MCQ
A number is selected at random from the numbers $1$ to $30$. The probability that it is a prime number is :
- A$\frac{2}{3}$
- B$\frac{1}{6}$
- ✓$\frac{1}{3}$
- D$\frac{11}{30}$
✓
Answer
Correct option: C.
$\frac{1}{3}$
Prime numbers between $1$ and $30$ are $\{2, 3, 5, 7, 11, 13, 17, 19, 23\},$ and $29.$
Total number of prime number $= 10$
The probability that the number chosen from $1$ to $30$ is a prime number $=\frac{10}{30}=\frac{1}{3}$
Total number of prime number $= 10$
The probability that the number chosen from $1$ to $30$ is a prime number $=\frac{10}{30}=\frac{1}{3}$
Need a full question paper?
Generate a complete, print-ready paper with questions like this in minutes — across 16+ boards, with answer keys.
Explore more
Similar questions
$\triangle D E F \sim \triangle A B C$; If $D E: A B=2: 3$ and $\operatorname{ar}(\triangle D E F)$ is equal to $44 $ square units, then area $(\triangle A B C)$ in square units is
→There are 16 observations arranged in increasing order of their values in a data. The median will be the value of:
The perpendicular bisector of a line segment $A (-8,0)$ and $B (8,0)$ passes through a point $(0, k )$. The value of $k$ is
→A lamp post $8 m$ high casts a shadow $8 \sqrt{3} m$ long on the ground. The sun's elevation at this moment is
→→
If the area of the triangle formed by the points $(x, 2 x),(-2,6)$ and $(3,1)$ is 5 square units, then $x=$
→The coordinates of a point on $x-$ axis which lies on the perpendicular bisector of the line segment joining the points $(7, 6)$ and $(-3, 4)$ are,
→In Fig. AD = 4cm, BD = 3cm and CB = 12cm, find the $\cot\theta.$
→In $\triangle\text{ABC},\text{AB}=6\sqrt{\text{3 cm}}=\text{AC}=12\text{ cm}$ and $\text{BC}=6\text{ cm}$ The angle $B$ is :
→Choose the correct answer from the given four options: The $4^{\text {th }}$ term from the end of the $AP: -11,-8,-5, \ldots, 49$ is:
→