# Probability

**Random Experiment:**An experiment is said to be a random experiment, if it's out-come can't be predicted with certainty.

If a coin is tossed, we can't say, whether head or tail will appear. So it is a random experiment.

**Sample Space:**The sent of all possible out-comes of an experiment is called the sample space. It is denoted by 'S' and its number of elements are n(s).

In throwing a dice, the number that appears at top is any one of 1, 2, 3, 4, 5, 6. So here:

S = {1, 2, 3, 4, 5, 6} and n(s) = 6

Similarly in the case of a coin, S = {Head, Tail} or {H, T} and n(s) = 2.

The elements of the sample space are called sample point or event point.

**Event:**Every subset of a sample space is an event. It is denoted by 'E'.

In throwing a dice S = {1, 2, 3, 4, 5, 6}, the appearance of an event number will be the event E = {2, 4, 6}.

Clearly E is a sub set of S.

**Simple event;**An event, consisting of a single sample point is called a simple event.

In throwing a dice, S = {1, 2, 3, 4, 5, 6}, so each of {1}, {2}, {3}, {4}, {5} and {6} are simple events.

**Compound event:**A subset of the sample space, which has more than on element is called a mixed event.

In throwing a dice, the event of appearing of odd numbers is a compound event, because E = {1, 3, 5} which has '3' elements.

**Equally likely events:**Events are said to be equally likely, if we have no reason to believe that one is more likely to occur than the other.

When a dice is thrown, all the six faces {1, 2, 3, 4, 5, 6} are equally likely to come up.

**Exhaustive events:**When every possible out come of an experiment is considered.

A dice is thrown, cases 1, 2, 3, 4, 5, 6 form an exhaustive set of events.

# Classical Definition of Probability

**In a throw of a dice, S = {1, 2, 3, 4, 5, 6}**

Let E1 = Event of getting a number less than '7'.

So 'E1' is a sure event.

So, we can say, in a sure event n(E) = n(S)

**Mutually exclusive or disjoint event:**If two or more events can't occur simultaneously, that is no two of them can occur together.

When a coin is tossed, the event of occurrence of a head and the event of occurrence of a tail are mutually exclusive events.

**Independent or mutually independent events:**Two or more events are said to be independent if occurrence or non-occurrence of any of them does not affect the probability of occurrence or non-occurrence of the other event.

When a coin is tossed twice, the event of occurrence of head in the first throw and the event of occurrence of head in the second throw are independent events.

**Difference between mutually exclusive an mutually independent events:**Mutually exclusiveness is used when the events are taken from the same experiment, where as independence is used when the events are taken from different experiments.

**Random Experiment:**An experiment is said to be a random experiment, if its out-come canâ€™t be predicted with certainty.

If a coin is tossed, we canâ€™t say, whether head or rail will appear. So it is a random experiment.

**Sample Space:**The set of all possible out-comes of an experiment is called the sample â€“ space.

If S = {H, T}, than â€˜Hâ€™ and â€˜Tâ€™ are sample points.

# Important

- The probability of an event lies between â€˜Oâ€™ and â€˜1â€™.
- The probability of an impossible event is â€˜Oâ€™ i.e. P(E) = O
- The probability of a sure event is 1. i.e. P(S) = 1. where â€˜Sâ€™ is the sure event.
- If two events â€˜Aâ€™ and â€˜Bâ€™ are such that A âŠ‚ B, then P(A) < = P(B).

- Sunday and Monday
- Monday and Tuesday
- Tuesday and Wednesday
- Wednesday and Thursday
- Thursday and Friday
- Friday and Saturday
- Saturday and Sunday

# Probability Using Principals of Permutation & Combination

- â€˜1â€™ is red and â€˜2â€™ are white,
- â€˜2â€™ are blue and 1 is red,
- none is red.

^{18}C

_{3}= 18! / (3! Ã— 15!) = (18 Ã— 17 Ã— 16)/(3 Ã— 2 Ã— 1) = 816

- Let E
_{1}= Event of getting â€˜1â€™ ball is red and â€˜2â€™ are white_{1}) =^{6}C_{1}Ã—^{4}C_{2}_{1}) = n(E_{1})/n(S) = 36/816 = 3/68 - Let E
_{2}= Event of getting â€˜2â€™ balls are blue and â€˜1â€™ is red._{2}) =^{8}C_{2}Ã—^{6}C_{1}_{2}) = 168/816 = 7/34 - Let E
_{3}= Event of getting â€˜3â€™ non â€“ red balls. So now we have to choose all the three balls from 4 white and 8 blue balls._{3}) =^{12}C_{3}= 12! / (3! Ã— 9!) = (12 Ã— 11 Ã— 10) / (3 Ã— 2 Ã— 1) = 220_{3}) = n(E_{3}) / n(S) = 220/816 = 55/204

**Remember:**

- If â€˜Aâ€™ and â€˜Bâ€™ are mutually exclusive events then
- If â€˜Aâ€™ and â€˜Bâ€™ by any two events, then the probability of occurrence of at least one of the events â€˜Aâ€™ and â€˜Bâ€™ is given by: