Dependent and Independent Events

In probability, an event is simply a set of outcomes from a random experiment. It can be anything that can happen when you perform an experiment.

For example, if you toss a coin:

1) Possible outcomes (sample space): {Heads, Tails}

2) Events could be:

• Getting a Head → {Heads}
• Getting a Tail → {Tails}

An event in probability falls under two categories

Dependent Events

Dependent events are events in which the outcome of one affects the probability of the next. In simple words, two or more events are dependent if the occurrence of one event changes the likelihood that another event will occur.

So, if the probability of Event B changes because Event A has already happened, the events are said to be dependent.

Example of Dependent Events

Suppose three cards are drawn from a deck without replacement.

• When the first card is drawn, the probability of getting a king is based on all 52 cards.
• For the second draw, the probability changes because one card has already been removed from the deck.
• By the third draw, the probability depends on the outcomes of both previous draws.

Since the deck becomes smaller after every draw, the chances of getting certain cards change each time. Therefore, these events are dependent.

Independent Events

Independent events are events whose outcomes do not affect each other. In other words, if the probability of Event A remains unchanged even after Event B occurs (and vice versa), then A and B are independent events.

Examples of Independent Events

1. Tossing a Coin

Sample space: S = {H, T}

Getting Heads does not affect the chance of getting Tails.
So, these outcomes are independent.

2. Rolling a Die

Sample space: S = {1, 2, 3, 4, 5, 6}

Each outcome is unaffected by the others.
Thus, each result is independent.

Independent Compound Events

Independent events can also involve two different experiments happening at the same time.

3. Tossing a Coin and Rolling a Die

Sample Space (S) = {(1, H), (2, H), (3, H), (4, H), (5, H), (6, H), (1, T), (2, T), (3, T), (4, T) (5, T) (6, T)}.

Here:

• The coin result does not influence the die result.
• The die result does not influence the coin result.

Since neither outcome affects the other, these events are independent.

Note: A and B are two events associated with the same random experiment, then A and B are known as independent events if P(A ∩ B) = P(B).P(A)

Independent Events vs Dependent Events

The difference between independent events and dependent events is discussed in the table below,

Also Check:

• Coin Toss Probability
• Events in Probability
• Types of Events

Solved Examples on Dependent and Independent Events

Example 1: An instructor has a question bank with 300 Easy T/F, 200 Difficult T/F, 500 Easy MCQ, and 400 Difficult MCQ. If a question is selected randomly from the question bank, What is the probability that it is an easy question given that it is an MCQ?

Solution:

Total question in the question bank = 300 + 200 + 500 + 400

P(Easy) = (300+500)/1400 = 800/1400 = 4/7

P(MCQ) = (400+500)/1400 = 900/1400 = 9/14

P(Easy ∩ MCQ) = (500)/1400 =5/14

P(Easy|MCQ) = P(Easy ∩ MCQ)/P(MCQ)

P(Easy|MCQ) = (5/14)/(9/14) = 5/9

Thus, probability of an easy question given it is an MCQ is 5/9.

Example 2: In a shipment of 20 apples, 3 are rotten. 3 apples are randomly selected. What is the probability that all three are rotten if the first and second are not replaced?

Solution:

Total Apple = 20

Rotten Apple = 3

• Possibility of the first apple being rotten = 3/20
• Possibility of the second apple being rotten = 2/19
• Possibility of the third apple being rotten = 1/18

Probability of all three apples being rotten = P(3 Rotten) = (3/20 ×  2/19 × 1/18) = 6/6840 = 1/1140

Thus, probability that all three apples are rotten is, 1/1140

Example 3: John has to select two students from a class of 10 girls and 15 boys. What is the probability that both students chosen are boys?

Solution:

Total number of students = 10 + 15 =25

Probability of choosing the first boy

P (Boy 1) = 15/25

P (Boy 2) = 14/24

P (Boy 1 and Boy 2) = P (Boy 1) and P (Boy 2)

P (Boy 1 and Boy 2) = (15/25) × (14/24) = 7/20

Thus, probability of choosing both boys is 7/20

Example 4: A multiple-choice test consists of two problems. Problem 1 has 5 options and Problem 2 has 4 options. Each problem has only one correct answer. What is the probability of randomly guessing the correct answer to both problems?

Solution: 

Here, the probability of the correct answer to Problem 1 = P(A) and the probability of the correct answer to Problem 2 = P(B) are independent events.

Thus the probability of a correct answer to Problem 1 and Problem 2 both = P(A ∩ B) = P(A).P(B)

• P(A) = 1/5
• P(B) = 1/4

P(A ∩ B) = (1/5) × (1/4) = 1/20

Thus, probability of getting both answers correct is 1/20.