Probability of Independent Events

The event of an occurrence which does not depend on any other event is called an Independent event. For example, if a coin flipped in the air and got the outcome as Head, then again flipped the coin and got the outcome as Tail. In both cases, the occurrence of both events does not depend on each other. It is one of the events in probability. Here is the complete definition of independent events in probability.

Independent Events

In Probability, the set of outcomes of an experiment or an activity is called events. There are various types of events such as independent events, dependent events, mutually exclusive events, etc.

If the probability of occurrence of event A is not dependent on the occurrence of another event B, then A and B are said to be independent events.

Consider an example of rolling a die. If A is the event, where 'the number appearing is odd’ and B is another event, where  ‘the number appearing is a multiple of 3’, then

P(A ∩ B) = \[\frac {1} {6}\]

P(A│B) = \[\frac {P(A∩B)} {P(B)}\]

= \[(\frac {1} {6})\] / \[(\frac {1} {3})\] 

= \[\frac {1} {2}\]

P(A) = P(A│B) = \[\frac {1} {2}\], which proves that event B has no relationship with the occurrence of event A.

Using the Multiplication rule of probability, P(A ∩ B) = P(B). P(A│B)

P(A ∩ B) = P(B) .P(A)

What are Mutually Exclusive Events?

An event is said to be mutually exclusive if both events cannot occur at the same time. Mutually exclusive events do not have an outcome in common.

Independent Events vs Mutually Exclusive Events

Solved Examples

Example 1:

P(B∩C) + P(C∩A) + P(A∩B) - 3P(A∩B∩C)

= 12.14+14.13+13.12−3(13.12.14)

= 14

Example 2:

If three coins are tossed together, what are the probability that the first shows head, the second shows tail and the third shows head?

Solution:

The probability that the first coin shows head = \[\frac {1} {2}\]

The probability that the second coin shows tail = \[\frac {1} {2}\]

The probability that the third coin shows head = \[\frac {1} {2}\]

P(A∩B∩C) = P(A).P(B).P(C) 

= 12.12.12

= 18

Example 3:

Solution:

(1−P(A))(1−P(B))(1−P(C))

Therefore, the required probability is:

= 1−[(1−12)(1−13)(1−14)]

= 1−[12.23.34]

= 1−14

= 34

Therefore, the probability that the sum gets solved is \[\frac {3} {4}\].

Example 4:

If a dice is rolled and a coin is tossed together, what is the probability that the coin gets a tail and the dice gets a 2?

Solution:

Probability of getting a tail =\[\frac {1} {2}\]

Probability of throwing a 2 = \[\frac {1} {6}\]

Therefore, getting a tail on a coin does not depend on getting a 2 on a dice and vice versa, the two events are independent. The probability that the two given events occur together is equal to the product of their respective probability.

Therefore, the required probability is: 12×16

= 112

Therefore, the probability that the coin shows tail and the dice shows 2 is \[\frac {1} {12}\].