6.2 Probability Rules: Independent, Dependent, Mutually Exclusive, and Complementary Events

6.2 Probability Rules: Independent, Dependent, Mutually Exclusive, and Complementary Events

Quick Summary: GMAT probability questions require identifying whether events are mutually exclusive (Addition Rule) or independent/dependent (Multiplication Rule). When a problem involves draws without replacement, always update both the numerator and denominator. When facing an 'at least one' question, immediately deploy the complementary shortcut: $P(\text{at least one}) = 1 - P(\text{none})$.

Probability on the GMAT Focus Edition is an extension of counting. Every probability problem can be framed as the ratio of two combinatorial values: the size of the restricted event set divided by the size of the total sample space.

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The Core Probability Framework and Bounds

For any experiment with a finite sample space $S$ consisting of equally likely outcomes, the probability of an event $E$ is defined as:

$P(E) = \frac{\text{Number of Favorable Outcomes } n(E)}{\text{Total Number of Equally Likely Outcomes } n(S)}$

Axiomatic Properties to Memorize

1. Strict Bounds: For any event $E$, $0 \le P(E) \le 1$.
   • If $P(E) = 0$, the event is mathematically impossible.
   • If $P(E) = 1$, the event is certain to occur.
2. The Complement Rule: The complement of event $A$ (denoted $A'$ or $A^c$) represents the event that $A$ does not occur. Since an event must either occur or not occur: $P(A) + P(A') = 1 \implies P(A') = 1 - P(A)$

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The Addition Rule: Compound "OR" Events

The probability that event $A$ or event $B$ occurs (the union, denoted $A \cup B$) depends on whether the events can happen at the same time.

General Addition Rule (Overlapping Events)

If events $A$ and $B$ can occur simultaneously, adding $P(A)$ and $P(B)$ double-counts their joint occurrence. We must subtract the intersection $P(A \cap B)$:

$P(A \cup B) = P(A) + P(B) - P(A \cap B)$

Mutually Exclusive (Disjoint) Events

Two events are mutually exclusive if they cannot occur at the same time ($A \cap B = \emptyset$). In this case, $P(A \cap B) = 0$, and the addition rule simplifies to:

$P(A \cup B) = P(A) + P(B)$

Example: When rolling a fair 6-sided die, rolling a 2 and rolling a 5 are mutually exclusive: $P(2 \cup 5) = P(2) + P(5) = \frac{1}{6} + \frac{1}{6} = \frac{2}{6} = \frac{1}{3}$

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The Multiplication Rule: Joint "AND" Events

The probability that event $A$ and event $B$ both occur (the intersection, denoted $A \cap B$) depends on whether the occurrence of $A$ affects the likelihood of $B$.

General Multiplication Rule (Dependent Events)

If the occurrence of event $A$ alters the probability of event $B$, the events are dependent:

$P(A \cap B) = P(A) \times P(B|A)$

where $P(B|A)$ represents the conditional probability of $B$ given that $A$ has already occurred.

Independent Events

Two events are independent if the occurrence of one does not affect the probability of the other: $P(B|A) = P(B)$. When independence holds, the multiplication rule simplifies to:

$P(A \cap B) = P(A) \times P(B)$

Example: Tossing a coin and rolling a die are independent. The probability of getting Heads and rolling a 4 is: $P(\text{Heads} \cap 4) = P(\text{Heads}) \times P(4) = \frac{1}{2} \times \frac{1}{6} = \frac{1}{12}$

The Independence vs. Mutual Exclusivity Conceptual Trap

Candidates frequently confuse independent events with mutually exclusive events. They are fundamentally different concepts:

• Mutually exclusive means events cannot happen together ($P(A \cap B) = 0$).
• Independent means the occurrence of one does not change the probability of the other ($P(A \cap B) = P(A) \times P(B)$).

Mathematical Consequence: If two events $A$ and $B$ both have non-zero probabilities ($P(A) > 0$ and $P(B) > 0$), they cannot be both independent and mutually exclusive. If they were mutually exclusive, $P(A \cap B) = 0$. But if they were independent, $P(A \cap B) = P(A) \times P(B) > 0$. Zero cannot equal a positive number!

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Sampling: Replacement vs. Non-Replacement

A substantial proportion of GMAT probability questions involve drawing items (marbles, cards, chips, or people) from a set.

1. Sampling With Replacement (Independent)

When an item is drawn, recorded, and returned to the pool before the next draw:

• The total number of items in the denominator remains constant.
• The number of favorable items in the numerator remains constant.
• Each draw is completely independent.

2. Sampling Without Replacement (Dependent)

When an item is drawn and set aside:

• The total number of available items decreases by 1 on each subsequent draw.
• If a favorable item was selected, the number of remaining favorable items decreases by 1.
• Each draw is dependent on previous draws.

Comparative Walkthrough

A bag contains 4 red chips and 6 blue chips (10 chips total). Two chips are drawn sequentially.

• With Replacement: Probability of drawing two red chips: $P(R_1 \cap R_2) = \frac{4}{10} \times \frac{4}{10} = \frac{2}{5} \times \frac{2}{5} = \frac{4}{25} = 0.16$
• Without Replacement: Probability of drawing two red chips: $P(R_1 \cap R_2) = \frac{4}{10} \times \frac{3}{9} = \frac{2}{5} \times \frac{1}{3} = \frac{2}{15} \approx 0.133$

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The "At Least One" Complementary Shortcut

When a GMAT question asks for the probability that an event occurs "at least once" in $n$ trials, direct computation is often inefficient. Direct calculation requires computing the probabilities of 1 success, 2 successes, 3 successes, ..., up to $n$ successes, and summing them.

Because the complement of "at least one" is "none at all" (zero occurrences), use the complement shortcut:

$P(\text{At least one success}) = 1 - P(\text{Zero successes})$

Step-by-Step Execution:

1. Identify the condition for "complete failure" in a single trial.
2. Calculate the probability that this failure occurs in every trial.
3. Subtract that joint failure probability from 1.

Classic Coin Example

If a fair coin is flipped 5 times, what is the probability of obtaining at least one Head?

• Direct method: Sum probabilities of 1, 2, 3, 4, and 5 Heads.
• Complement shortcut: The complement of "at least one Head" is "zero Heads" (all 5 flips are Tails): $P(\text{Zero Heads}) = \left(\frac{1}{2}\right)^5 = \frac{1}{32}$ $P(\text{At least one Head}) = 1 - \frac{1}{32} = \frac{31}{32}$

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Conditional Probability in Problem Solving

Conditional probability evaluates the chance of an event happening given that another event has already occurred. Mathematically:

$P(B|A) = \frac{P(A \cap B)}{P(A)}$

On GMAT Problem Solving questions, conditional probability is best solved by restricting the sample space: identify the specific subset of outcomes where condition $A$ is true, and count how many of those outcomes satisfy condition $B$.

Example: Two fair 6-sided dice are rolled, and the sum of the dice is 8. What is the probability that at least one of the dice shows a 3?

1. Restrict the sample space to Sum = 8: The pairs that sum to 8 are: $(2, 6), (3, 5), (4, 4), (5, 3), (6, 2)$. There are exactly 5 equally likely outcomes.
2. Count favorable outcomes within this restricted set: Pairs showing at least one 3 are $(3, 5)$ and $(5, 3)$. There are 2 favorable outcomes.
3. Calculate conditional probability: $P(\ge 1 \text{ three} \mid \text{Sum} = 8) = \frac{2}{5}$

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Worked Problem Solving Examples

Example 1: Multi-Stage Draw Without Replacement

Problem: A jar contains 5 red marbles, 4 green marbles, and 3 blue marbles. If 3 marbles are drawn at random one after another without replacement, what is the probability that at least one of the marbles drawn is red?

Solution:

1. Total marbles in jar = $5 + 4 + 3 = 12$.
2. Non-red marbles = $4 + 3 = 7$.
3. Apply the complement rule: $P(\text{at least one red}) = 1 - P(\text{no red marbles in 3 draws})$.
4. For "no red marbles," all 3 draws must be chosen from the 7 non-red marbles:
   • 1st draw non-red: $\frac{7}{12}$
   • 2nd draw non-red: $\frac{6}{11}$
   • 3rd draw non-red: $\frac{5}{10} = \frac{1}{2}$
5. Multiply the sequential probabilities: $P(\text{no red}) = \frac{7}{12} \times \frac{6}{11} \times \frac{1}{2} = \frac{7 \times 6 \times 1}{12 \times 11 \times 2} = \frac{42}{264} = \frac{7}{44}$
6. Subtract from 1: $P(\text{at least one red}) = 1 - \frac{7}{44} = \frac{37}{44}$

Example 2: Overlapping Event Demographics

Problem: In an executive seminar of 60 participants, 35 speak Spanish, 25 speak French, and 12 speak both Spanish and French. If a participant is selected at random, what is the probability that the participant speaks neither Spanish nor French?

Solution:

1. Let $S$ be the event of speaking Spanish, and $F$ be the event of speaking French.
2. Given: $n(\text{Total}) = 60$, $n(S) = 35$, $n(F) = 25$, $n(S \cap F) = 12$.
3. Use the Addition Rule to find the number of participants who speak Spanish or French: $n(S \cup F) = n(S) + n(F) - n(S \cap F) = 35 + 25 - 12 = 48$
4. The number of participants speaking neither language is the complement of the union: $n(\text{Neither}) = n(\text{Total}) - n(S \cup F) = 60 - 48 = 12$
5. Compute the probability: $P(\text{Neither}) = \frac{12}{60} = \frac{1}{5} = 0.20$

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High-Frequency GMAT Traps

• Trap 1: The Order Trap in Compound Selections
  If a question asks for the probability of drawing "one red and one blue marble" in two draws, candidates often compute $P(\text{Red 1st}) \times P(\text{Blue 2nd})$. But the order could be Red-then-Blue OR Blue-then-Red! You must either multiply by 2 or use combinations: $\frac{\binom{R}{1}\binom{B}{1}}{\binom{Total}{2}}$.

• Trap 2: Forgetting Non-Replacement Denominator Reduction
  Failing to reduce the sample size denominator on successive draws without replacement converts a dependent problem into an incorrect independent calculation.

• Trap 3: Exhaustive Summation on "At Least One" Items
  Attempting to calculate $P(1) + P(2) + P(3) + \dots$ on multi-trial problems consumes precious minutes and introduces compounding arithmetic error. When you see "at least one," immediately set up $1 - P(\text{none})$.