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Ex.15.1 Q13 Probability Solution - NCERT Maths Class 10

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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.

Video Solution
Probability
Ex 15.1 | Question 13

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

$=\frac{\begin{bmatrix} \text { Number of}\\ \text{ possible outcomes }\end{bmatrix} }{ \begin{bmatrix}\text { Total no of} \\ \text{favorable outcomes} \end{bmatrix} }$

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}

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