Probability Formulas

Experiment:

An operation which can produce some well-defined outcomes is called an experiment.

Random Experiment:

An experiment in which all possible outcomes are known and the exact output cannot be predicated in advance, is called a random experiment.

The total number of possible outcomes of an experiment in any trial is known as the exhaustive number of events.

For example

• In tin-owing a die, the exhaustive number of cases is 6 since any one of the six faces marked with 1, 2, 3, 4, 5, 6 may come uppermost.
• In tossing a coin, the exhaustive number of cases is 2, since either head or tail may turn over.
• If a pair of dice is thrown, then the exhaustive number of cases is 6 x 6 = 36
• In drawing four cards from a well-shuffled pack of cards, the exhaustive number of cases is ⁵²C₄.

Events are said to be mutually exclusive if no two or more of them can occur simultaneously in the same trial.

For example,

• In tossing of a coin the events head (H) and tail (T) are mutually exclusive.
• In throwing of a die all the six faces are mutually exclusive.
• In throwing of two dice, the events of the face marked 5 appearing on one die and face 5 (or other) appearing on the other are not mutually exclusive.

Outcomes of a trial are equally likely if there is no reason for an event to occur in preference to any other event or if the chances of their happening are equal.

For example,

• In throwing of an unbiased die, all the six faces are equally likely to occur.
• In drawing a card from a well-shuffled pack of 52 cards, there are 52 equally likely possible outcomes.

The favourable cases to an event are the outcomes, which entail the happening of an event.

For example,

• In the tossing of a die, the number of cases which are favourable to the “ appearance of a multiple of 3” is 2, viz, 3 and 6.
• In drawing two cards from a pack, the number of cases favourable to “drawing 2 aces” is ⁴C _V
• In throwing of two dice, the number of cases favourable to “getting 8 as the sum” is 5,: (2,6), (6,2), (4,4), (3, 5) (5,3).

Events are said to be independent if the happening (or non ­happening) of one event is not affected by the happening or non­ happening of others.

Sample space:

When we perform an experiment then the set S of all possible outcomes is called the sample space.

Examples of Sample Space:

• In tossing a coin, S = (H,T).
• If two coins are tossed, then S = ( HH, HT, TH, TT )
• In rolling a dice, we have, S = {1,2,3,4,5,6}

Event:

Any subset of a sample space is called an event.

Probability of occurrence of an event:

Let S be the sample space and let E be an event. Then, ECS.

\(P ( E ) = \frac { n ( E ) } { n ( S ) }\)

n ( E ) = no. of desired events.

n ( S ) = Total no. of events. i.e. no of sample spaces

Results on Probability:

1. P ( S ) = 1
2. \(O \leq P ( E ) \leq 1\)
3. \(P ( \phi ) = 0\)
4. For any events A and B, we have: \(P ( A \cup B ) = P ( A ) + P ( B ) – P ( A \cap B )\)
5. \(\frac { 1 } { A } \text { denotes (not-A), then } P ( \overline { A } ) = 1 – P ( A )\)

Probability Formulas | Problems based on leap year

• A leap year has 366 days, hence it has 52 complete weeks I and 2 more days. When the year is not a leap year, it has 52 complete weeks and 1 more day.

Probability Formulas | Problems based on dice

 Possible outcomes by rolling dice are as tabulated:

 Sum of Possible outcomes by rolling dice are as tabulated:

Probability Formulas | Problems Based on Cards

2, 3, 4, 5, 6, 7, 8, 9, 10 – no. cards = 9 x 4 = 36.
Ace, King, Queen; Jack – Honour cards = 4 x 4 = 16.
King, Queen; Jack – Face cards = 3 x 4 =12.

DEFINITION OF PROBABILITY

If there are n-mutually exclusive, exhaustive and equally likely outcomes to a random experiment and ‘m’ of them are favourable to an event A, then the probability of happening of A is denoted by P (A) and is defined by

\(\mathrm { P } ( \mathrm { A } ) = \frac { \mathrm { m } } { \mathrm { n } }\) \(P ( A ) = \frac { \text { No. of elementary events favourable to } A } { \text { Total no. of equally likely elementary events } }\)

P(A) can never be negative.
Since, the number of cases in which the event A will not happen is ‘n

\(\begin{array} { l } { – \mathrm { m } ^ { \prime } , \text { then the probability } \mathrm { P } ( \overline { \mathrm { A } } ) \text { of not happening of } \mathrm { A } \text { is given by } } \\ { \mathrm { P } ( \overline { \mathrm { A } } ) = \frac { \mathrm { n } – \mathrm { m } } { \mathrm { n } } = 1 – \frac { \mathrm { m } } { \mathrm { n } } = 1 – \mathrm { P } ( \mathrm { A } ) } \\ { \Rightarrow \quad \mathrm { P } ( \mathrm { A } ) + \mathrm { P } ( \overline { \mathrm { A } } ) = 1 } \end{array}\)

The ODDS IN FAVOUR of occurrence of A are given by

\(\mathrm { m } : ( \mathrm { n } – \mathrm { m } ) \text { or } \mathrm { P } ( \mathrm { A } ) : \mathrm { P } ( \overline { \mathrm { A } } ) \)

The ODDS AGAINST the occurrence of A are given by

\(( \mathrm { n } – \mathrm { m } ) : \mathrm { m } \text { or } \mathrm { P } ( \overline { \mathrm { A } } ) : \mathrm { P } ( \mathrm { A } ) \)

ALGEBRA OF EVENTS

Let A and B be two events related to a random experiment. We define

• The event “A or B” denoted by “A \(\cup B\)”, which occurs when A or B or both occur. Thus,
  P(A \(\cup B\)) Probability that at least one of the events occur
• The event “A and B”, denoted by “A\( ( A \cap B ) \) B” , which occurs when A and B both occur. Thus, P ( \( A \cap B \))= Probability of simultaneous occurrence of A and B.
• The event “ Not – A” denoted by \(\overline{A}\) , which occurs when and only when A does not occur. Thus
  P ( \(\overline{A}\) ) = Probability of non-occurrence of the event A.
• \(\overline { \mathrm { A } } \cap \overline { \mathrm { B } }\) denotes the “ non-occurrence of both A and B”.
• \(^ { ” } A \subset B ^ { \prime \prime }\) denotes the “ occurrence of A implies the occurrence of B”.
• For example :
  Consider a single throw of die and following two events
  A = the number is even = {2, 4, 6}
  B = the number is a multiple of 3 = {3, 6}
  \(\mathrm { P } ( \mathrm { A } \cup \mathrm { B } ) = \frac { 4 } { 6 } = \frac { 2 } { 3 } , \quad \mathrm { P } ( \mathrm { A } \cap \mathrm { B } ) = \frac { 1 } { 6 }\)
  \( P ( \overline { A } ) = \frac { 1 } { 2 } , P ( \overline { A } \cap \overline { B } ) = P ( \overline { A \cup B } ) = 1 – \frac { 2 } { 3 } = \frac { 1 } { 3 } \)