Ex.15.1 Q13 Probability Solution - NCERT Maths Class 10

Go back to  'Ex.15.1'

Question

A die is thrown once. Find the probability of getting

(i) a prime number;

(ii) a number lying between \(2\) and \(6\);

(iii) an odd number.

   

Text Solution

What is known?

A die is thrown once. So \(1, 2, 3, 4, 5, 6\) are equally likely outcomes.

Totally there are \(6\) outcomes.

What is unknown?

The probability of getting

(i) a prime number;

(ii) a number lying between \(2\) and \(6\);

(iii) an odd number.

Reasoning:

This question can be solved easily by using the formula

Probability of an event\(\begin{align}=\frac{\text{ Number of possible outcomes }}{\text{Total no of favourable outcomes}}\end{align}\)

Steps:

No of outcomes when you throw a die\( = (1,2,3,4,5,6 )= 6\)

No of  prime numbers on dice are \(1,3\) and \(5 = 3\)

(i) Probability of getting a prime number

\[\begin{align} \text{ }\,\,\text{}&={}\frac{\text{Number of prime numbers}}{\text{total no of outcomes}} \\&=\frac{3}{6} \\ &=\frac{1}{2} \\\end{align}\]

(ii)Numbers lying between \(2\) and \(6\) are \(3,4,5 = 3\)

Probability of getting a number lying between \(2\) and \(6\)

\[\begin{align}&=\frac{\text{ Number lying between }2\text{ and }6}{\text{total no of outcomes}} \\  &=\frac{3}{6} \\\end{align}\]

(iii)Total number of odd numbers are \(1,3\) and \(5 = 3\)

Probability of getting a odd number 

\[\begin{align}& =\frac{\text{Number of odd numbers}}{\text{total no of outcomes}} \\ {} & =\frac{3}{6} \\ \end{align}\]