Baye’s Theorem Of Probability explains the probability of an event based on the prior information about the conditions that might be associated with the event.

The topic plays an important role in the preparation for upcoming JEE examinations. In this article, we cover of Baye’s Theorem Of Probability, its examples and also answers the question – What Are Mutually Exclusive Events? Both the topics are curated here so that the students can easily access them.

**Statement: **

Consider E_{1}, E_{2}, …, E_{n} to be a set of events with a sample space S, where the events E_{1}, E_{2}, …, E_{n} possess a probability of occurrence that is not 0. If A is any event related with S, then Bayes Theorem Of Probability is mathematically represented as,

**P(E _{i} / A) = [P(E_{i}) P (A / E_{i}**

**)] / [∑**

_{k=1}

^{n}**P(E**

_{k}) P (A / E_{k})]** **for any k = 1, 2, 3, …., n

**Proof of Baye’s Theorem Of Probability **

The conditional probability formula is given by

P (E_{i}│A) = P (E_{i }∩ A) / P (A) ⋯⋯⋯⋯⋯⋯⋯⋯(1)

Using the multiplication rule of probability,

P (E_{i }∩ A) = P (E_{i}) P (A│E_{i}) ⋯⋯⋯⋯⋯⋯⋯⋯(2)

Using the total probability theorem,

P (A) = ∑_{k=1}^{n} P (E_{k}) P (A | E_{k}) ⋯⋯⋯⋯⋯⋯⋯⋯⋯⋯⋯(3)

Putting the values from equations (2) and (3) in equation 1,

P (E_{i} / A) = [P (E_{i}) P (A / E_{i})] / [∑_{k=1}^{n} P (E_{k}) P (A / E_{k})]

Another form of Baye’s theorem is given by

P (A / B) = [P (B / A) P (A)] / P (B)

Here A & B are two different events and P (B) ≠ 0.

**Example 1: **For two events A and B, if P (A) = P (A / B) = 1 / 4 and P (B / A) = 1 / 2, then

**Solution: **

P (B / A) = 1 / 2

⇒ P (B ∩ A) P (A) = 1 / 2

⇒P (B ∩ A) = 1 / 8

P (A / B) = 1 / 4

⇒ P (A ∩ B) P (B) = 1 / 4

⇒ P (B) = 1 / 2

P (A ∩ B) = 1 / 8 = P (A) . P (B)

∴ Events A and B are independent.

Now, P (A′ / B) = P (A′ ∩ B) / P (B) = P (A′) P (B) / P (B) = 3 / 4 and

P (B′ / A′) = P (B′ ∩ A′) / P (A′) = P (B′) P (A′) / P (A′) = 1 / 2

**Example 2: **Let 0 < P (A) < 1, 0 < P (B) < 1 and P (A ∪ B) = P (A) + P (B) − P (A) P (B). Then

**Solution: **

Since P (A ∩ B) = P (A) / P (B)

It means A and B are independent events so A^{c} and B^{c} will also be independent.

Hence

P (A ∪ B)^{c }= P (A^{c} ∩ B^{c}) = P (A^{c}) P (B^{c}) (Demorgan’s law)

As A is independent of B, hence

P (A / B) = P (A), {∵P (A ∩ B) = P (B) P (A / B)}.

**What Are Mutually Exclusive Events?**

The events that cannot occur at the same time can be termed as mutually exclusive events. For example, a coin can’t be tossed to obtain a head and a tail at once. Hence tossing a head and tossing a tail are mutually exclusive events. If A and B are two events, then it is represented as

**P (A and B) = 0**

It means that the probability of event A and event B happening together is 0.