MCQ
A box contains $90$ discs, numbered from $1$ to $90$. If one disc is drawn at random from the box, the probability that it bears a prime $-$ number less than $23,$ is :
- A$\frac{7}{20}$
- B$\frac{10}{90}$
- ✓$\frac{4}{45}$
- D$\frac{9}{89}$
✓
Answer
Correct option: C.
$\frac{4}{45}$
Total number of possibility $= 90.$
Favorable condition possibility of a prime number less than $23 = \{2, 3, 5, 7, 11, 13, 17, 19\} = 8.$
Therefore probability of a prime number less than $23=\frac{8}{90}=\frac{4}{45}$
Favorable condition possibility of a prime number less than $23 = \{2, 3, 5, 7, 11, 13, 17, 19\} = 8.$
Therefore probability of a prime number less than $23=\frac{8}{90}=\frac{4}{45}$
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