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1299 LearnersLast updated on November 25, 2025

Independent events are the events that occur together, but the outcome of one event does not affect the outcome of the other in any manner. In this topic, we are going to talk about independent events and how they are different from dependent events.
Two events are independent when the outcome of one event does not affect the result of the other. In simple terms, if one event occurs and does not change the probability of the second event, the events are independent. For example, winning a lottery and going for a picnic have no connection; one does not influence the other, so they are independent events.
Here, we represent the probability of independent events using the formula:
\(P(A ∩ B) = P(A) × P(B)\)
This means that if events A and B are independent, the probability that both events happen is the product of their individual probabilities.
For example,
If event A is rolling the die and getting a 4
\(P(A) = \frac{1}{6}\)
If event B is rolling the coin and getting heads
\(P(A) = \frac{1}{6}\)
Since rolling a die does not affect a coin toss, these events are independent.
\(P(A∩B) = \frac{1}{6} × \frac{1}{2} = \frac{1}{12}\)
So, the probability of getting a four on a die and a head on a coin is \(\frac{1}{12}\).
Probability has two types of events: Independent events and dependent events. Here are a few differences between the two events:
| Independent Events | Dependent Events |
| When two events do not impact the probability of each other's outcome, we call it independent events. | The occurrence of one event affects the probability of another event. |
| Example: Watching a movie on your laptop and booking a cab. | Example: Drawing twp cards from a deck without replacement |
| Formula: P(A and B) = \(P(A) × P(B)\) |
Formula: P(A and B) = P(A) × P(B after A) |
When learning about independent events there are a few certain rules that must be followed such as:
1. Rule of multiplication
This rule is used to find the probability that two independent events co-occur.
For independent events A and B:
\(P(A∩B) = P(A) P(B)\)
For example,
Event A is rolling a die and getting a 6 → \(P(A) = \frac{1}{6}\)
Event B is tossing a coin and getting heads → \(P(B) = \frac{1}{2}\)
Since the die roll does not affect the coin toss, the two events are independent.
P(A ⋂ B) = \(\frac{1}{6}\)\(×\frac{1}{2}\) = \(\frac{1}{12}\)
2. Rule of addition
This rule is used to find the probability that at least one of the two events occurs. It is written as:
\(P(A ∪ B)=P(A)+P(B)-P(A ∩ B)\)
For example,
Let the event A be a student passes math → \(P(A) = 0.7\)
Event B: A student passes science → \(P(B) = 0.6\)
If the events are independent:
\(P(A∩B) = 0.7 × 0.6 = 0.42\)
Now using the addition rule:
\(P(A∪B) = 0.7 + 0.6 – 0.42 = 0.88\)
This means there is an 88% chance the student passes either math, science, or both.
To find the probability of independent events, we use a formula:
P(A and B) = \(P(A) × P(B)\)
Where:
P(A and B) is the probability that both events occur.
P(A) is the probability of event A.
P(B) is the probability of event B.
If these statements are true, then A and B are independent events.
Step 1: Calculate the individual probabilities so first, we find the probability of P(A) and then the probability of P(B)
Step 2: Then we calculate the joint probability, which is P(A and B)
Step 3: Determine whether it is independent or not by using the formula P(A and B) = \(P(A) × P(B)\). If it is not true, then events are dependent.
Before using probability formulas, it’s essential to check whether the events are independent or dependent. You can follow these steps.
Step 1: Check whether the events can happen in any order.
If yes, go to step 2. If no, go to step 3.
Step 2: Check if one event changes or affects the outcome of the other.
If yes, go to step 4. If no, go to step 3.
Step 3: If one event does not affect the other, the events are independent. Then, use the formula for independent events.
Step 4: If one event affects the other, the events are dependent. Then, use the formula for dependent events.
Independent events are those where the outcome of one event does not affect the other. Mastering this concept helps in solving probability problems more accurately and confidently.
When understanding the concept of independent events, students tend to make mistakes. Here, are some common mistakes and their solutions.
There are many uses of independent events. Let us now see the uses and applications of independent events in different fields:
Coin Tossing in Sports: We use independent events in sports where in football, the coin toss at the start of the game is independent of the match’s outcome. Each toss has a 50% chance of landing heads or tails, no matter what happened before.
Rolling Dice in Board Games: We use the concept of independent events in games like Monopoly or Yahtzee, where rolling a die is independent of the previous rolls. The probability of rolling 6 remains 1/6 on each throw.
Lottery Draws and Gambling: We use the concept of independent events in lottery draws and gambling, where each lottery ticket has an equal chance of winning, independent of the previous draws. In a roulette game, the ball landing on red is independent of past spins.
Weather forecasting: Meteorologists consider each day's weather as an independent event when predicting rainfall or sunshine probabilities.
Coin tossing: Each flip of a coin is an independent event, since the result of one toss doesn't affect the outcome of the next.
What is the probability of getting heads on the first toss and tails on the second toss?
0.25
P(Heads) = 0.5
P(Tails) = 0.5
We use the formula: P(A and B) = \( P(A) × P(B)\)
P(Heads and Tails) = \(0.5 × 0.5 = 0.25\)
The probability of getting heads on the first toss and tails on the second toss is 0.25
What is the probability of getting tails on a coin toss and a 4 on a die roll
1/12 or 0.0833 (approximately).
P(Tails) = ½ = 0.5
P(4 on a die) = \(\frac{1}{6}\) \( ≈ 0.1667\)
We use the formula: P(A and B) = \(P(A) × P(B)\)
P (Tails and 4 on a die) = \(\frac{1}{2}\) × \(\frac{1}{6}\) = \(\frac{1}{12}\).
What is the probability of drawing an Ace from a standard 52-card deck and rolling a 6 on a die?
\(\frac{1}{78}\)
P(Ace) = \(\frac{4}{52}\) = \(\frac{1}{13}\)
P(6 on a die) = \(\frac{1}{6}\)
We use the formula: P(A and B) = \(P(A) × P(B)\)
P (Ace and 6 on a die) = \(\frac{1}{13}\) × \(\frac{1}{6}\) = \(\frac{1}{78}\).
What is the probability of rolling a 2 on the first roll and a 4 on the second roll with a die?
\(\frac{1}{6}\)
P(2 on a die) = \(\frac{1}{6}\)
P(4 on a die) = \(\frac{1}{6}\)
We use the formula: P(A and B) = \(P(A) × P(B)\)
P(2 on a die and 4 on a die) = \(\frac{1}{6}\) × \(\frac{1}{6}\) = \(\frac{1}{36}\).
What is the probability of rolling an even number and drawing a red card from a standard deck (there are 26 red cards out of 52)?
0.25
P(even number on die) = \(\frac{3}{6}\) = 0.5
P(Red card) = \(\frac{26}{52}\) = 0.5
We use the formula: P(A and B) = \(P(A) × P(B)\)
P (even number on die and red card) = \(0.5 × 0.5 = 0.25\).
Jaipreet Kour Wazir is a data wizard with over 5 years of expertise in simplifying complex data concepts. From crunching numbers to crafting insightful visualizations, she turns raw data into compelling stories. Her journey from analytics to education ref
: She compares datasets to puzzle games—the more you play with them, the clearer the picture becomes!






