Independent literally means that the things which are not dependent on each other. It means that things hold an individual stand without any concern of what the other is up to. In probability, independent events mean that the probability of occurrence of one event does not predict the probability of occurrence of other event.

P(A $\cap$ B) = P(A) . P(B)
If two events are independent, the occurrence or nonoccurrence of one does not depend on the occurrence or non occurrence of the other.
Let A and B be two events with respective probabilities P(A) and P(B). Let P$(\frac{B}{A})$ be the conditional probability of event B, given that event A has happened.
Then, the probability of simultaneous occurrence of A and B is
P(A $\cap$ B) = P(A) . P($\frac{B}{A}$)
If the events are independent, the statement reduces to
P(A $\cap$ B) = P(A) . P(B) → Read More A compound event consists of two or more simple events. When one event does not affect the other, it is said to be an independent event. If the outcome of an event affect the outcome of another event, then it is known as dependent event. Probability of compound events is found by multiplying the probability of each of the events together.
Given below are some of the examples on independent events.
Solved Examples
Question 1: Two fair dice are rolled. If the sum of the numbers obtained is 4, find the probability that the numbers obtained on both the dice are even.
Solution:
Solution:
Let the events A and B be
A: The sum of the numbers is 4.
B: The numbers on both the dice are even.
We need to find P$(\frac{B}{A})$ = $\frac{P(A\cap B)}{P(A)}$
Event A has 3 favorable outcomes, namely, (1, 3), (2, 2) and (3, 1) and P(sample space) = 36, total number of outcomes.
P[Sum 4] = P(A) = $\frac{3}{36}$
Event (A $\cap$ B) has 1 favorable outcome, namely, (2, 2).
Therefore, P[Sum 4 and numbers even] = P(A$\cap$B) = $\frac{1}{36}$
Therefore, the probability that the numbers obtained on both the dice are even is $\frac{1}{36}$.
Thus, P[Numbers even given sum 4] = P($\frac{B}{A}$)
= $\frac{P(A\cap B)}{P(A)}$
= $\frac{\frac{1}{36}}{\frac{3}{36}}$
= $\frac{1}{3}$
A: The sum of the numbers is 4.
B: The numbers on both the dice are even.
We need to find P$(\frac{B}{A})$ = $\frac{P(A\cap B)}{P(A)}$
Event A has 3 favorable outcomes, namely, (1, 3), (2, 2) and (3, 1) and P(sample space) = 36, total number of outcomes.
P[Sum 4] = P(A) = $\frac{3}{36}$
Event (A $\cap$ B) has 1 favorable outcome, namely, (2, 2).
Therefore, P[Sum 4 and numbers even] = P(A$\cap$B) = $\frac{1}{36}$
Therefore, the probability that the numbers obtained on both the dice are even is $\frac{1}{36}$.
Thus, P[Numbers even given sum 4] = P($\frac{B}{A}$)
= $\frac{P(A\cap B)}{P(A)}$
= $\frac{\frac{1}{36}}{\frac{3}{36}}$
= $\frac{1}{3}$
Question 2: A box has 1 red and 3 white balls. Balls are drawn one after one from
the box. Find the probability that the two balls drawn would be red if
Solution:
 The balls are drawn with replacement
 The balls are drawn without replacement.
Solution:
Let A: First ball drawn is red.
B: Second ball drawn is red.
Draw with replacement:
Here, P(A) = $\frac{1}{4}$. Also, since the first ball is returned before the second draw is made,
P($\frac{B}{A}$) = $\frac{1}{4}$
P[Two balls are red] = P(A $\cap$ B)
= P(A) . P($\frac{B}{A}$)
= $\frac{1}{4}$ . $\frac{1}{4}$
= $\frac{1}{16}$
Draw without replacement:
Since the first ball is not returned before the second draw is made, P($\frac{B}{A}$) = $\frac{0}{4}$
P[Two balls are red] = P(A $\cap$ B)
= P(A). P($\frac{B}{A}$)
= $\frac{1}{4}$ $\times$ $\frac{0}{4}$
= 0
B: Second ball drawn is red.
Draw with replacement:
Here, P(A) = $\frac{1}{4}$. Also, since the first ball is returned before the second draw is made,
P($\frac{B}{A}$) = $\frac{1}{4}$
P[Two balls are red] = P(A $\cap$ B)
= P(A) . P($\frac{B}{A}$)
= $\frac{1}{4}$ . $\frac{1}{4}$
= $\frac{1}{16}$
Draw without replacement:
Since the first ball is not returned before the second draw is made, P($\frac{B}{A}$) = $\frac{0}{4}$
P[Two balls are red] = P(A $\cap$ B)
= P(A). P($\frac{B}{A}$)
= $\frac{1}{4}$ $\times$ $\frac{0}{4}$
= 0
Question 3: A machine has two parts. The probability of failure of one of the parts in a given period of time is 0.06. The probability of failure of the other part in the same period is 0.08. What is the probability that the machine fails in that period of time?
Solution:
Solution:
Let the events A and B be
A: First part fails in the given period.
B: Second part fails in the same period.
Given P(A) = 0.06 and P(B) = 0.08
The machine fails if any of the part fails.
Therefore, P[Machine fails] = P(A $\cup$ B)
By Addition theorem of probability we have,
= P(A) + P(B)  P(A $\cap$ B)
= P(A) + P(B)  P(A) $\times$ P(B) (A and B are independent events)
= 0.06 + 0.08  0.06 $\times$ 0.08
= 0.06 + 0.08  0.0048
= 0.1352
Therefore, the probability that the machine fails in that period of time is 0.1352
A: First part fails in the given period.
B: Second part fails in the same period.
Given P(A) = 0.06 and P(B) = 0.08
The machine fails if any of the part fails.
Therefore, P[Machine fails] = P(A $\cup$ B)
By Addition theorem of probability we have,
= P(A) + P(B)  P(A $\cap$ B)
= P(A) + P(B)  P(A) $\times$ P(B) (A and B are independent events)
= 0.06 + 0.08  0.06 $\times$ 0.08
= 0.06 + 0.08  0.0048
= 0.1352
Therefore, the probability that the machine fails in that period of time is 0.1352