# Ex.15.1 Q13 Probability Solution - NCERT Maths Class 10

## 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

\[=\frac{\begin{Bmatrix} \text { Number of}\\ \text{ possible } \\ \text{outcomes }\end{Bmatrix} }{ \begin{Bmatrix}\text { Total no of} \\ \text{favorable} \\ \text{outcomes} \end{Bmatrix} }\]

**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 \(6\)

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

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

Probability of getting a odd number

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