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\( P \)

Multi-Event
Probability

Master the rules for combining probabilities across multiple events: independent, dependent, mutually exclusive, and overlapping.

ISEE Upper Level

What Is a Probability Event?

An event is any outcome or set of outcomes from a random experiment. A multi-event probability problem asks: what is the chance that two or more events occur, either together or in sequence?

Single Event
Rolling a \(5\) on one die: \(P = \frac{1}{6}\)
Multi-Event
Rolling a \(5\) then a \(3\): combines two separate outcomes

Visualizing the Sample Space

When you flip a coin twice, the sample space contains every possible combination. A tree diagram maps them all out.

Start
\(\frac{1}{2}\)
\(\frac{1}{2}\)
 H T H T H T HH
\(\frac{1}{4}\)
 HT
\(\frac{1}{4}\)
 TH
\(\frac{1}{4}\)
 TT
\(\frac{1}{4}\)

Independent vs. Dependent Events

The most important distinction in multi-event probability: does the first event change the conditions for the second?

Independent
Outcome of Event A has no effect on Event B. Example: rolling a die, then flipping a coin.
Dependent
Outcome of Event A changes the probability of Event B. Example: drawing two cards without replacement.

The Multiplication Rule

When you need the probability that Event A and Event B both occur, you multiply.

\( P(A) \)
 ×
\( P(B) \)
 =
\( P(A \cap B) \)

The Addition Rule

When you need the probability that Event A or Event B occurs (at least one), you add: but subtract the overlap to avoid double-counting.

Sample Space A B
\( A \cap B \)

Mutually Exclusive vs. Overlapping

Mutually Exclusive Overlapping
Can both happen? No Yes
\( P(A \text{ and } B) \) \( 0 \) \( > 0 \)
"Or" formula \( P(A) + P(B) \) \( P(A) + P(B) - P(A \cap B) \)
Example Rolling a 2 or a 5 Drawing a red card or a king

The Complement Shortcut

Sometimes it is easier to find the probability that something does not happen, then subtract from 1.

Example: What is the probability of getting at least one head in 3 coin flips?

1
\( P(\text{no heads}) = P(\text{all tails}) = \left(\frac{1}{2}\right)^3 = \frac{1}{8} \)
2
\( P(\text{at least one head}) = 1 - \frac{1}{8} = \frac{7}{8} \)

Step-by-Step Strategy

1
Identify the events. List every event the question involves.
2
Classify the relationship. Are events independent or dependent? Mutually exclusive or overlapping?
3
Choose the operation. "And" → multiply. "Or" → add (subtract overlap). "At least one" → complement.
4
Compute each probability. Adjust the denominator for dependent events (e.g., \(51\) cards after drawing one).
5
Simplify. Reduce fractions to lowest terms: the ISEE answer choices are always simplified.

Example 1: Independent Events

A bag has 3 red and 5 blue marbles. You draw one marble, replace it, then draw again. What is the probability of drawing red both times?

Draw 1
\( \frac{3}{8} \)
 8 marbles × Draw 2
\( \frac{3}{8} \)
 replaced → still 8 =
\( \frac{9}{64} \)

Example 2: Dependent Events

Same bag: 3 red, 5 blue. This time you draw without replacement. What is the probability of drawing red both times?

Draw 1
\( \frac{3}{8} \)
 8 marbles × Draw 2
\( \frac{2}{7} \)
 not replaced → 7 left =
\( \frac{6}{56} \)

Example 3: "Or" with Overlap

A standard 52-card deck. What is the probability of drawing a heart or a queen?

1
\( P(\text{heart}) = \frac{13}{52} \) : there are 13 hearts
2
\( P(\text{queen}) = \frac{4}{52} \) : there are 4 queens
3
\( P(\text{heart and queen}) = \frac{1}{52} \) : the Queen of Hearts is in both groups
4
\( P(\text{heart or queen}) = \frac{13}{52} + \frac{4}{52} - \frac{1}{52} = \frac{16}{52} = \frac{4}{13} \)

Common Mistakes

✕ Adding Instead of Multiplying
For "A and B," you must multiply. Adding gives a number that is far too large.
✕ Forgetting to Adjust for Dependence
Without replacement, the denominator shrinks. Using the original total gives the wrong second probability.
✕ Ignoring the Overlap
For "A or B" with overlapping events, you must subtract \( P(A \cap B) \) to avoid double-counting.

The Counting Principle Connection

The Fundamental Counting Principle says: if event A can happen in \( m \) ways and event B can happen in \( n \) ways, then A followed by B can happen in \( m \times n \) ways.

Example: Spinner A has 4 sectors, Spinner B has 6 sectors. Total outcomes: \( 4 \times 6 = 24 \). If A has 1 winning sector and B has 2 winning sectors, the probability of both winning: \( \frac{1}{4} \times \frac{2}{6} = \frac{2}{24} = \frac{1}{12} \).

Formula Summary

Situation Keyword Formula
Both events (independent) and \( P(A) \times P(B) \)
Both events (dependent) and \( P(A) \times P(B \mid A) \)
Either event (exclusive) or \( P(A) + P(B) \)
Either event (overlap) or \( P(A) + P(B) - P(A \cap B) \)
At least one complement \( 1 - P(\text{none}) \)

Question 1 of 5

A fair die is rolled twice. What is the probability of rolling a 6 on both rolls?

Question 2 of 5

A jar has 4 green and 6 yellow candies. You pick two without replacement. What is the probability both are green?

Question 3 of 5

From a standard 52-card deck, what is the probability of drawing a diamond or an ace?

Question 4 of 5

A coin is flipped 4 times. What is the probability of getting at least one head?

Question 5 of 5

Spinner A has sections numbered 1 to 5. Spinner B has sections numbered 1 to 4. If both spinners are spun, what is the probability that both land on an even number?