Adding Probabilities When Events Cannot Overlap
In Spotting Disjoint Events in Context, you practiced checking whether two events can occur on the same outcome. Now we use that idea to calculate the probability of one event or another. For a single card drawn from a standard deck, the events “the card is a diamond” and “the card is a club” cannot both happen. Their probabilities can be added directly.
In probability, “A or B” includes outcomes in A, outcomes in B, and any outcomes in both. It means that at least one of the two events occurs. When A and B are mutually exclusive, there are no outcomes in both, so adding the probabilities counts every outcome in \(A\cup B\) exactly once.
The condition matters: the formula in this box is for mutually exclusive events. As in What Mutually Exclusive Events Mean, disjoint events have no outcomes in common, so \(A\cap B=\varnothing\). If the events overlap, simply adding their probabilities counts their shared outcomes twice. This tutorial focuses on recognizing and using the disjoint-events rule; the general addition rule addresses overlapping events.
A Card-Draw Checklist
For a card problem, first be precise about the chance process: one card is selected at random from a well-shuffled standard 52-card deck. One outcome is one particular card, such as the queen of hearts. Because each of the 52 cards is equally likely to be selected, a probability can be found by dividing a favorable-card count by 52, as in Equally Likely Outcomes and Counting Probability.
Translate the wording into clear descriptions of which cards belong to A and B.
Ask whether one card could satisfy both descriptions. If so, the events are not mutually exclusive, and this addition shortcut does not apply.
For a standard deck, divide the number of cards in each event by 52.
If the events are disjoint, add their probabilities and state what the result means for the card draw.
In each example, the check for overlap comes before the arithmetic. Two descriptions that sound different are not automatically disjoint: a card can have both a rank and a suit. For instance, the queen of hearts is both a queen and a heart. The event definitions—not just their labels—determine whether direct addition is valid.
Worked Example: A Diamond or a Club
One card is selected at random from a well-shuffled standard deck. What is the probability that it is a diamond or a club?
State: Let \(D\) be the event that the selected card is a diamond, and let \(C\) be the event that it is a club. We want \(P(D\text{ or }C)\).
Plan and check: The 52 individual cards are equally likely outcomes because the card is selected at random from a well-shuffled deck. A single card has exactly one suit, so it cannot be both a diamond and a club. Thus, \(D\) and \(C\) are mutually exclusive, and the addition rule applies.
Do: There are 13 diamonds and 13 clubs. Therefore,
As a direct check, the favorable cards are the 13 diamonds plus the 13 clubs, or 26 distinct cards. Dividing \(26\) by \(52\) also gives \(1/2\). The decimal \(0.5\) is equivalent to one-half.
Conclude: The probability that the selected card is a diamond or a club is \(1/2\), or 0.5. In this model, half of the 52 possible cards have one of those two suits.
Adding Events Defined by Rank
The addition rule is not limited to suits. It also works when the events describe different ranks, provided one card cannot belong to both. Be careful to use the same definition throughout: in the examples here, a face card means a jack, queen, or king. A 10 is not a face card.
Worked Example: A 10 or a Face Card
One card is selected at random from a well-shuffled standard deck. What is the probability of selecting a 10 or a face card?
Define and check: Let \(T\) be the event that the card is a 10, and \(F\) the event that it is a face card (a jack, queen, or king). One card cannot be both a 10 and a jack, queen, or king, so \(T\) and \(F\) are mutually exclusive. Each of the 52 cards is equally likely.
Count each event: There are four 10s, one in each suit. There are \(3\times4=12\) face cards: three face-card ranks in each of four suits.
Apply the rule:
To check the count another way, the favorable cards consist of four 10s and twelve face cards, with no card in both groups: \(4+12=16\) distinct cards. Thus \(16/52\) is the direct favorable-outcomes calculation. Reducing the fraction by 4 gives \(4/13\); dividing 4 by 13 gives approximately 0.3077, rounded to four decimal places.
Conclude: The probability of selecting a 10 or a face card is \(4/13\), or about 0.3077. In repeated selections from the same model, this probability represents the long-run proportion of draws that produce a 10 or a face card.
If the question had said “a 10 or a king,” the events would still be disjoint, because no card is both a 10 and a king. The relevant check is always whether a single possible card could satisfy both event definitions.
Use the Event Definitions to Check Overlap
Sometimes the challenge is not counting the cards but deciding whether the simple addition rule is appropriate. A useful check is to name a card that might belong to both events. If you can find one, the events overlap and are not mutually exclusive. If the card properties make such an outcome impossible, the events are disjoint.
Worked Example: A Red Queen or a Black King
One card is selected at random from a well-shuffled standard deck. What is the probability that it is a red queen or a black king?
Define and check: Let \(R\) be the event that the card is a red queen, and \(K\) the event that it is a black king. A card cannot be both red and black, and it cannot be both a queen and a king. Therefore, \(R\) and \(K\) are mutually exclusive. All 52 cards are equally likely.
Count and calculate: There are two red queens, the queen of hearts and the queen of diamonds. There are two black kings, the king of spades and the king of clubs. The addition rule gives
For a direct verification, the four favorable cards are the two red queens and the two black kings. These are four distinct cards, so the probability is \(4/52\). Reducing by 4 gives \(1/13\), and \(1\div13\approx0.0769\), rounded to four decimal places.
Conclude: The probability of drawing a red queen or a black king is \(1/13\), or about 0.0769. The addition is valid because the two sets of favorable cards have no card in common.
When Direct Addition Is Not Valid
Consider a different question: What is the probability of selecting a queen or a club? Let \(Q\) be the event that the card is a queen, and \(C\) the event that it is a club. These events overlap: the queen of clubs is both a queen and a club. Therefore, they are not mutually exclusive, and \(P(Q)+P(C)\) would count the queen of clubs twice.
This overlap check prevents a common mistake: seeing the word “or” and immediately adding. First identify the events, then check whether one outcome can satisfy both. The phrase “queen or club” includes the queen of clubs because probability “or” includes outcomes in either event and outcomes in both. The disjoint-events formula is appropriate only when the events have no shared outcomes.
Common Mistakes and AP Exam Tips
- Adding before checking disjointness. Start by testing whether one card could satisfy both event descriptions. A queen can also be a club, but a card cannot be both a diamond and a club.
- Treating “or” as “one but not both.” In probability, “A or B” includes outcomes in either event and outcomes in both. When the events are mutually exclusive, there simply are no outcomes in both.
- Counting event probabilities with the wrong denominator. For a single card drawn from a standard deck, use 52 possible cards. Make sure each probability describes the same card-draw process.
- Confusing a rank with a suit. One card has both a rank and a suit, so an event about rank may overlap with an event about suit. Check actual cards rather than assuming these categories are disjoint.
- Stopping at a numerical answer. A strong response identifies the events, explains why they are mutually exclusive, shows the addition, and interprets the result in context. The overlap check justifies the method.
A concise, full-credit explanation might say: “A card cannot be both a diamond and a club, so these events are mutually exclusive. There are 13 cards of each suit, giving \(P(D\text{ or }C)=13/52+13/52=1/2\). Thus, the probability of drawing a diamond or a club is one-half.” This communicates the condition, calculation, and conclusion.
Check Your Understanding
Assume one card is selected at random from a well-shuffled standard deck. For each question, identify whether the events are mutually exclusive before calculating.
- What is the probability of drawing a heart or a spade? State why the addition rule for mutually exclusive events applies.
- What is the probability of drawing an ace or a 10? Show the event counts and simplify the result.
- Are the events “drawing a king” and “drawing a heart” mutually exclusive? Give one card outcome that supports your answer.
- What is the probability of drawing a red 7 or a black queen? Explain why the events are disjoint and show the calculation.
- A student adds the probability of drawing a club to the probability of drawing a queen to answer “club or queen.” What should the student check before deciding whether this calculation is valid?