Calculate Independent Event Probabilities

Enter probability of Event A P(A) and probability of Event B P(B).

Probability $P(A)$ (e.g. 0.50 = 50%).
Probability $P(B)$ (e.g. 0.30 = 30%).

Calculation Results

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Calculated using independent event multiplication rules: P(A \cap B) = P(A) \cdot P(B), \quad P(A \cup B) = P(A) + P(B) - P(A) \cdot P(B)

*Note: Events A and B are independent if the occurrence of Event A does not affect the probability of Event B.

Quick Summary

The Probability of Independent Events Calculator evaluates joint probability $P(A \cap B) = P(A) \cdot P(B)$, union $P(A \cup B)$, neither event, and exactly one event.

Formula Explanation

P(A \cap B) = P(A) \cdot P(B)
P(A \cup B) = P(A) + P(B) - P(A \cap B)
P(\text{Neither}) = (1 - P(A)) \cdot (1 - P(B))

How It Works

Independent events do not influence each other's occurrence. The Probability of Independent Events Calculator multiplies individual probabilities to compute joint intersection $P(A \cap B)$ and applies the addition rule for union $P(A \cup B)$.

Step-by-Step Worked Example

Practical Problem: Flipping a coin ($P(A) = 0.50$ for heads) and rolling a die ($P(B) = 0.1667$ for rolling a 6). Calculate probability of both happening.

  1. Step 1: Verify event independence: Coin flip outcome does not affect die roll.
  2. Step 2: Calculate joint probability $P(A \cap B)$: $P(A \cap B) = 0.50 \times 0.166667 = \mathbf{0.083333\text{ (8.33\%)}}.$
  3. Step 3: Calculate union $P(A \cup B)$ (At least one): $0.50 + 0.166667 - 0.083333 = \mathbf{0.583333\text{ (58.33\%)}}.$
  4. Step 4: Calculate neither event occurring: $(1 - 0.50) \times (1 - 0.166667) = 0.50 \times 0.833333 = \mathbf{0.416667\text{ (41.67\%)}}.$
  5. Step 5: Interpretation: The chance of getting both heads and a 6 is 8.33% (1 in 12).

Real-World Calculation Examples

Scenario 1: Coin Flip & Die Roll (P(A)=0.50, P(B)=0.1667)

Parameters: P(A) = 0.50, P(B) = 0.1667
Result: P(A∩B) = 0.0833 (8.33% joint chance).

Scenario 2: Dual System Hardware Redundancy

Parameters: Server A uptime = 99%, Server B uptime = 99%
Result: System Failure P(Neither) = 0.0001 (99.99% system reliability!).

Scenario 3: Weather Rain (50%) & Traffic Jam (30%)

Parameters: P(Rain) = 0.50, P(Traffic) = 0.30
Result: P(Both) = 0.1500.

Scenario 4: Two Independent Student Exam Passes

Parameters: P(Student 1) = 0.80, P(Student 2) = 0.90
Result: P(Both Pass) = 0.7200.

Key Benefits of Using This Calculator

Complete 4-Event Probability Breakdown

Calculates Both $P(A \cap B)$, Union $P(A \cup B)$, Neither, and Exactly One event.

Multiplication Rule Verification

Applies exact multiplication rule $P(A \cap B) = P(A) \cdot P(B)$.

Hardware Redundancy Modeling

Evaluates dual-system parallel reliability and failure probabilities.

100% Free & Client-Side

Executes locally in your browser with zero latency or web server transmission.

Frequently Asked Questions (FAQ)

What are independent events in probability?

Two events A and B are independent if the occurrence of A gives no information about whether B occurs ($P(A|B) = P(A)$).

What is multiplication rule for independent events?

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

What is general addition rule for any two events?

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

What is difference between independent and mutually exclusive events?

Mutually exclusive events CANNOT occur together ($P(A \cap B) = 0$); independent events CAN occur together ($P(A \cap B) = P(A) P(B) > 0$).

How do I calculate probability of N independent events all occurring?

Multiply all individual probabilities together: $P(A_1 \cap A_2 \dots \cap A_N) = P(A_1) \times P(A_2) \dots \times P(A_N)$.

How do I calculate probability of at least one event in N independent trials?

Use complement rule: $P(\text{At least one}) = 1 - (1 - p)^N$.

How do I calculate independent probabilities in Excel?

Use multiplication formula =P_A * P_B.

What is Gambler's Fallacy?

Gambler's fallacy incorrectly assumes that past independent trial outcomes influence future independent outcomes (e.g. "red is due" on roulette).

Can two non-empty events be both independent and mutually exclusive?

No — if events are mutually exclusive, $P(A \cap B) = 0$, but if independent, $P(A \cap B) = P(A)P(B) > 0$.

What is pairwise independence vs mutual independence?

Pairwise independence means every pair of events is independent; mutual independence means all subsets of events are independent.