Probability

Probability means the event that something will happen.
Probability happening of event = Number of favourable outcomes /Total number of possible outcomes
P(E) = n(E)/n(s)
Where, P(E) = Probability of an event
n(E) = number of favourable outcomes
n(S) = total number of possible outcomes
The probability of any event (E) is a number between 0 and 1.
0 ≤ P(E) ≤ 1

Ex- There are 7 balls in a bag. 4 of them are red and 3 of them are blue. What is the probability that a blue ball will be picked?
Solution:- given, number of favourable outcomes = 3
Total number of outcomes = 7
Required probability = Number of favourable outcomes /Total number of possible outcomes
= 3/7

Terms Related to Probability

Experiment

An action or operation resulting in two or more outcomes is called an experiment.
For example- Tossing of a coin, Throwing a coin.

Sample Space

The set of all possible outcomes of an experiment is called the sample space, denoted by S.
For example- If we throw a die, then sample space
S = { 1, 2, 3, 4, 5, 6 }
If we toss a coin, then sample space S = { H, T }

Possible outcomes

All possibilities related to an event are known as possible outcomes.
For example- When a coin is tossed, these are two possible outcomes, Probability of getting H = 1/2
Probability of getting T = 1/2.
When a single die is thrown, there are six possible outcomes 1, 2, 3, 4, 5, 6. Probability of getting any one of these number is 1/6.

Event

Any subset of a sample space is an event.
For example- Getting a head is an event related to tossing of a coin.

Impossible Event

Impossible event has no chance of occurring. the null set ( φ ) is called impossible event or null event.
For example- Getting 7 when a dice is thrown is an impossible or a null event.

Ex- A bag contains 15 black balls, if a ball is picked at random from the bag. Find the probability that ball picked is of Red color.
Solution:- The bag contains 15 black balls and there is no red ball in the bag.
So, number of favorable even = 0
Total outcomes = 15
∴ Probability = 0/15 = 0

Sure Event

The entire sample space is called sure or certain event.
For example- Getting an odd or event number on throwing a dice is a sure event.

Equally Likely Events

A number of simple events are said to be equally likely if there is no reason for one event to occur in preference to any other event.
For example- When a dice is rolled the possible outcome of getting an odd number = 3, possible outcome of getting an even number = 3
So, getting a even number or odd number are equally likely events.

Compound Event

It is joint occurrence of two or more simple events
For example- The event of at least one head appears when two fair coins are tossed is a compound event, A{ HT, TH, HH }

Mutually Exclusive

Two events E1 and E2 related to an experiment E, having sample space S are known as mutually exclusive.
If the probability of occurrence of both events simultaneously is zero.
P ( E1 ∩ E2 ) = 0
For example- When a coin is tossed either head or tail will appear. Head and tail coin not occur simultaneously. Therefor occurrence of a head or a tail are two mutually exclusive events

Exhaustive Events

Two events E1 and E2 related to an experiment E, having sample space S are known as mutually exhaustive, if the probability of occurrence of event E1 or E2 is 1.
P ( E1 ∩ E2 ) = 1
For example-Let A be probability of getting an even number when a dice is rolled and B be the probability of getting an odd number. The probability the probability of occurrence of event A or B is i.e., any of the even can occur so they are mutually exhanstive.

Dependent Events

If the outcome or occurrence of the first affects the outcome or occurrence of the second, then two events are called dependent.

Independent Events

Two or more events are said to be independent if occurrence or non-occurrence of any of them does not influence the probability of occurrence or non-occurrence of other events.
If A and B are independent events, then
P( A and B ) = P( A ∩ B ) = P(A) . P(B)
For example- Getting head after tossing a coin and getting a 5 on a rolling single 6-sided die are independent events.

Ex- A fair coin is tossed repeatedly. If the tail appears on first four tosses, then the probability of the head appearing on the fifth toss equals.
Solution:- The event that the fifth toss results a head is independent of the event that the first four tosses results tails.
∴ Probability of the required event = 1/2.

Law of Total Probability

If E1, E2, E3……..En be n mutually exclusive events related to an experiment, then probability of an event A which occurs with E1 or E2 or E3……..En is given by
P(A) = P(E1) P( A/E1) + P(E2) P( A/E2) + …….. P( A/En)

Conditional Probability

The conditional probability of an event B in relationship to an event A is the probability that event B occurs given that event A has already been occurred. The notation for conditional probability is P (B/A). It is pronounced as the probability of happening of an event B given that A has already been happened.
P(A/B) = P( A ∩ B )/ P(B) and
P(B/A) = P( A ∩ B )/ P(A)

Ex- A English teacher conducted two tests in her class. 35% of the students passed both tests and 70% of the students passed the first test . What percent of the students passed the second test given that they have already passed the first test?
Solution:-
P(B/A) = P( A and B )/ P(A)
= 35/70
= 1/2
= 50%

Rules Related to Probability

When two events A and B are mutually exclusive, the probability that A or B will occur, is the sum of the probability of each event.
P ( A or B ) = P(A) + P(B) and P( A ∩ B ) = P(A) + P(B)
But when two events A and B are non-mutually exclusive, the probability that A or b will occur, is
P ( A or B ) = P(A) + P(B) - P( A and B )
P ( A ∪ B ) = P(A) + P(B) - P( A ∩ B )

Ex- From a well shuffled pack of 52 cards, a card is frawer at random, find the probability that it is either a heart or a queen.
Solution:- Let A be the probability of getting a heart card and B be the probability of getting a queen card
P(A) = 13/52 P(B) = 4/52 P( A ∩ B ) = 1/52
∴ Required probability = P ( A ∪ B ) = P(A) + P(B) - P( A ∩ B )
= 13/52 + 4/52 - 1/52
= 16/52 = 4/13

Multiplication Theorem of Probability

When two events A and B are mutually exclusive, the probability that A and B will occur simultaneously is given as
P ( A ∩ B ) = P(A) . P(B) ( A and B are independent event )

Ex- Four persons P, Q, R and S appeared for an interview. Find the probability that both P and S are selected in the interview?
Solution:-
Probability that P is selected = P (P) = 1/4
Probability that S is selected = P (S) = 1/4
∴ Required probability that both are selected = 1/4 × 1/4 = 1/16

Types of Questions

Question Based on Coins

This types of question are based on tossing of coins and obtaining a particular face (Head/Tail) or obtaining same face on two or more coins.

Ex- If a coin is tossed what is the probability of each outcome?
Solution:- Here, total outcomes = ( H, T ) = 2
favourable outcome = 1
Required probability = 1/2

Ex- If a coin is tossed twice, then find the probability that a Tail is obtained atleast once.
Solution:- Total outcomes = ( HH, TT, HT, TH ) = 4
Favourable outcome = ( TT, HT, TH ) = 3
∴ Required probability = 3/4

Question Based on Dice

This type of questions are based on rolling of one or more dice and getting a particular number on the face or a particular sum on faces of the dice etc.

Ex- A single 6-sided die is rolled. What is the probability of getting prime number and getting composite number?
Solution:- Total outcomes = ( 1, 2, 3, 4, 5, 6 ) = 6
Prime numbers = ( 2, 3, 5 ) = 3
Composite number = ( 4, 6 ) = 2
Probability of getting prime number = 3/6 = 1/2
Probability of getting composite number = 2/6 = 1/3

Ex- A single 6-sided die is rolled. What is the probability of getting a 3 or a 6?
Solution:- Total outcomes = ( 1, 2, 3, 4, 5, 6 ) = 6
Probability of getting a number in a single throw of die = 1/6
Probability of getting 2 = 1/6
Probability of getting 6 = 1/6
∴ Required probability of getting either 3 or 6 = 1/6 + 1/6 = 2/6 = 1/3

Question Based on Playing Cards

There are total 52 cards in a deck of playing cards.
There are 26 red and 26 black cards.
There are 13 cards of each suit Clubs, Diamonds, Hearts and Spades.
There are 4 Aces, 4 Jacks, 4 Queens and 4 Kings.
There are 12 face cards.

Ex- A total of six cards are chosen at random from a standard deck of 52 playing cards. What is the probability of choosing 6 kings?
Solution:- There are only 4 kings, so we cannot choose 6 kings.
Thus, it is impossible event.
∴ Probability = 0/52 = 0

Ex- A single card is chosen at random from a standard deck of 52 playing cards. What is the probability of choosing a jack or a diamond?
Solution:- There are 4 jack in a standard deck and 13 diamond.
Also 1 jack is of diamond. so probability of getting a jack = 4/52
Probability of getting a jack of diamond = 13/52
Probability of getting a jack of diamond = 1/52
∴ Required probability of getting a jack or diamond
= 4/52 + 13/52 - 1/52 = 16/52 = 4/13

Question Based on Marbles or Balls

This types of questions are based on choosing a ball or a marble of particular color from one or more bag containing different colored balls or marbles.

Ex- A bag contains 3 red, 4 green and 5 black balls. If a single ball is chosen at random from the jar, what is the probability that it is red or black?
Solution:- Total outcomes = 3 + 4 + 5 = 12
Probability of getting a red ball = 3/12
Probability of getting a black ball = 5/12
∴ Probability of getting red or black = 3/12 + 5/12 = 8/12 = 2/3