### Theorems of probability

##
__Addition theorem on probability__

__Addition theorem on probability__

The probability of happening an event can easily be found using the definition of probability. But just the definition cannot be used to find the probability of happening at least one of the given events. A theorem known as “Addition theorem” solves these types of problems. The statement and proof of “Addition theorem” and its usage in various cases is as follows.

**Mutually exclusive events:**

Two or more events are said to be mutually exclusive if they don’t have any element in common. i.e. if, the occurrence of one of the events prevents the occurrence of the others then those events are said to be mutually exclusive.

**Example:**

The event of getting 2 heads,

**and the event of getting 2 tails,***A***when two coins are tossed are mutually exclusive.***B*
Because A = {HH}; B = {TT}.

**Mutually exhaustive events:**

Two events are said to be mutually exhaustive if there is a certainty of occurring at least one of those two events. i.e. one of those events will definitely happen.

If

**and***A***are two mutually exhaustive then the probability of their union is 1.***B*
i.e. P(AUB)=1.

**Example:**

The event of getting a head and the event of getting a tail when a coin is tossed are mutually exhaustive.

**Addition theorem on probability:**

If

**and***A***are any two events then the probability of happening of at least one of the events is defined as P(AUB) = P(A) + P(B)- P(A∩B).***B***Proof:**

Since events are nothing but sets,

From set theory, we have

n(AUB) = n(A) + n(B)- n(A∩B).

Dividing the above equation by n(S), (where

**is the sample space)***S*
n(AUB)/ n(S) = n(A)/ n(S) + n(B)/ n(S)- n(A∩B)/ n(S)

Then by the definition of probability,

P(AUB) = P(A) + P(B)- P(A∩B).

**Example:**

If the probability of solving a problem by two students George and James are 1/2 and 1/3 respectively then what is the probability of the problem to be solved.

**Solution:**

Let

**and***A***be the probabilities of solving the problem by George and James respectively.***B*
Then P(A)=1/2 and P(B)=1/3.

The problem will be solved if it is solved at least by one of them also.

So, we need to find P(AUB).

By addition theorem on probability, we have

P(AUB) = P(A) + P(B)- P(A∩B).

P(AUB) = 1/2 +.1/3 – 1/2 * 1/3 = 1/2 +1/3-1/6 = (3+2-1)/6 = 4/6 = 2/3

**Note:**

If

**and***A***are any two mutually exclusive events then P(A∩B)=0.***B*
Then P(AUB) = P(A)+P(B).

##
*Multiplication Theorem on Probability*

*Multiplication Theorem on Probability*
The probability of happening an event can easily be found using the definition of probability. But just the definition cannot be used to find the probability of happening of both the given events. A theorem known as “Multiplication theorem” solves these types of problems. The statement and proof of “Multiplication theorem” and its usage in various cases is as follows.

Multiplication theorem on probability:

If

**and***A***are any two events of a sample space such that P(A) ≠0 and P(B)≠0, then***B*
P(A∩B) = P(A) * P(B|A) = P(B) *P(A|B).

Example: If P(A) = 1/5 P(B|A) = 1/3 then what is P(A∩B)?

Solution: P(A∩B) = P(A) * P(B|A) = 1/5 * 1/3 = 1/15

##### Independent events:

Two events

**and***A***are said to be independent if there is no change in the happening of an event with the happening of the other event.***B*
i.e. Two events

**and***A***are said to be independent if***B*
P(A|B) = P(A) where P(B)≠0.

P(B|A) = P(B) where P(A)≠0.

i.e. Two events

**and***A***are said to be independent if***B*
P(A∩B) = P(A) * P(B).

**Example:**

While laying the pack of cards, let A be the event of drawing a diamond and B be the event of drawing an ace.

Then P(A) = 13/52 = 1/4 and P(B) = 4/52=1/13

Now, A∩B = drawing a king card from hearts.

Then P(A∩B) = 1/52

Now, P(A/B) = P(A∩B)/P(B) = (1/52)/(1/13) = 1/4 = P(A).

So, A and B are independent.

[Here, P(A∩B) = = = P(A) * P(B)]

**Note:**

(1) If 3 events A,B and C are independent the

P(A∩B∩C) = P(A)*P(B)*P(C).

(2) If A and B are any two events, then P(AUB) = 1-P(A’)P(B’).

##
__Conditional Probability__

__Conditional Probability__

In many situations, once more information becomes available, we are able to revise our estimates for the probability of further outcomes or events happening. For example, suppose you go out for lunch at the same place and time every Friday and you are served lunch within 15 minutes with probability 0.9. However, given that you notice that the restaurant is exceptionally busy, the probability of being served lunch within 15 minutes may reduce to 0.7. This is the conditional probability of being served lunch within 15 minutes given that the restaurant is exceptionally busy.

The usual notation for "event A occurs given that event B has occurred" is "A | B" (A given B). The symbol | is a vertical line and does not imply division. P(A | B) denotes the probability that event A will occur given that event B has occurred already.

- A rule that can be used to determine a conditional probability from unconditional probabilities is:
- where:
- P(A | B) = the (conditional) probability that event A will occur given that event B has occured already
- = the (unconditional) probability that event A and event B both occur
- P(B) = the (unconditional) probability that event B occurs
- Using the multiplication rule, gives Bayes' Theorem in its simplest form:
- Using the Law of Total Probability:
P(A | B) = P(B | A).P(A)

P(B | A).P(A) + P(B | A').P(A')- where:
- P(A) = probability that event A occurs
- P(B) = probability that event B occurs
- P(A') = probability that event A does not occur
- P(A | B) = probability that event A occurs given that event B has occurred already
- P(B | A) = probability that event B occurs given that event A has occurred already
- P(B | A') = probability that event B occurs given that event A has not occurred already

##

**Bayes' Theorem**

**Bayes' Theorem**

Bayes' Theorem is a result that allows new information to be used to update the conditional probability of an event.