** Probability **In an experiment if ‘ n ‘ is the number of exhaustive cases and ‘m’ is the number of favourable cases of an event A. Then the probability of event

**A**is denoted by

**P(A)**.

**P(A) = **

## Probability Theory

**Random Experiment**: An experiment in which all possible out comes are known and the exact output cannot be predicated in advance is called Random Experiment.

**EXAMPLE :**

(i) Tossing a fair coin

(ii) Rolling an unbiased dice

(iii) Drawing a card from a pack of well shuffled cards**Trail :**Conducting a Random Experiment once is known as a Trail**Outcome**: The result of a Trail in the random experiment is called as Outcome

**EXAMPLE :**In tossing a single coin outcomes are**H**and**T****Sample Space :**A set of all possible outcomes of a random experiments is known as Sample Space

**EXAMPLE :**

( i ) In the experiment of tossing a coin the sample space**S H,T**( ii ) If Two coins are tossed then

**S HH,HT,TH,TT**( iii ) In throwing a die S

**1,2,3,4,5,6****Event :**Any non-empty subset of a Sample Space is called an Event

**EXAMPLE :**( i ). in Tossing a single coin getting Head or Tail is an Event

( ii )Getting an Ace (or) Diamond from a pack of 52 cards is an event**Exhaustive Events :**The total number of possible outcomes of an experiment is known as Exhaustive Events

**EXAMPLE :**In the experiment of throwing a die the total number of possible outcomes = 6**Favourable Events :**The number of event which favour the happing of the events are known as Favourable Cases (or) Events

**EXAMPLE :**In tossing two dice the number of cases favourable to getting the sum 3 is (2, 1), (1, 2) i.e., 2**Mutually Exclusive Events :**If two events have no common outcomes then they are called Mutually Exclusive

**EXAMPLE:**In tossing a coin the events Head & Tail are mutually exclusive because Head & Tail cannot happen at the same time.**Independent Events :**Two events are said to be independent if the happening of one event does not affect the happening of the other.**Dependent Events**: Two events are said to be dependent if the happening of an event will affect the happening of the other event .

**EXAMPLE:**If we draw a card from a pack of 52 cards and replace it before we draw a second card. The second card is independent of first one. If we don’t replace the first card before the second draw. The second draw depends on the first one.**Axioms of Probability :**Let S be the sample space and A be the event i.e.,

**1**.

**2.**P(S) = 1

**3**. If A and B are mutually exclusive ( Disjoint i.e., )

= P(A)+P(B)**Results of Probability :**. for any two events A & B ,

1

**2**. If denotes compliment of event A then , = 1-P(A)

**3.**For any two events A & B ,

**4**. For any three events A, B & C ,

## Probability Examples and Probability Solutions

**Question 1:** A die is rolled, find the probability that an even number is obtained.

**Solution : **Let us first write the sample space S of the experiment.

S = {1,2,3,4,5,6}

Let E be the event "an even number is obtained" and write it down.

E = {2,4,6}

We now use the formula of the classical probability.

P(E) =

**Question 2:** Two coins are tossed, find the probability that two heads are obtained.

**Note:** Each coin has two possible outcomes H (heads) and T (Tails).

**Solution**: The sample space S is given by.

S = {(H,T),(H,H),(T,H),(T,T)}

Let E be the event "two heads are obtained".

E = {(H,H)}

We use the formula of the classical probability.

P(E) =

**Question 3:** You toss a coin AND roll a die. What is the probability of getting a tail and a on the die?

**Solution 3: **Probability of getting a tail when a single coin is tossed =

Probability of getting 4 when a die is thrown =

Required probability =

**Question 4:** A number is selected from the numbers 1, 2, 3 and then a second number y is randomly selected from the numbers 1, 4, 9. What is the probability that the product xy of the two numbers will be less than 9 ways

**Solution 4:** Number X can be selected in three ways and corresponding to each such way there are three ways of selecting number y.

Therefore, two numbers can be selected in 9 ways.

The favourable number of elementary events for which the product xy of the two numbers will less than 9 = (1,1),(1,4),(2,1),(2,4),(3,1) = 5

Hence the required probability =

**Question 5:** Find the probability of having 53 Sundays in

( i ) a leap year (ii) a non leap year

**Solution 5:** An ordinary year has 365 days i.e., 52 weeks and 1 odd day.

This day can be any one of the 7 days of the week,

P(that this day is Sunday) =

Hence, P(an Ordinary Year has 53 Sundays) =

A leap year has 366 days. i.e., 52 weeks and 2 odd days

This day can be any one of the 7 days of the week

P(that this day is Sunday) =

Hence P(a leap Year has 53 Sundays) =

**Question 6:** Two cards are drawn at random from a pack of 52 cards. What is the probability that either both are black and both are queen.

**Solution 6:** Let S be the Sample space. Then

n(S) = Number of ways of drawing 2 balls out of (6+4) =

Let E = Event of getting both balls of the same colour . Then

N(E) = Number of ways of drawing ( 2 balls out 6 )or ( 2 balls out of 4)

=

P(E) =

## Probability Questions from Previous Year Exams

Probability Aptitude

## Probability Video lectures

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