Choose the correct answer from the given four options : Someone i… - Vidyadip

MCQ

Choose the correct answer from the given four options : Someone is asked to take a number from $1$ to $100$. The probability that it is a prime is :

A
$\frac{1}{5}$
B
$\frac{6}{25}$
✓
$\frac{1}{4}$
D
$\frac{13}{50}$

✓

Answer

Correct option: C.

$\frac{1}{4}$

Total Number of outcomes $= 100$
So, the prime number between $1$ to $100$ are $\{2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43, 47, 53, 56, 61, 67, 71, 73, 79, 83, 89,\}$ and $97.$
$\therefore$ Total number of possible outcomes $= 25$
$\therefore\ \ \text{Reqired probability}=\frac{25}{100}=\frac{1}{4}$

Need a full question paper?

Generate a complete, print-ready paper with questions like this in minutes — across 16+ boards, with answer keys.

Start Generating Free

Explore more

More "M.C.Q (1 Marks)" from Probability→Probability — all question types→Maths STD 10 question papers→CBSE Board STD 10 question papers→CBSE Board question paper generator→

Similar questions

If $\tan \theta=\frac{3}{4}$, then $\cos ^2 \theta-\sin ^2 \theta=$

The system $x -2 y =3$ and $3 x + Ky =1$ has a unique solution only when

If in two triangles $A B C$ and $P Q R, \frac{A B}{Q R}=\frac{B C}{P R}=\frac{C A}{P Q}$, then

The mode of a frequency distribution can be determined graphically from:

If the system of equations 2x + 3y = 5, 4x + ky = 10 has infinitely many solutions, then k =

Mark the correct alternative in the following : The common difference of the $A.P. \ \frac{1}{3},\frac{1-3\text{b}}{3},\frac{1-6\text{b}}{3}, ....$ is

Mark the correct alternative in the following:
If Sn denote the sum of n terms of an A.P. with first term a and common difference d such that $\frac{\text{S}_\text{x}}{\text{Sk}_\text{x}}$ is independent of x, then

$\frac{\tan35^\circ}{\cot55^\circ}+\frac{\cot78^\circ}{\tan12^\circ}=?$

If the median of the data : $\{24, 25, 26, x + 2, x + 3, 30, 31, 34\}$ is $27.5,$ then $x =$

If the zeroes of a quadratic polynomial $ax^2+ bx + c$, $\text{C}\neq{0}$ are equal, then :