A$ and $B$ are independent can you replace $P(B\mid A)$ with $P(B)$ and write $P(A\cap B)=P(A)P(B)$.

Common Mistakes and AP Exam Tips

Adding the event probabilities without subtracting the overlap. An outcome in both events would be counted twice in $P(A)+P(B)$. Subtract $P(A\cap B)$ once to count that outcome once.
Using $P(A)P(B)$ automatically. This product gives the intersection only for independent events. For dependent events, use the general multiplication rule and find the conditional probability.
Confusing a marginal probability with a conditional probability. $P(B)$ describes the chance of $B$ before restricting attention to $A$. $P(B\mid A)$ describes the chance of $B$ after $A$ is known to have occurred.
Updating the deck incorrectly. After one card is drawn without replacement, there are 51 cards left. If the first card matches the event’s description, reduce the relevant card count too.
Leaving “or” ambiguous. In probability, $A\cup B$ includes the case in which both events occur. It does not mean “one or the other, but not both.”
Giving a number without context. State what the probability measures and identify which draws the events describe. A clear answer makes the meaning of “at least one” explicit.

AP Exam Tip: For dependent events, show the conditional probability used in the multiplication rule, such as $P(B\mid A)$, and explain how the chance process changes after $A$. Then write the general addition rule, subtract the overlap once, and conclude with the probability in context. Do not claim independence just because both marginal probabilities are easy to find.

Key Takeaway

For dependent events, find the overlap with a conditional probability and then use the general addition rule. In card draws without replacement, update the deck after the first draw; the conditional probability for the second draw may differ from its marginal probability.

Key takeaway: For $P(A\text{ or }B)$ with dependent events, calculate $P(A\cap B)=P(A)P(B\mid A)$, then subtract that overlap from $P(A)+P(B)$. Use the product of marginal probabilities for the overlap only when the events are independent.

Check Your Understanding

For each question, show how you find the overlap before applying the addition rule.

Two cards are drawn without replacement. Let $A$ mean the first card is a diamond and $B$ mean the second card is a diamond. Write $P(B\mid A)$ and use it to find $P(A\cap B)$.
Using the events in Question 1, find $P(A\cup B)$. Give the answer as a fraction and a decimal rounded to four places.
Two cards are drawn without replacement. Let $A$ mean the first card is an ace and $B$ mean the second card is an ace. Find $P(A\cup B)$.
Explain why $P(A\cap B)=P(A)P(B)$ is not the appropriate way to find the overlap for two hearts drawn without replacement.
In one or two sentences, explain why the general addition rule applies to both independent and dependent events.

What You'll Learn