Probability RS Aggarwal Class-8 ICSE Maths Goyal Brothers

Probability RS Aggarwal Class-8 ICSE Maths Goyal Brothers Prakashan Solutions Chapter-27. We provide step by step Solutions of Exercise / lesson-27 Graphical Representations of Statistical Data for ICSE Class-8  RS Aggarwal  Mathematics.

Our Solutions contain all type Questions of Exe-27 with Notes to develop skill and confidence. Visit official Website CISCE for detail information about ICSE Board Class-8 Mathematics

Probability RS Aggarwal Class-8 ICSE Maths Goyal Brothers Prakashan Solutions Chapter-27

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Notes on Probability   

Exe-27

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Notes on Probability   

Probability means possibility / Chance. It is a branch of mathematics that deals with the occurrence of a random event. The value is expressed from zero to one.

Definition of Probability:
Probability is defined as the numerical method of measuring uncertainty involved in a situation.
It is widely used in the study of mathematics, statistics, gambling, physical science, biological science, weather forecasting, finance etc. to draw conclusions.

Experiment:
An experiment is defined as an action or process that results in well defined outcomes

Formula for Probability

The probability formula is defined as the possibility of an event to happen is equal to the ratio of the number of favourable outcomes and the total number of outcomes.

Formula for Probability

Probability of event to happen P(E) = Number of favourable outcomes/Total Number of outcomes

Random experiment:

An experiment, in which we know all the results, but cannot predict them, is called a random experiment.

Outcomes:

The possible results of an experiment are called the outcomes

Probability of an Event

A combination of outcomes is called an event

Assume an event E can occur in r ways out of a sum of n probable or possible equally likely ways. Then the probability of happening of the event or its success  is expressed as

P(E) = r/n

The probability that the event will not occur or known as its failure is expressed as:

P(E’) = (n-r)/n = 1-(r/n)

E’ represents that the event will not occur.

Therefore, now we can say;

P(E) + P(E’) = 1

This means that the total of all the probabilities in any random test or experiment is equal to 1

Probability of an event = $\frac{\text{No.of favourable outcomes /}}{\text{Total number of outcomes}}$.

Probability Terms and Definition

Some of the important probability terms are discussed here:

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Exe-27 , Probability RS Aggarwal Class-8 ICSE Maths Goyal Brothers Prakashan Solutions

Question 1:

A coin is tossed once. What is the probability of getting a head ?

Answer :

On tossing a coin once,

Number of possible outcome = 2

Similarly, favorable outcome getting a head = 1

So, P(E) = Number of favorable outcome/Number of all possible outcome = 1/2

Question 2:

A die is thrown once. What is the probability of getting :

(i) the number 4

(ii) an odd number ?

Answer :

A die has six numbers: 1, 2, 3, 4, 5, 6

Number of possible outcomes = 6

(i) the number = 4

Number of favorable outcome of getting a number 4 = 1

So, P(E) = Number of favorable outcome/Number of all possible outcome = 1/6

(ii) an odd number

Number of favorable outcomes = 3

So, P(E) = Number of favorable outcome/Number of all possible outcome = 3/6

= 1/2   Ans..

Question 3:

A die is thrown once, What is the probability of getting :

(i) a prime number

(ii) a number greater than 2

(iii) a number other than 2 and 5.

Answer :

A die has six numbers: 1, 2, 3, 4, 5, 6

Number of possible outcomes = 6

(i) a prime number

Number of favorable outcomes = a prime number = 1, 3, 5 which are 3 in numbers

So, P(E) = Number of favorable outcome/Number of all possible outcome = 3/6

= 1/2

(ii) Number of favorable outcome = greater than 2 i.e. four numbers 3, 4, 5 and 6

So, P(E) = Number of favorable outcome/Number of all possible outcome = 4/6

= 2/3

(iii) Number of favorable outcome = Number other than 2 and 5 i.e. 1, 3, 4, 6

So, P(E) = Number of favorable outcome/Number of all possible outcome = 4/6

= 2/3

Question 4:

Two coins are tossed simultaneously. Find the probability of getting :

(i) exactly one head

(ii) at least one tail

(iii) no tail

(iv) at most one head

Answer :

Total number of possible outcomes 4 ex. (HH), (HT), (TT) and (TH)

(i) Exactly one head

Possible number of favorable outcomes = 2

(ex. TH and HT)

So, P(E) = Number of favorable outcome/Number of all possible outcome = 2/4

= 1/2

(ii) At least one tail

Possible number of favorable outcomes = 3

Total number of possible outcomes = 4

So, P(E) = Number of favorable outcome/Number of all possible outcome = 3/4

(iii) No tails

Possible number of favorable  outcomes = 1

Total number of possible outcomes = 4

So, P(E) = Number of favorable outcome/Number of all possible outcome = 1/4

(iv) At most one heads

Possible number of favorable outcomes = 3
(ex. TT, HT and TH)

Total numbers of possible outcomes = 4

So, P(E) = Number of favorable outcome/Number of all possible outcome = 3/4

–: End of Probability RS Aggarwal Class-8 Solutions :–

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