5.4 Probability & Independent/Dependent Events

5.4 Probability & Independent/Dependent Events

Probability measures the mathematical likelihood that a specific event will occur. On the GRE Quantitative Reasoning section, probability questions range from basic single-stage outcome counts to multi-stage independent and dependent trials, conditional probability, and complement rules.

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Basic Principles & Scale of Probability

The classical probability of an event $E$, denoted $P(E)$, assumes all outcomes in a sample space are equally likely:

$P(E) = \frac{\text{Number of Favorable Outcomes}}{\text{Total Number of Possible Outcomes}}$

The Probability Scale

Probability is strictly bounded between $0$ and $1$ inclusive: $0 \le P(E) \le 1$

• $P(E) = 0$: The event is impossible.
• $P(E) = 1$: The event is certain.
• Probabilities can be expressed as simplified fractions, decimals, or percentages.

Complementary Probability Rule

The complement of event $E$ (denoted $E^c$ or $\text{not } E$) represents all outcomes where $E$ does not occur. Because $E$ and $E^c$ partition the entire sample space:

$P(E) + P(E^c) = 1 \implies P(E^c) = 1 - P(E)$

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Addition Rules: Combining "OR" Events

When calculating the probability that Event A OR Event B occurs ($P(A \cup B)$), the addition rule depends on whether the events can happen simultaneously.

1. Mutually Exclusive (Disjoint) Events

Events $A$ and $B$ are mutually exclusive if they cannot occur at the same time ($P(A \cap B) = 0$).

$P(A \text{ or } B) = P(A) + P(B)$

Example: Rolling a fair 6-sided die once. What is the probability of rolling a 2 OR a 5?

• $P(2) = \frac{1}{6}$, $P(5) = \frac{1}{6}$.
• $P(2 \text{ or } 5) = \frac{1}{6} + \frac{1}{6} = \frac{2}{6} = \frac{1}{3}$.

2. General Addition Rule (Non-Disjoint Events)

If events $A$ and $B$ can overlap, simple addition double-counts their intersection:

$P(A \text{ or } B) = P(A) + P(B) - P(A \text{ and } B)$

Worked Example: Drawing a single card from a standard 52-card deck. What is the probability of drawing a Heart OR a King?

• $P(\text{Heart}) = \frac{13}{52}$.
• $P(\text{King}) = \frac{4}{52}$.
• $P(\text{King of Hearts}) = \frac{1}{52}$.
• $P(\text{Heart or King}) = \frac{13}{52} + \frac{4}{52} - \frac{1}{52} = \frac{16}{52} = \frac{4}{13}$.

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Multiplication Rules: Combining "AND" Events

When calculating the probability that both Event A AND Event B occur ($P(A \cap B)$), the multiplication rule depends on whether the outcome of $A$ influences $B$.

1. Independent Events

Events $A$ and $B$ are independent if the occurrence of $A$ does not affect the probability of $B$.

$P(A \text{ and } B) = P(A) \times P(B)$

Example (With Replacement): Drawing a marble from a bag with 4 red and 6 blue marbles, recording its color, replacing it, and drawing a second marble.

• $P(\text{Red}_1) = \frac{4}{10} = \frac{2}{5}$.
• $P(\text{Red}_2) = \frac{4}{10} = \frac{2}{5}$.
• $P(\text{Both Red}) = \frac{2}{5} \times \frac{2}{5} = \frac{4}{25}$.

2. Dependent Events & Sampling Without Replacement

If the outcome of event $A$ alters the sample space for event $B$, the events are dependent.

$P(A \text{ and } B) = P(A) \times P(B|A)$

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

Worked Example (Without Replacement): Drawing 2 red marbles sequentially from 4 red and 6 blue marbles WITHOUT replacement.

• $P(\text{Red}_1) = \frac{4}{10}$.
• $P(\text{Red}_2 | \text{Red}_1) = \frac{3}{9}$ (since 3 red marbles and 9 total marbles remain).
• $P(\text{Both Red}) = \frac{4}{10} \times \frac{3}{9} = \frac{12}{90} = \frac{2}{15}$.

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Conditional Probability & Tree Diagrams

Conditional Probability Formula

The probability of event $A$ occurring given that event $B$ has already occurred is:

$P(A|B) = \frac{P(A \text{ and } B)}{P(B)} \quad \text{for } P(B) > 0$

Multi-Stage Probabilities & Tree Diagrams

A tree diagram visualizes multi-stage probabilistic events. Each branch represents an outcome weighted by its conditional probability.

                             TREE DIAGRAM (Marble Sampling)
                                     +--- Red (3/9)  --> P(R1, R2) = (4/10)*(3/9) = 12/90
                                    / 
                +--- Red (4/10) ----+
               /                    
  Start Stage -                      +--- Blue (6/9) --> P(R1, B2) = (4/10)*(6/9) = 24/90
               
                +--- Blue (6/10) ---+--- Red (4/9)  --> P(B1, R2) = (6/10)*(4/9) = 24/90
                                    
                                     +--- Blue (5/9) --> P(B1, B2) = (6/10)*(5/9) = 30/90

• Path Multiplication Rule: Multiply probabilities along a single path to get the probability of that joint sequence.
• Branch Addition Rule: Add probabilities of disjoint paths that satisfy your favorable condition.

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Strategic "At Least One" Rule

Calculating "at least one" directly often requires computing and adding multiple multi-stage probabilities. The shortcut relies on complementary probability:

$P(\text{At least one success}) = 1 - P(\text{All failures})$

Worked Example: A fair six-sided die is rolled 3 times. What is the probability of rolling at least one 6?

1. Identify Failure: Failing to roll a 6 on a single roll has probability $P(\text{No 6}) = \frac{5}{6}$.
2. Compute All Failures across 3 Trials: $P(\text{No 6 on all 3 rolls}) = \left(\frac{5}{6}\right)^3 = \frac{125}{216}$.
3. Apply Complement Rule: $P(\text{At least one 6}) = 1 - \frac{125}{216} = \frac{91}{216}$.