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If the occurrence of one event does not affect the occurrence or non-occurrence of other event, the events are said to be independent. Tossing a coin is the most common example of an independent event. The result of each toss is not determined by the result of the previous toss. Another example is that the probability of drawing a card from the deck is independent of the probability of drawing another card from the deck if the first card is replaced. If A and B are two independent events then, the conditional probability of A occurring, given that B occurs is equal to the unconditional probability of A, that is,

P(A / B) =P(A)

Similarly, the conditional probability of B occurring, given that A occurs is equal to the unconditional probability of B, that is,

P(B / A) =P(B)

If A1,A2…..An are independent events then P(A1,A2…..An) = P(A1)P(A2)…..P(An).

Therefore, the probability that two or more independent events occur simultaneously is equal to the product of their respective probabilities. In other words, for two independent events A and B,

P(A∩B) = P(A).P(B)

If A and B are two independent events,

P(A∩B) = P(A).P(B/A), P(A)>0

Or

P(A∩B) = P(B).P(A/B), P(B)>0

Proof:

For independent events, A and B, P(A/B) = P(A) and P(B/A) = P(B) Therefore,

P(A∩B) = P(A).P(B)

Now,

P(A/B) = P(A)

P(A/B) = \[\frac{P(A).P(B)}{P(B)}\]

P(A/B) = \[\frac{P(A∩B)}{P(B)}\]

P(A∩B) = P(B).P(A/B)

Similarly,

P(B/A) = P(B)

P(B/A) = \[\frac{P(A).P(B)}{P(A)}\]

P(B/A) = \[\frac{P(A∩B)}{P(A)}\]

P(A∩B) = P(A).P(B/A)

Therefore, the probability that any two independent events occur simultaneously is equal to the product of product of probability of occurrence of one event and the conditional probability of the other.

If p1,p2,p3…..pn are the probabilities of n independent events A1,A2…..An, then the probability of occurrence of at least one of these events is given by,

[1- (1-p1)(1-p2)....(1-p3)]

Therefore,

P(A1+A2+A3...An) = 1-\[P(\overline{A_{1}})\]\[P(\overline{A_{2}})\]...\[P(\overline{A_{n}})\]

Example 1:

Suppose A, B and C be three independent events. If P(A) = 1/3, P(B) = 1/2 and P(C) = 1/4, then what is the probability of exactly 2 events occurring out of 3 events.

Solution:

The probability that exactly two out of three events occur can be calculated as:

P(exactly two of A, B and C occur)

= P(B∩C) + P(C∩A) + P(A∩B) - P(A∩B∩C)

Since, A, B and C are independent events, the probability of two or more events occurring simultaneously can be calculated as the product of their respective probabilities. Therefore,

P(B∩C) + P(C∩A) + P(A∩B) - 3P(A∩B∩C)

= P(B).P(C) + P(C).P(A) + P(A).P(B) - 3P(A).P(B).P(C)

=\[\frac{1}{2}.\frac{1}{4} + \frac{1}{4}.\frac{1}{3} + \frac{1}{3}.\frac{1}{2}-3(\frac{1}{3}.\frac{1}{2}.\frac{1}{4})\]

=\[\frac{1}{4}\]

Hence, probability that exactly 2 events occur out of 3 is 1/4.

Example 2:

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

Solution:

Probability that the first coin shows head = 1/2

Probability that the second coin shows tail = 1/2

Probability that the third coin shows head = 1/2

Now, the results of the toss of three coins are independent of each other, hence, this is an independent event. Therefore, the probability that the three given events occur simultaneously is equal to the product of their respective probability.

Let A, B, C denote the three given events, therefore,

P(A∩B∩C) = P(A).P(B).P(C)

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

= \[\frac{1}{8}\]

Hence, probability that the first shows head, second shows tail and third shows head is 1/8.

Example 3:

Three students A, B and C are asked to solve a sum. The chances of A, B and C solving the sum is 1/2, 1/3 and 1/4 respectively. What is the probability that the sum is solved?

Solution:

The events of solving the problem by A, B and C are independent of each other. The probability that the sum gets solved by at least one student is given by,

1- [(1-P(A))(1-P(B))(1-P(C))]

Therefore, the required probability is:

= \[1-[(1-\frac{1}{2})(1-\frac{1}{3})(1-\frac{1}{4})]\]

= \[1-[\frac{1}{2}.\frac{2}{3}.\frac{3}{4}]\]

= \[1-\frac{1}{4}\]

= \[\frac{3}{4}\]

Hence, the probability that the sum gets solved is 3/4.

Example 4:

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

Solution:

Probability of getting a tail =1/2

Probability of throwing a 2 = 1/6

Since, 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 simultaneously is equal to the product of their respective probability.

Therefore, required probability is:

\[\frac{1}{2} × \frac{1}{6}\] = \[\frac{1}{12}\]

Hence, probability that the coin shows tail and the dice shows 2 is 1/12.

If A and B are two independent events then A and \[\overline{B}\] are also independent events. Conversely, B and \[\overline{A}\] are also independent events.

If A, B and C are three mutually independent events then, A and B∪C are independent. A and B∩C are also independent.

If P(A1∪A2) = \[1 - P(\overline{A_{1}})P(\overline{A_{2}})\] then A1 and A2 are independent events.

For any two independent events, A1 and A2, P{(A1∪A2)∩(\[\overline{A_{1}}\]∩\[\overline{A_{2}}\])} is less than 1/4.

For independent events, A1,A2…..An, where P(Ai) = \[\frac{1}{i+1}\], where i = 1, 2, ……n .Then the probability that none of the events will occur is \[\frac{1}{n+1}\]

Even though the occurrence of two or more independent events are not affected by the occurrence or non-occurrence of the other, they can have the same outcomes. For example: The result of every toss even though independent of the result of any other toss, can be the same.

An event is said to be independent of itself if P(A) = P(A∩A) = P(A).P(A) = 1

If one event influences the occurrence or non-occurrence of other event, they are said to be dependent events.

FAQ (Frequently Asked Questions)

1. What is the difference between Mutually Exclusive and Independent Events?

Ans. Mutually exclusive events are those events that cannot occur simultaneously while the occurrence or non-occurrence of one event does not affect the other if they are independent events. For mutually exclusive events, P(A∩B) = ф, unlike independent events. Some examples of mutually exclusive events are:

In a single throw of coin, either heads or tails will appear and not both.

In throw of a dice, either an odd or an even number will appear and not both.

2. How are Independent Events related to Conditional Probability?

Ans. Conditional probability is the probability of the occurrence of one event in the case that a second event occurs. Conditional probability that an event A occurs, given that event B occurs is given by,

P(A/B) = P(A∩B) / P(B)

However, if two events are independent, the occurrence of one event will not affect the occurrence of other. Hence the conditional probability of occurrence of event A, provided that event B will occur is equal to the probability of occurrence of event A, that is,

P(A/B) = P(A)