What Are Dependent Events in Probability?

In probability theory, certain events are interconnected such that the occurrence or non-occurrence of one event has a direct effect on the likelihood of another. These are referred to as dependent events, and their analysis necessitates formal treatment using conditional probability. The mathematical structure underpinning dependent events enables accurate computation of combined probabilities in scenarios involving successive steps without replacement.

Formal Definition of Dependent Events Using Probability Theory

Let $A$ and $B$ be two events defined on a finite sample space $S$. The events $A$ and $B$ are termed dependent events if the probability of occurrence of $B$ is affected by the occurrence (or non-occurrence) of $A$, and vice versa. Formally, $A$ and $B$ are dependent if \[ P(B|A) \neq P(B) \] or equivalently, \[ P(A|B) \neq P(A), \] where $P(B|A)$ denotes the conditional probability of $B$ given that $A$ has occurred.

If, instead, $P(B|A) = P(B)$, the events $A$ and $B$ are said to be independent. For dependent events, the computation of joint probabilities must employ conditional probability to reflect the altered sample space after the first event.

Conditional Probability and Dependent Events: Algebraic Structure

Given events $A$ and $B$ with $P(A) > 0$, the probability of $B$ conditioned on $A$ is defined by \[ P(B|A) = \frac{P(A \cap B)}{P(A)}. \] This formula represents the likelihood that both $A$ and $B$ occur, relative to the probability that $A$ occurs.

Rearranging the above equation yields the multiplication rule for dependent events: \[ P(A \cap B) = P(A) \cdot P(B|A). \] Here, $P(A \cap B)$ is the probability that both $A$ and $B$ occur in succession, with the outcome of $A$ generally affecting the probability of $B$ if the events are dependent.

Differentiating Dependent and Independent Events Mathematically

For independent events, the relationship simplifies to $P(B|A) = P(B)$. Therefore, \[ P(A \cap B) = P(A) \cdot P(B). \] For dependent events, $P(B|A) \neq P(B)$, and the adjustment for the first event directly modifies the sample space for the second. Deciding whether events are dependent or independent relies on analyzing whether one event changes the underlying set of possible outcomes for the other. For further comparison, see Dependent vs Independent Events.

Sequential Computation of Probability for Dependent Events

Consider two events $A$ and $B$ occurring in succession without replacement. After event $A$ occurs, the sample space for event $B$ is altered, resulting in \[ P(A \cap B) = P(A) \cdot P(B|A). \] If more than two dependent events $A_1,\,A_2,\,\ldots,\,A_n$ are considered, the formula generalizes as \[ P(A_1 \cap A_2 \cap \ldots \cap A_n) = P(A_1) \cdot P(A_2|A_1) \cdot P(A_3|A_1 \cap A_2) \cdots P(A_n|A_1 \cap \ldots \cap A_{n-1}). \] Each step requires evaluating the conditional probability given all prior outcomes.

In practical terms, dependent events frequently arise in problems involving selections made without replacement, hierarchical sampling, or successive actions where each stage depends on the outcome of the previous one. Foundational discussion on these topics can be found at Understanding Probability.

Worked Example: Drawing Colored Balls Without Replacement

Given: A bag contains $5$ red balls and $3$ green balls. Two balls are drawn successively without replacement. Compute the probability that the first ball is red and the second ball is green.

Step 1: The probability that the first ball is red is \[ P(\text{Red}_1) = \frac{5}{8}. \]

Step 2: After removing one red ball, there are $4$ red and $3$ green balls remaining $(7$ balls left in total$)$.

Step 3: The probability that the second ball is green, given that the first was red, is \[ P(\text{Green}_2 | \text{Red}_1) = \frac{3}{7}. \]

Step 4: The probability that the first is red and second is green is \[ P(\text{Red}_1 \cap \text{Green}_2) = P(\text{Red}_1) \cdot P(\text{Green}_2 | \text{Red}_1) = \frac{5}{8} \times \frac{3}{7} = \frac{15}{56}. \]

Final result: The probability that the first ball is red and the second is green is $\displaystyle\frac{15}{56}$.

Worked Example: Selecting Cards Without Replacement

Given: Two cards are drawn sequentially from a well-shuffled deck of $52$ playing cards, without replacement. Find the probability that the first card is a spade and the second is also a spade.

Step 1: Total spades in a standard deck $= 13$.

Step 2: Probability that the first card is a spade: \[ P(\text{Spade}_1) = \frac{13}{52} = \frac{1}{4}. \]

Step 3: After removing one spade, $12$ spades remain among $51$ cards. Probability that the second card is a spade given the first was a spade: \[ P(\text{Spade}_2 | \text{Spade}_1) = \frac{12}{51}. \]

Step 4: Therefore, \[ P(\text{Spade}_1 \cap \text{Spade}_2) = P(\text{Spade}_1) \cdot P(\text{Spade}_2 | \text{Spade}_1) = \frac{1}{4} \times \frac{12}{51} = \frac{12}{204} = \frac{3}{51} = \frac{1}{17}. \]

Final result: The probability that both cards drawn are spades is $\displaystyle\frac{1}{17}$.

Algebraic Criterion for Identifying Dependent Events

Given any two events $A$ and $B$, the events are dependent if \[ P(B|A) \neq P(B). \] Checking for this discrepancy allows the determination of dependency. In most practical selection problems without replacement, the sample space evolves, ensuring dependence between successive outcomes. In contrast, with replacement typically produces independent events. Detailed discussion is available at Experimental vs Theoretical Probability.

Venn Diagram Representation of Dependent Events

In the context of set theory, $A$ and $B$ are subsets of the sample space $S$. The intersection $A \cap B$ represents the event in which both $A$ and $B$ occur. For dependent events, \[ P(A \cap B) = P(A) \cdot P(B|A) \] can be visualized as the area where $A$ and $B$ overlap, with $P(B|A)$ indicating the proportion of outcomes in $A$ that also belong to $B$. The size of $A \cap B$ relative to $A$ is not generally the same as its size relative to $S$ due to dependency.

Key Remarks and Frequently Confused Points

For mutually exclusive events, $A \cap B = \varnothing$, so $P(A \cap B) = 0$. This case is distinct from dependency and independence, as mutually exclusive events cannot occur together. For more clarification on these distinctions, refer to Independent Events In Probability.

For conditional probability to be defined, the event on which one is conditioning (i.e., $P(A)$ when computing $P(B|A)$) must have $P(A) > 0$.

Advanced Example: Multiple Dependent Draws

Given: A box contains $4$ red balls, $3$ green balls, and $2$ blue balls. Three balls are drawn in succession without replacement. What is the probability that they are drawn in the order red, green, blue?

Step 1: Probability that the first ball is red: \[ P(R_1) = \frac{4}{9} \] Because total balls $= 4 + 3 + 2 = 9$.

Step 2: After a red ball is removed, $3$ red, $3$ green, $2$ blue balls remain $(8$ balls total$)$. Probability that the second ball is green: \[ P(G_2|R_1) = \frac{3}{8} \]

Step 3: After red and green balls are removed, $3$ red, $2$ green, $2$ blue balls remain $(7$ balls total$)$. Probability that the third ball is blue: \[ P(B_3|R_1 \cap G_2) = \frac{2}{7} \]

Step 4: The required probability is \[ P(R_1 \cap G_2 \cap B_3) = P(R_1) \cdot P(G_2|R_1) \cdot P(B_3|R_1 \cap G_2) = \frac{4}{9} \times \frac{3}{8} \times \frac{2}{7} \] Compute sequentially: \[ \frac{4}{9} \times \frac{3}{8} = \frac{12}{72} = \frac{1}{6}, \] \[ \frac{1}{6} \times \frac{2}{7} = \frac{2}{42} = \frac{1}{21}. \]

Final result: The probability that the first is red, the second is green, and the third is blue (in order) is $\displaystyle\frac{1}{21}$.

Summary of Dependent Event Probabilities in Outcomes Without Replacement

For events where selections or trials are performed without replacement, the events are inherently dependent. The probability of joint occurrence must be calculated using the multiplication rule involving conditional probability; each selection alters the sample space for subsequent steps. For further studies in statistics and probability theory, visit Statistics And Probability Overview.